Environmental Effects of Population Growth

The increasing population results into increasing demands, needs and services. Their needs are unlimited but country has limited resources to fulfill their needs and desire. This note provides information with the effects of rapid population growth.

Summary

The increasing population results into increasing demands, needs and services. Their needs are unlimited but country has limited resources to fulfill their needs and desire. This note provides information with the effects of rapid population growth.

Things to Remember

  • The increasing population results into increasing demands, needs and services. Their needs are unlimited but country has limited resources to fulfill their needs and desire.
  • Population growth have adversely effect forest and vegetation, aquatic and terrestrial animals, air, water, land etc.
  • Different problems arise by over growth population are flood, pollution, disasters etc.

MCQs

No MCQs found.

Subjective Questions

Q1:

Write down the condition for the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 to the perpendicular and parallel to each other.


Type: Short Difficulty: Easy

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Answer: <p>Given:</p> <p>a<sub>1</sub>x + b<sub>1</sub>y + c<sub>1</sub> = 0..............................(1)</p> <p>a<sub>2</sub>x + b<sub>2</sub>y + c<sub>2</sub> = 0..............................(2)</p> <p>Slope of eq<sup>n</sup> (1), m<sub>1</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {a_1}{b_1}\)</p> <p>Slope of eq<sup>n</sup> (2), m<sub>2</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {a_2}{b_2}\)</p> <p>when the lines are parallel, then:</p> <p>m<sub>1</sub> = m<sub>2</sub></p> <p>or,-\(\frac {a_1}{b_1}\) =-\(\frac {a_2}{b_2}\)</p> <p>&there4; a<sub>1</sub>b<sub>2</sub> = a<sub>2</sub>b<sub>1 Ans</sub></p> <p>when the lines are perpendicular, then:</p> <p>m<sub>1</sub>&times; m<sub>2</sub>= - 1</p> <p>or,-\(\frac {a_1}{b_1}\)&times;-\(\frac {a_2}{b_2}\) = - 1</p> <p>&there4; a<sub>1</sub>a<sub>2</sub> = - b<sub>1</sub>b<sub>2</sub><sub>Ans</sub></p>

Q2:

Prove that the two straight lines 5x + 4y - 10 = 0 and 15x + 12y - 7 = 0 are parallel to each other.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>5x + 4y - 10 = 0............................(1)</p> <p>15x + 12y - 7 = 0.........................(2)</p> <p>Slope of eq<sup>n</sup> (1), m<sub>1</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 54\)</p> <p>Slope of eq<sup>n</sup> (2), m<sub>2</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {15}{12}\) = -\(\frac 54\)</p> <p>m<sub>1</sub> = m<sub>2</sub> = - \(\frac 54\)</p> <p>&there4; The given two lines are parallel to each other. <sub>Proved</sub></p>

Q3:

Prove that the pair of straight lines x + 3y = 2 and 6x - 2y = 9 are perpendicular to each other.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>x + 3y = 2..................................(1)</p> <p>6x - 2y = 9................................(2)</p> <p>Slope of eq<sup>n</sup> (1), m<sub>1</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 13\)</p> <p>Slope of eq<sup>n</sup> (2), m<sub>2</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 6{-2}\) =3</p> <p>We have,</p> <p>m<sub>1</sub>&times; m<sub>2</sub> =- \(\frac 13\)&times; 3 = - 1</p> <p>&there4;m<sub>1</sub>&times; m<sub>2</sub> = - 1</p> <p>Hence, the given two lines are perpendicular each other. <sub>Proved</sub></p> <p></p>

Q4:

If two lines 3x - 2y - 5 = 0 and 2x + py - 3 = 0 are parallel to each other, find the value pf p.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>3x - 2y - 5 = 0..............................(1)</p> <p>2x + py - 3 = 0.............................(2)</p> <p>Slope of eq<sup>n</sup> (1), m<sub>1</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 3{-2}\) = \(\frac 32\)</p> <p>Slope of eq<sup>n</sup> (2), m<sub>2</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2{p}\)</p> <p>When two lines are parallel lines, then:</p> <p>m<sub>1</sub> = m<sub>2</sub></p> <p>or,\(\frac 32\) = - \(\frac 2{p}\)</p> <p>or, p = \(\frac {-2 &times; 2}3\)</p> <p>&there4; p = -\(\frac 43\) <sub>Ans</sub></p>

Q5:

If two lines 4x + ky - 4 = 0 and 2x - 6y = 5 are perpendicular to each others find the value of k.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>4x + ky - 4 = 0..............................(1)</p> <p>2x - 6y = 5.............................(2)</p> <p>Slope of eq<sup>n</sup> (1), m<sub>1</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 4k\)</p> <p>Slope of eq<sup>n</sup> (2), m<sub>2</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2{-6}\) = \(\frac 13\)</p> <p>when two line are perpendicular to each other,</p> <p>m<sub>1</sub>&times; m<sub>2</sub> = - 1</p> <p>or, - \(\frac 4k\) &times; \(\frac 13\) = - 1</p> <p>or, - \(\frac 43\)&times; - 1 = k</p> <p>&there4; k = \(\frac 43\) <sub>Ans</sub></p>

Q6:

Write down the formula for the angle between the pair of lines y = m1x + c1 and y = m2x + c2, stating also the condition when they are perpendicular and parallel to each other.


Type: Short Difficulty: Easy

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Answer: <p>The formulae of angle between y = m<sub>1</sub>x + c<sub>1</sub> andy = m<sub>2</sub>x + c<sub>2</sub>is:</p> <p>tan\(\theta\) = &plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>when m<sub>1</sub>&times; m<sub>2</sub> = -1, the two lines are perpendicular to each other.</p> <p>when m<sub>1</sub> = m<sub>2</sub>, the two lines are parallel to each other.</p>

Q7:

If the line passing through (3, -4) and (-2, a) is parallel to the line given by the equation y + 2x + 3 = 0. Find the value of a.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>Slope of points (3, -4) and (-2, a)</p> <p>m<sub>1</sub> = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {a + 4}{-2 - 3}\) = \(\frac {-(a + 4)}5\)</p> <p>Given eq<sup>n</sup> is y + 2x + 3 = 0</p> <p>Slope of above eq<sup>n</sup> (m<sub>2</sub>) = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 21\) = - 2</p> <p>when lines are parallel then,</p> <p>m<sub>1</sub> = m<sub>2</sub></p> <p>or,\(\frac {-(a + 4)}5\) = - 2</p> <p>or, a + 4 = 10</p> <p>or, a = 10 - 4</p> <p>&there4; a = 6 <sub>Ans</sub></p>

Q8:

Prove that the line joining the points (3, -4) and (-2, 6) is parallel to the line y + 2x + 3 = 0.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>Slope of the points (3, -4) and (-2, 6) is:</p> <p>m<sub>1</sub> = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {6 + 4}{-2 - 3}\) = \(\frac {10}{-5}\) = -2</p> <p>Slope of the eq<sup>n</sup> y + 2x + 3 = 0 is:</p> <p>m<sub>2</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 21\) = -2</p> <p>From above,</p> <p>m<sub>1</sub>= m<sub>2</sub> = - 2</p> <p>Hence, the lines are parallel. <sub>Proved</sub></p>

Q9:

For what the value of k, the line kx - 3y + 6 = 0 is perpendicular to the lie joining (4, 3) and (5, -3)?


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>Given eq<sup>n</sup> is kx - 3y + 6 = 0</p> <p>Slope of above eq<sup>n</sup> is:</p> <p>m<sub>1</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac k{-3}\) = \(\frac k3\)</p> <p>Slope ofthe point (4, 3) and (5, -3) is:</p> <p>m<sub>2</sub> = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {-3 - 3}{5 - 4}\) = - \(\frac 61\) = -6</p> <p>If lines are perpendicular then:</p> <p>m<sub>1</sub>&times; m<sub>2</sub> = -1</p> <p>or, \(\frac k3\)&times; -6 = -1</p> <p>or, k = \(\frac {-1}{-2}\)</p> <p>&there4; k = \(\frac 12\) <sub>Ans</sub></p> <p></p>

Q10:

Write down the formula for the angle between the pair of lines y = m1x + c1 and y = m2x + c2, stating also the condition when they are perpendicular to each other.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>Given equations of the lines are:</p> <p>y = m<sub>1</sub>x + c<sub>1</sub>...............................(1)</p> <p>y = m<sub>2</sub>x + c<sub>2</sub>...............................(2)</p> <p>If \(\theta\) be the angle between two lines (1) and (2);</p> <p>The formula of angle between the given lines is:</p> <p>tan\(\theta\) =&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>&there4; \(\theta\) = tan<sup>-1</sup>(&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\))</p> <p>If two lines are perpendicular (\(\theta\) = 90&deg;)</p> <p>tan 90&deg; =&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or,&infin;=&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, \(\frac 10\) =&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, 1 + m<sub>1</sub>m<sub>2</sub> = 0</p> <p>&there4; m<sub>1</sub>m<sub>2</sub> = -1 <sub>Ans</sub></p>

Q11:

If two lines 2x + ay + 3 = 0 and 3x - 2y = 5 are perpendicular to each other find the value of a.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>The given equations are:</p> <p>2x + ay + 3 = 0....................(1)</p> <p>3x - 2y = 5.............................(2)</p> <p>Slope of equation (1), m<sub>1</sub> = -\(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2a\)</p> <p>Slope of equation (2), m<sub>2</sub> = -\(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 3{-2}\) = \(\frac 32\)</p> <p>If equation (1) and equation (2) are perpendicular to each other:</p> <p>m<sub>1</sub>&times; m<sub>2</sub> = -1</p> <p>or,- \(\frac 2a\)&times;\(\frac 32\) = -1</p> <p>or, -6 = - 2a</p> <p>or, a = \(\frac 62\)</p> <p>&there4; a = 3 <sub>Ans</sub></p> <p></p>

Q12:

Prove that 2x + 4y - 7 = 0 and 6x + 12y + 4 = 0 are parallel to each other.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>Given equations of the lines are:</p> <p>2x + 4y - 7 = 0...........................(1)</p> <p>6x + 12y + 4 = 0.......................(2)</p> <p>Slope of equation (1) is: m<sub>1</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 24\) = -\(\frac 12\)</p> <p>Slope of equation (2) is: m<sub>2</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 6{12}\) = -\(\frac 12\)</p> <p>&there4; m<sub>1</sub> = m<sub>2</sub> = \(\frac {-1}2\)</p> <p>Since, the slope of these equations are equal, the lines are parallel to each other. <sub>Proved</sub></p>

Q13:

Find the actual angle between the lines 3x + 5y = 7 and 3y = 2x + 4.


Type: Short Difficulty: Easy

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Answer: <p>Given lines are:</p> <p>3x + 5y = 7 i.e. 3x + 5y - 7 = 0........................(1)</p> <p>3y = 2x + 4 i.e. 2x - 3y + 4 = 0........................(2)</p> <p>Slope of eq<sup>n</sup> (1), m<sub>1</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-2}{-3}\) = \(\frac 23\)</p> <p>Slope of eq<sup>n</sup> (2), m<sub>2</sub> = - \(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-3}{5}\)</p> <p>Let \(\theta\) be the angle between the equation (1) and (2):</p> <p>tan\(\theta\) =&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, tan\(\theta\) =&plusmn; (\(\frac {\frac 23 + \frac 35}{1 - \frac 23 &times; \frac 35}\))</p> <p>or, tan\(\theta\) =&plusmn; (\(\frac {\frac {10 + 9}{15}}{\frac {5 - 2}5}\))</p> <p>or, tan\(\theta\) =&plusmn; (\(\frac {19}{15}\) &times; \(\frac 53\))</p> <p>or, tan\(\theta\) =&plusmn; \(\frac {19}9\)</p> <p>For acute angle,</p> <p>tan\(\theta\) = \(\frac {19}9\)= 2.11</p> <p>&there4; \(\theta\) = 65&deg;</p> <p>&there4; The acute angle between two lines is 65&deg;. <sub>Ans</sub></p>

Q14:

Find the acute angle between lines 3y - x - 6 = 0 and y = 2x + 5.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>Given equation are:</p> <p>3y - x - 6 = 0..............................(1)</p> <p>y = 2x + 5 i.e. -2x + y = 5...................(2)</p> <p>Slope of eq<sup>n</sup> (1), m<sub>1</sub>= -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {(-1)}3\) = \(\frac 13\)</p> <p>Slope of eq<sup>n</sup> (2), m<sub>2</sub>= -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {(-2)}1\) = 2</p> <p>If \(\theta\) be the angle between the eq<sup>n</sup> (1) and (2),</p> <p>tan\(\theta\) =&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, tan\(\theta\) =&plusmn; \(\frac {\frac 13 - 2}{1 + \frac 13 &times; 2}\)</p> <p>or, tan\(\theta\) =&plusmn; \(\frac {\frac {1 - 6}3}{\frac {3 + 2}3}\)</p> <p>or, tan\(\theta\) =&plusmn; \(\frac {-5}3\)&times; \(\frac 35\)</p> <p>&there4; tan\(\theta\) =&plusmn; (-1)</p> <p>Taking -ve sign,</p> <p>tan\(\theta\) = +1</p> <p>tan\(\theta\) = 45&deg;</p> <p>&there4;\(\theta\) = 45&deg; <sub>Ans</sub></p>

Q15:

Find the acute angle between the lines x - 3y = 4 and 2x - y = 3.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>x - 3y = 4......................(1)</p> <p>2x - y = 3......................(2)</p> <p>Slope of eq<sup>n</sup> (1), m<sub>1</sub> = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-1}{-3}\) = \(\frac 13\)</p> <p>Slope of eq<sup>n</sup> (2), m<sub>2</sub> = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-2}{-1}\) = 2</p> <p>If \(\theta\) be the angle between the eq<sup>n</sup> (1) and (2),</p> <p>tan\(\theta\) =&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, tan\(\theta\) =&plusmn; \(\frac {\frac 13 - 2}{1 + \frac 13 &times; 2}\)</p> <p>or, tan\(\theta\) =&plusmn; \(\frac {\frac {1 - 6}3}{\frac {3 + 2}3}\)</p> <p>or, tan\(\theta\) =&plusmn; \(\frac {-5}3\)&times; \(\frac 35\)</p> <p>&there4; tan\(\theta\) =&plusmn; (-1)</p> <p>Taking -ve sign,</p> <p>tan\(\theta\) = +1</p> <p>tan\(\theta\) = 45&deg;</p> <p>&there4;\(\theta\) = 45&deg; <sub>Ans</sub></p>

Q16:

In the given figure, find the value of \(\theta\).


Type: Short Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 192px;"><img src="/uploads/a26.jpg" alt="fdgs" width="118" height="163"><figcaption><br></figcaption></figure><p>Here,</p> <p>Given equation are:</p> <p>y - 3x - 2 = 0</p> <p>or, -3x + y - 2 = 0..............................(1)</p> <p>y = 2x + 5</p> <p>or, - 2x + y = 5...................................(2)</p> <p>Slope of eq<sup>n</sup> (1), m<sub>1</sub> = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-3}{-1}\) = 3</p> <p>Slope of eq<sup>n</sup> (2), m<sub>2</sub> = -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {-2}{1}\) =2</p> <p>If \(\theta\) be the angle between two lines,</p> <p>tan\(\theta\) =&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, tan\(\theta\) =&plusmn; \(\frac {3 - 2}{1 + 3 &times; 2}\)</p> <p>or, tan\(\theta\) =&plusmn; \(\frac 1{1 + 6}\)</p> <p>&there4; tan\(\theta\) =&plusmn; \(\frac 17\)</p> <p>Taking +ve sign,</p> <p>\(\theta\) = tan<sup>-1</sup>(\(\frac 17\))</p> <p>&there4; \(\theta\) = 8.13&deg; <sub>Ans</sub></p>

Q17:

Find the obtuse angle between the lines x = 3y + 8 and 2x + 11 = 7y.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>Given lines are:</p> <p>x = 3y + 8</p> <p>i.e. x - 3y - 8 = 0..................................(1)</p> <p>2x + 11 = 7y</p> <p>i.e. 2x - 7y + 11 = 0............................(2)</p> <p>Slope of eq<sup>n </sup>(1),<sup></sup>m<sub>1</sub>= - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 1{-3}\) = \(\frac 13\)</p> <p>Slope of eq<sup>n </sup>(2),<sup></sup>m<sub>2</sub>= - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2{-7}\) = \(\frac 27\)</p> <p>Let \(\theta\) be the angle between the equation (1) and (2),</p> <p>tan\(\theta\) =&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, tan\(\theta\) = &plusmn; \(\frac {\frac 13 - \frac 27}{1 + \frac 13 &times; \frac 27}\)</p> <p>or, tan\(\theta\) =&plusmn; \(\frac {\frac {7 - 6}{21}}{\frac {21 + 2}{21}}\)</p> <p>or, tan\(\theta\) =&plusmn; \(\frac 1{21}\)&times; \(\frac {21}{23}\)</p> <p>&there4; tan\(\theta\) =&plusmn; \(\frac 1{23}\)</p> <p>For obtuse angle,</p> <p>tan\(\theta\) = -\(\frac 1{23}\)</p> <p>or, tan\(\theta\) = tan (180&deg; -2&deg;)</p> <p>&there4; \(\theta\) = 178&deg;</p> <p>&there4; The obtuse angle between two lines is 178&deg;. <sub>Ans</sub></p>

Q18:

Find the equation of the lines passing the point (2, -1) and perpendicular to the line 5x - 7y + 10 = 0.


Type: Long Difficulty: Easy

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Answer: <p>Given equation is:</p> <p>5x - 7y + 10 = 0............................(1)</p> <p>Equation (1) changes in perpendicular form</p> <p>7x + 5y + k = 0.............................(2)</p> <p>The point (2, -1) passes through the equation (2)</p> <p>7&times; 2 + 5&times; -1 + k = 0</p> <p>or, 14 - 5 + k = 0</p> <p>or, 9 + k = 0</p> <p>&there4; k = -9</p> <p>Putting the value of k in equation (2)</p> <p>7x + 5y - 9 = 0</p> <p>&there4; Required equation is:7x + 5y - 9 = 0 <sub>Ans</sub></p>

Q19:

Find the equation of the line passing through the point (2, -3) and perpendicular to the line 5x - 4y + 19 = 0.


Type: Long Difficulty: Easy

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Answer: <p>Given line is:</p> <p>5x - 4y + 19 = 0...........................(1)</p> <p>Equation (1) changes into the perpendicular form:</p> <p>4x + 5y + k = 0............................(2)</p> <p>The point (2, -3) passes through the equation (2)</p> <p>4&times; 2 + 5&times; -3 + k = 0</p> <p>or, 8 - 15 + k = 0</p> <p>or, -7 + k = 0</p> <p>&there4; k = 7</p> <p>Putting the value of k in equation (2)</p> <p>4x + 5y + 7 = 0</p> <p>&there4; The required equation is 4x + 5y + 7 = 0. <sub>Ans</sub></p>

Q20:

Find the equation of the perpendicular bisector of the line joining point (3, -7) and (-5, 3).


Type: Long Difficulty: Easy

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Answer: <p>The slope of the line joining (3, - 7) and (-5, 3) is:</p> <p>m<sub>1</sub> = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {3 + 7}{-5 - 3}\) = \(\frac {10}{-8}\) = -\(\frac 54\)</p> <p>The mid-pointof the line joining (3, - 7) and (-5, 3) is:</p> <p>P(x, y) = (\(\frac {3 - 5}2\), \(\frac {-7 + 3}2\)) = (\(\frac {-2}2\), \(\frac {-4}2\)) = (-1, -2)</p> <p>The equation of the line passing through (-1, -2) is:</p> <p>y + 2 = m (x + 1)............................(1)</p> <p>Equation (1) is the perpendicular bisector of the line joining given two points,</p> <p>m<sub>1</sub>&times; m<sub>2</sub> = -1</p> <p>or,-\(\frac 54\)&times; m<sub>2</sub> = -1</p> <p>&there4; m<sub>2</sub> = \(\frac 45\)</p> <p>Putting the value of m<sub>2</sub> in equation (1)</p> <p>y + 2 = \(\frac 45\)(x + 1)</p> <p>or, 5y + 10 = 4x + 4</p> <p>or, 4x - 5y - 10 + 4 = 0</p> <p>&there4; 4x - 5y - 6 = 0 <sub>Ans</sub></p>

Q21:

Find the equation of the straight line, which passes through the point (-6, 4) and perpendicular to 3x - 4y + 9 = 0.


Type: Long Difficulty: Easy

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Answer: <p>Here,</p> <p>Given line is:</p> <p>3x - 4y + 9 = 0..........................(1)</p> <p>The line perpendicular to the equation (1) is:</p> <p>4x + 3y + k = 0.........................(2)</p> <p>The equation (2) passes through the point (-6, 4)</p> <p>4&times; -6 + 3&times; 4 + k = 0</p> <p>or, -24 + 12 + k = 0</p> <p>or, -12 + k = 0</p> <p>&there4; k = 12</p> <p>Putting the value of k in equation (2)</p> <p>4x + 3y + 12 = 0 <sub>Ans</sub></p>

Q22:

The points A and B have co-ordinates (3, -1) and (7, 1) respectively. Find the equation of the perpendicular bisector of AB.


Type: Long Difficulty: Easy

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Answer: <p>Given:</p> <p>The co-ordinatesof A and B are: (3, -1) and (7, 1) respectively.</p> <p>Equation of A(3, -1) and B(7, 1):</p> <p>y + 1 = \(\frac {1 + 1}{7 - 3}\) (x - 3) [\(\because\) y - y<sub>1</sub> = \(\frac {y_2 - y_1}{x_2 - x_1}\) (x - x<sub>1</sub>)]</p> <p>or, 4y + 4 = 2x - 6</p> <p>or, 2x - 4y - 6 - 4 = 0</p> <p>or, 2x - 4y - 10 = 0</p> <p>or, 2(x - 2y - 5) = 0</p> <p>&there4; x - 2y - 5 = 0...................................(1)</p> <p>Mid-point of A(3, -1) and B(7, 1)</p> <p>= (\(\frac {3 + 7}2\), \(\frac {-1 + 1}2\)) [\(\because\) (x, y) = (\(\frac {x_1 + x_2}2\), \(\frac {y_1 + y_2}2\))]</p> <p>= (\(\frac {10}2\), \(\frac 02\))</p> <p>= (5, 0)</p> <p>The perpendicular form of the equation x - 2y - 5 = 0 is:</p> <p>2x + y + k = 0........................(2)</p> <p>The point (5, 0) passes through the equation (2)</p> <p>2x + y + k = 0</p> <p>or, 2&times; 5 + 0 + k = 0</p> <p>or, 10 + k = 0</p> <p>&there4; k = -10</p> <p>Putting the value of k in equation (2),</p> <p>2x + y - 10 = 0 <sub>Ans</sub></p>

Q23:

Find the equation of the straight line which passes through the point (-2, -3) and perpendicular to the line 5x + 7y = 14.


Type: Long Difficulty: Easy

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Answer: <p>Here,</p> <p>Given equation is:</p> <p>5x + 7y - 14 = 0.........................(1)</p> <p>Equation (1) changes in perpendicular form</p> <p>7x - 5y + k = 0............................(2)</p> <p>Point (-2, -3) passes through the equation (2)</p> <p>7&times; (-2) - 5&times; (-3) + k = 0</p> <p>or, -14 + 15 + k = 0</p> <p>or, 1 + k = 0</p> <p>&there4;k = -1</p> <p>Putting the value of k in equation (2)</p> <p>7x - 5y - 1 = 0</p> <p>&there4; Required equation is:7x - 5y - 1 = 0 <sub>Ans</sub></p>

Q24:

Find the equation of the line passing through (2, 3) and perpendicular to the line 4x - 3y = 10.


Type: Long Difficulty: Easy

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Answer: <p>Given equation is:</p> <p>4x - 3y = 10</p> <p>i.e. 4x - 3y - 10 = 0..................................(1)</p> <p>Equation (1) changes in the perpendicular form</p> <p>3x + 4y + k = 0.........................................(2)</p> <p>Point (2, 3) passes through the equation (2)</p> <p>3&times; 2 + 4&times; 3 + k = 0</p> <p>or, 6 + 12 + k = 0</p> <p>or, 18 + k = 0</p> <p>&there4; k = -18</p> <p>Substituting the value of k in equation (2)</p> <p>3x + 4y - 18 = 0</p> <p>&there4; Required equation is: 3x + 4y - 18 = 0 <sub>Ans</sub></p>

Q25:

Find the equation of the line passing through the point (2, -3) and perpendicular to the line 3x + 4y + 18 = 0.


Type: Long Difficulty: Easy

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Answer: <p>Given equation is: 3x + 4y + 18 = 0.........................................(1)</p> <p>Equation (1) changes in perpendicular form:</p> <p>4x - 3y + k = 0.....................................(2)</p> <p>Point (2, -3) passes through the equation (2)</p> <p>4&times; 2 - 3&times; (-3) + k = 0</p> <p>or, 8 + 9 + k = 0</p> <p>or, 17 + k = 0</p> <p>&there4; k = -17</p> <p>Putting the value of k in eq<sup>n</sup> (2)</p> <p>4x - 3y - 17 = 0</p> <p>&there4; The required equation is:4x - 3y - 17 = 0 <sub>Ans</sub></p>

Q26:

Find the equation of the line which passes through the point (4, 6) and is perpendicular to the line x - 2y - 2 = 0.


Type: Long Difficulty: Easy

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Answer: <p>Given equation is: x - 2y - 2 = 0........................................(1)</p> <p>Equation (1) changes in perpendicular form:</p> <p>2x + y + k = 0.....................................(2)</p> <p>Point (4, 6) passes through the equation (2)</p> <p>2 &times; 4 + 6 + k = 0</p> <p>or, 8 + 6+ k = 0</p> <p>or, 14 + k = 0</p> <p>&there4; k = -14</p> <p>Putting the value of k in eq<sup>n</sup> (2)</p> <p>2x + y - 14 = 0</p> <p>&there4; The required equation is: 2x + y - 14 = 0<sub>Ans</sub></p>

Q27:

Find the equation of a straight line passing through the point (2, 3) and perpendicular to the straight line x - 3y - 2 = 0.


Type: Long Difficulty: Easy

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Answer: <p>Given equation is: x - 3y - 2 = 0............................(1)</p> <p>Equation (1) changes in perpendicular form:</p> <p>3x + y + k = 0.....................................(2)</p> <p>Point (2, 3) passes through the equation (2)</p> <p>3&times; 2+ 3+ k = 0</p> <p>or, 6+ 3+ k = 0</p> <p>or, 9+ k = 0</p> <p>&there4; k = -9</p> <p>Putting the value of k in eq<sup>n</sup> (2)</p> <p>3x + y -9 = 0</p> <p>&there4; The required equation is: 3x + y -9 = 0 <sub>Ans</sub></p>

Q28:

\(\triangle\)PQR has its vertices P(-2, 1), Q(2, 3) and R(-2, -4). Find the equation of the perpendicular drawn from the vertex P to QR.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 300px;"><img src="/uploads/b12.jpg" alt="bbz" width="142" height="152"><figcaption><br></figcaption></figure><p>Vertices of \(\triangle\)PQR are: P(-2, 1), Q(2, 3) and R(-2, -4).</p> <p>We know that:</p> <p>y - y<sub>1</sub> = \(\frac {y_2 - y_1}{x_2 - x_1}\)(x- x<sub>1</sub>)</p> <p>Eq<sup>n</sup> of QR,</p> <p>y - 3 = \(\frac {-4 - 3}{-2 - 2}\)(x - 2)</p> <p>or, y - 3 = \(\frac {-7}{-4}\)(x - 2)</p> <p>or, y - 3 = \(\frac {7}{4}\)(x - 2)</p> <p>or, 4y - 12 = 7x - 14</p> <p>or, 7x - 14 - 4y +12 = 0</p> <p>or, 7x - 4y - 2 = 0..................................(1)</p> <p>The eq<sup>n</sup> (1) change in perpendicular form,</p> <p>4x + 7y + k = 0.......................................(2)</p> <p>The point (-2, 1) passes through eq<sup>n</sup> (1),</p> <p>4&times; (-2) + 7&times; 1 + k = 0</p> <p>or, -8 + 7 + k = 0</p> <p>or, -1 + k = 0</p> <p>&there4; k = 1</p> <p>Putting the value of k in eq<sup>n</sup> (2)</p> <p>4x + 7y + 1 = 0 <sub>Ans</sub></p>

Q29:

A(-1, 5), B(-4, -1) and C(3, -2) are the vertices of \(\triangle\)ABC. Find the equation of the line which passes through the vertex A and parallel to the side BC.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 305px;"><img src="/uploads/c16.jpg" alt="xcx" width="143" height="134"><figcaption><br></figcaption></figure><p>Given:</p> <p>A(-1, 5), B(-4, -1) and C(3, -2) are the vertices of \(\triangle\)ABC.</p> <p>We know,</p> <p>Equation of the line passing through B(-4, -1) and C(3, -2) is:</p> <p>y - y<sub>1</sub> = \(\frac {y_2 - y_1}{x_2 - x_1}\)(x - x<sub>1</sub>)</p> <p>or, y + 1 = \(\frac {-2 + 1}{3 + 4}\)(x + 4)</p> <p>or, y + 1 = \(\frac {-1}7\)(x + 4)</p> <p>or, 7 (y + 1) = -1 (x + 4)</p> <p>or, 7y + 7 = -x - 4</p> <p>or, x + 7y + 4 +7 = 0</p> <p>&there4; x + 7y + 11 = 0..................................(1)</p> <p>The line parallel to the equation (1)</p> <p>x + 7y + k = 0..........................................(2)</p> <p>The equation (2) passes through the point A(-1, 5)</p> <p>-1 + 7&times; 5 + k = 0</p> <p>or, -1 + 35 + k = 0</p> <p>or, 34 + k = 0</p> <p>&there4; k = -34</p> <p>Putting the value of k in the equation (2)</p> <p>x + 7y - 34 = 0</p> <p>&there4; The required equation is:x + 7y - 34 = 0 <sub>Ans</sub></p>

Q30:

In the given figure D is the mid-point of BC. Prove that: AD ⊥ BC. Find the equation of AD also.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 400px;"><img src="/uploads/d8.jpg" alt="xc" width="147" height="147"><figcaption><br></figcaption></figure><p>Here,</p> <p>ABC is a triangle. The vertices of \(\triangle\)ABC are A(2, 8), B(-3, 5) and C(5, 3).</p> <p>D is the mid-point of BC.</p> <p>Co-ordinates of D</p> <p>= (\(\frac {x_1 + x_2}2\), \(\frac {y_1 + y_2}2\))</p> <p>= (\(\frac {-3 + 5}2\), \(\frac {5 + 3}2\))</p> <p>= (\(\frac 22\), \(\frac 82\))</p> <p>= (1, 4)</p> <p>Slope of BC (m<sub>1</sub>)</p> <p>= \(\frac {y_2 - y_1}{x_2 - x_1}\)</p> <p>= \(\frac {3 - 5}{5 + 3}\)</p> <p>= \(\frac {-2}8\)</p> <p>= \(\frac {-1}4\)</p> <p>Slope of AD (m<sub>2</sub>)</p> <p>= \(\frac {y_2 - y_1}{x_2 - x_1}\)</p> <p>= \(\frac {4 - 8}{1 - 2}\)</p> <p>= \(\frac {-4}{-1}\)</p> <p>= 4</p> <p>Now,</p> <p>m<sub>1</sub>m<sub>2</sub> = \(\frac {-1}4\)&times; 4 = -1</p> <p>The product of two slope is equal to -1 so these are perpendicular.</p> <p>&there4; AD&perp; BC <sub>Proved</sub></p> <p>Again,</p> <p>Equation of AD;</p> <p>y - y<sub>1</sub> =\(\frac {y_2 - y_1}{x_2 - x_1}\)(x - x<sub>1</sub>)</p> <p>or, y - 8 = \(\frac {4 - 8}{1 - 2}\)(x - 2)</p> <p>or, y - 8 = \(\frac {-4}{-1}\)(x - 2)</p> <p>or, y - 8 = 4x - 8</p> <p>or, 4x - 8 - y + 8 = 0</p> <p>&there4; 4x - y = 0</p> <p>Hence, the required equation is:4x - y = 0 <sub>Ans</sub></p>

Q31:

The vertices of A(3, 4), B(-2, 2) and C(3, -3) of a \(\triangle\)ABC. AD is perpendicular drawn from the vertex A on the opposite side BC. Find the equation of the AD.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 400px;"><img src="/uploads/e7.jpg" alt="xz" width="146" height="146"><figcaption><br></figcaption></figure><p>Here,</p> <p>The vertices of \(\triangle\)ABC are: A(3, 4), B(-2, 2) and C(3, -3).</p> <p>Equation of BC is:</p> <p>y - y<sub>1</sub>= \(\frac {y_2 - y_1}{x_2 - x_1}\)(x - x<sub>1</sub>)</p> <p>or, y - 2 = \(\frac {-3 - 2}{3 + 2}\)(x + 2)</p> <p>or, y - 2 = \(\frac {-5}5\)(x + 2)</p> <p>or, y - 2 = -x - 2</p> <p>or, x + 2 + y - 2 = 0</p> <p>&there4; x + y = 0.......................................(1)</p> <p>The eq<sup>n</sup> (1) changes in perpendicular form,</p> <p>x - y + k = 0......................................(2)</p> <p>The point A(3, 4) passes through eq<sup>n</sup> (1)</p> <p>3 - 4 + k = 0</p> <p>or, -1 + k = 0</p> <p>&there4; k = 1</p> <p>Putting the value of k in eq<sup>n</sup> (1)</p> <p>x - y + 1 = 0</p> <p>&there4; The required equation is:x - y + 1 = 0 <sub>Ans</sub></p>

Q32:

From the point P(-2, 4), if PQ is drawn perpendicular to the line 7x - 24y + 10 = 0. Find the equation of the line PQ. Also determine the length of PQ.


Type: Long Difficulty: Easy

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Answer: <p>Given equation is: 7x - 24y + 10 = 0......................(1)</p> <p>Equation (1) change in perpendicular form</p> <p>24x + 7y + k = 0..............................(2)</p> <p>Equation (2) passes through the point (-2, 4)</p> <p>24&times; (-2) + 7&times; 4 + k = 0</p> <p>or, -48 + 28 + k = 0</p> <p>or, -20 + k = 0</p> <p>&there4; k = 20</p> <p>Putting the value of k in equation (2)</p> <p>24x + 7y + 20 = 0</p> <p>Equation of the line PQ is: 24x + 7y + 20 = 0</p> <p>We know,</p> <p>d = \(\begin {vmatrix} \frac {Ax + By + C}{\sqrt {A^2 + B^2}}\\ \end {vmatrix}\)</p> <p>Perpendicular length of PQ;</p> <p>=\(\begin {vmatrix} \frac {7 &times; (-2) - 24 &times; 4 + 10}{\sqrt {7^2 + (24)^2}}\\ \end {vmatrix}\)</p> <p>=\(\begin {vmatrix} \frac {-14 - 96 + 10}{\sqrt {49 + 576}}\\ \end {vmatrix}\)</p> <p>=\(\begin {vmatrix} \frac {-100}{\sqrt {625}}\\ \end {vmatrix}\)</p> <p>=\(\begin {vmatrix} \frac {-100}{25}\\ \end {vmatrix}\)</p> <p>=\(\begin {vmatrix} -4\\ \end {vmatrix}\)</p> <p>= 4 units</p> <p>&there4; The perpendicular length of PQ = 4 units and equation of PQ is:24x + 7y + 20 = 0 <sub>Ans</sub></p>

Q33:

Find the equation of two lines which passes through (2, -1) and make an angle of 45° with the lines 6x + 5y - 1 = 0.


Type: Long Difficulty: Easy

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Answer: <p>Let: m be the slope of the required line, so that its equation is:</p> <p>y - y<sub>1</sub> = m(x - x<sub>1</sub>).............................(1)</p> <p>Point (2, -1) passes through equation (1):</p> <p>y + 1 = m(x - 2)...............................(2)</p> <p>Given equation is:</p> <p>6x + 5y - 1 = 0.................................(3)</p> <p>Slope of eq<sup>n</sup> (1) is:</p> <p>slope (m<sub>1</sub>) = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 65\)</p> <p>Now,</p> <p>Using angle formula,</p> <p>tan\(\theta\) =&plusmn; (\(\frac {m_1 - m_2}{1 + m_1m_2}\))</p> <p>or, tan 45&deg; =&plusmn; (\(\frac {m + \frac 65}{1 - m\frac 65}\))</p> <p>or, 1 =&plusmn; (\(\frac {\frac {5m + 6}5}{\frac {5 - 6m}5}\))</p> <p>or, 1 =&plusmn; (\(\frac {5m + 6}{5 - 6m}\))</p> <p>Taking +ve sign,</p> <p>5 - 6m = 5m + 6</p> <p>or, -6m - 5m = 6 - 5</p> <p>or, -11m = 1</p> <p>&there4; m = -\(\frac 1{11}\)</p> <p>Taking -ve sign,</p> <p>5 - 6m = - (5m + 6)</p> <p>or, 5 - 6m = -5m - 6</p> <p>or, -6m + 5m = - 6 - 5</p> <p>or, - m = - 11</p> <p>&there4; m = 11</p> <p>Putting the value of m = -\(\frac 1{11}\) in equation (2)</p> <p>y + 1 = m (x - 2)</p> <p>or, y + 1 = -\(\frac 1{11}\)(x - 2)</p> <p>or, 11y + 11 = -x + 2</p> <p>or, x + 11y + 11 - 2 = 0</p> <p>&there4; x + 11y + 9 = 0</p> <p>Putting the value of m = 11 in equation (2)</p> <p>y + 1 = m(x - 2)</p> <p>or, y + 1= 11 (x - 2)</p> <p>or, y + 1 = 11x - 22</p> <p>or, 11x - y - 22 - 1 = 0</p> <p>&there4; 1 1x - y - 23 = 0</p> <p>Hence, the required equations are: x + 11y + 9 = 0 and 11x - y - 23 = 0 <sub>Ans</sub></p>

Q34:

In the given figure the equation of AB is 12(x + 3) = 5y, find the equation of AC.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 400px;"><img src="/uploads/f4.jpg" alt="cx" width="143" height="143"><figcaption><br></figcaption></figure><p>Given:</p> <p>The eq<sup>n</sup> of AB is:</p> <p>12(x + 3) = 5y</p> <p>or, 12x + 36 - 5y = 0</p> <p>or, 12x - 5y + 36 = 0........................(1)</p> <p>Slope of eq<sup>n</sup> (1) is: m<sub>1</sub> = \(\frac {12}5\)</p> <p>The eq<sup>n</sup> of the line passes through (2, 3) is:</p> <p>y - y<sub>1</sub> = m (x - x<sub>1</sub>)</p> <p>or, y - 3 = m (x - 2)..........................(2)</p> <p>Slope of eq<sup>n</sup> (2) is: m<sub>2</sub> = m</p> <p>The angle between the lines (1) and (2) is 45&deg;.</p> <p>tan\(\theta\) =&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, tan 45&deg; =&plusmn; \(\frac {\frac {15}2 - m}{1 + \frac {12}5m}\)</p> <p>or, 1 =&plusmn; \(\frac {\frac {12 - 5m}5}{\frac {5 + 12m}5}\)</p> <p>or, 1 =&plusmn; \(\frac {12 - 5m}{5 + 12m}\)</p> <p>Taking +ve sign,</p> <p>5 + 12m = 12 - 5m</p> <p>or, 12m + 5m = 12 - 5</p> <p>or, 17m = 7</p> <p>&there4; m = \(\frac 7{17}\)</p> <p>Taking -ve sign,</p> <p>5 + 12m = - 12 + m</p> <p>or, 12m - 5m = - 12 - 5</p> <p>or, 7m = - 17</p> <p>&there4; m = -\(\frac {17}7\)</p> <p>Substituting the value of m = \(\frac 7{17}\) in eq<sup>n</sup> (2)</p> <p>y - 3 = \(\frac 7{17}\)(x - 2)</p> <p>or, 17(y - 3) = 7(x - 2)</p> <p>or, 17y - 51 = 7x - 14</p> <p>or, 7x - 17y + 51 - 14 = 0</p> <p>&there4; 7x - 17y + 37 = 0</p> <p>Substituting the value of m = \(\frac {-17}7\) in eq<sup>n</sup> (2)</p> <p>y - 3 = \(\frac {-17}7\)(x - 2)</p> <p>or, 7 (y - 3) = -17 (x - 2)</p> <p>or, 7y - 21 = - 17x + 34</p> <p>or, 17x - 34 + 7y - 21 = 0</p> <p>&there4; 17x + 7y - 55 = 0</p> <p>&there4; The required equations are:7x - 17y + 37 = 0 and17x + 7y - 55 = 0 <sub>Ans</sub></p>

Q35:

Find the equation of the straight lines passing through the point (4, 3) and making an angle of 60° to the line \(\sqrt 3\)x - y = 4.


Type: Long Difficulty: Easy

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Answer: <p>Here,</p> <p>The eq<sup>n</sup> of line is:</p> <p>\(\sqrt 3\)x - y = 4.............................(1)</p> <p>The slope of eq<sup>n</sup> (1) is: m<sub>1</sub> =\(\sqrt 3\)</p> <p>The eq<sup>n</sup> of the line passes through (4, 3) is:</p> <p>y - 3 = m(x - 4)..........................(2)</p> <p>Slope of eq<sup>n</sup> (2) is m<sub>2</sub> = m</p> <p>The angle between eq<sup>n</sup> (1) and (2) is 60&deg;.</p> <p>tan\(\theta\) =&plusmn; \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, tan 60&deg; =&plusmn; \(\frac {\sqrt 3 -m}{1 + \sqrt3m}\)</p> <p>or, \(\sqrt 3\) =&plusmn; \(\frac {\sqrt 3 - m}{1 + \sqrt 3m}\)</p> <p>Taking +ve sign,</p> <p>\(\sqrt 3\)(1 + \(\sqrt 3\)m) = \(\sqrt 3\) - m</p> <p>or, \(\sqrt 3\) + 3m = \(\sqrt 3\) - m</p> <p>or, 3m + m = \(\sqrt 3\) - \(\sqrt3\)</p> <p>or, 4m = 0</p> <p>&there4; m = 0</p> <p>Taking -ve sign,</p> <p>\(\sqrt 3\) + 3m = -\(\sqrt 3\) + m</p> <p>or, 3m - m = -\(\sqrt 3\) - \(\sqrt 3\)</p> <p>or, 2m = - 2\(\sqrt 3\)</p> <p>or, m = \(\frac {-2\sqrt 3}2\)</p> <p>&there4; m = -\(\sqrt 3\)</p> <p>Putting the value of m = 0 in eq<sup>n</sup> (2)</p> <p>y - 3 = m(x - 4)</p> <p>or, y - 3 = 0(x - 4)</p> <p>&there4; y - 3 = 0</p> <p>Putting the value of m = -\(\sqrt 3\) in eq<sup>n</sup> (2)</p> <p>y - 3 = m (x - 4)</p> <p>or, y - 3 = -\(\sqrt 3\) (x - 4)</p> <p>or, y - 3 = -\(\sqrt 3\)x + 4\(\sqrt 3\)</p> <p>&there4; \(\sqrt 3\)x + y = 4\(\sqrt 3\) + 3</p> <p>&there4; The required equations are: y - 3 = 0 and \(\sqrt 3\)x + y = 4\(\sqrt 3\) + 3 <sub>Ans</sub></p>

Q36:

Find the equation of straight line which passes through the point (1, -4) and make an angle of 45° with the straight line 2x + 3y + 5 = 0.


Type: Long Difficulty: Easy

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Answer: <p>Here,</p> <p>The given eq<sup>n</sup> is:</p> <p>2x + 3y + 5 = 0............................(1)</p> <p>Slope of equation (1) is: m<sub>1</sub> = \(\frac {-2}3\)</p> <p>The equation of the line passes through (1, -4) is:</p> <p>y + 4 = m(x - 1)...........................(2)</p> <p>Slope of equation (2) is: m<sub>2</sub> = m</p> <p>The angle between the lines (1) and (2) is 45&deg;.</p> <p>tan\(\theta\) =&plusmn; (\(\frac {m_1 - m_2}{1 + m_1m_2}\))</p> <p>or, tan 45&deg; =&plusmn; (\(\frac {\frac {-2}3 - m}{1 + m(\frac {-2}3)})\)</p> <p>or, 1 =&plusmn; (\(\frac {\frac {-2 - 3m}3}{\frac {3 - 2m}3})\)</p> <p>or, 1 =&plusmn; (\(\frac {-2 - 3m}{3 - 2m})\)</p> <p>Taking +ve sign,</p> <p>1 = \(\frac {-2 - 3m}{3 - 2m}\)</p> <p>or, -2 - 3m = 3 - 2m</p> <p>or, -3m + 2m = 3 + 2</p> <p>or, -m = 5</p> <p>&there4; m = -5</p> <p>Taking -vesign,</p> <p>1 = \(\frac {-(-2 - 3m)}{3 - 2m}\)</p> <p>or, 3 - 2m = 2 + 3m</p> <p>or, 3m + 2m = 3 - 2</p> <p>or, 5m = 1</p> <p>&there4; m = \(\frac 15\)</p> <p>&there4; m = -5, \(\frac 15\)</p> <p>Substituting the value of m = -5 in equation (2)</p> <p>y + 4 = - 5 (x - 1)</p> <p>or, y + 4 = - 5x + 5</p> <p>or, 5x + y + 4 - 5 = 0</p> <p>&there4; 5x + y - 1 = 0</p> <p>Substituting the value of m = \(\frac 15\) in equation (2)</p> <p>y + 4 = \(\frac 15\)(x - 1)</p> <p>or, 5y + 20 = x - 1</p> <p>or, x - 1 - 5y - 20 = 0</p> <p>&there4; x - 5y - 21 = 0</p> <p>Hence, the required equations are:5x + y - 1 = 0 andx - 5y - 21 = 0 <sub>Ans</sub></p>

Q37:

Find the equation of the straight lines passing through the point (2, 3) and making an angle of 45° with the line x - 3y = 2.


Type: Long Difficulty: Easy

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Answer: <p>Here,</p> <p>The given equation is:</p> <p>x - 3y = 2.............................(1)</p> <p>The slope of eq<sup>n</sup> (1) is: m<sub>1</sub> = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-1}{-3}\) = \(\frac 13\)</p> <p>Let, the equation of the line passes through point (2, 3) is;</p> <p>y - 3 = m(x - 2)....................(2) [\(\because\)y - y<sub>1</sub> = m(x - x<sub>1</sub>)]</p> <p>The slope of equation (2) is: m<sub>2</sub> = m</p> <p>If the angle between the lines (1) and (2) is 45&deg;, then:</p> <p>tan\(\theta\) =&plusmn; (\(\frac {m_1 - m_2}{1 + m_1m_2})\)</p> <p>or, tan 45&deg; =&plusmn; (\(\frac {\frac 13 - m}{1 + \frac 13m})\)</p> <p>or, 1 =&plusmn; \(\frac {\frac {1 - 3m}3}{\frac {3 + m}3}\)</p> <p>or, 1 =&plusmn; \(\frac {1 - 3m}{3 + m}\)</p> <p>Taking +ve sign,</p> <p>1 = \(\frac {1 - 3m}{3 + m}\)</p> <p>3 + m = 1 - 3m</p> <p>or, 3m + m = 1 - 3</p> <p>or, 4m = - 2</p> <p>or, m = -\(\frac 24\)</p> <p>&there4; m = - \(\frac 12\)</p> <p>Taking -ve sign,</p> <p>1 = -\(\frac {1 - 3m}{3 + m}\)</p> <p>or, 3 + m = - (1 - 3m)</p> <p>or, 3 + m = -1 + 3m</p> <p>or, 3m - m = 3 + 1</p> <p>or, 2m = 4</p> <p>or, m = \(\frac 42\)</p> <p>&there4; m = 2</p> <p>Putting the value of m = - \(\frac 12\) in eq<sup>n</sup> (2)</p> <p>y - 3 = -\(\frac 12\)(x - 2)</p> <p>or, 2y - 6 = - x + 2</p> <p>or, x + 2y - 2 - 6 = 0</p> <p>&there4; x + 2y - 8 = 0</p> <p>Putting the value of m = 2 in eq<sup>n</sup> (2)</p> <p>y - 3 = 2(x - 2)</p> <p>or, y - 3 = 2x - 4</p> <p>or, 2x - 4 - y + 3 = 0</p> <p>&there4; 2x - y - 1 = 0</p> <p>&there4; The required equations are: x + 2y - 8 = 0 and2x - y - 1 = 0 <sub>Ans</sub></p>

Q38:

Find the angle between two lines whose equations are y = m1x + c1 and y = m2x + c2.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 150px;"><img src="/uploads/g4.jpg" alt="vb" width="150" height="150"><figcaption><br></figcaption></figure><p>Let: AB and AC be two lines whose equations are: y = m<sub>1</sub>x + c<sub>1</sub> and y = m<sub>2</sub>x + c<sub>2</sub>, these lines meet OX in the point B and C respectively.</p> <p>Let: \(\angle\)ABX = \(\theta_1\) and \(\angle\) = \(\theta_2\)</p> <p>Let: tan\(\theta_1\) = m<sub>1</sub> and tan\(\theta_2\) = m<sub>2</sub></p> <p>Let: \(\angle\)CAB = \(\theta\), then;</p> <p>\(\theta_1\) = \(\theta\) + \(\theta_2\)</p> <p>&there4; \(\theta\) = \(\theta_1\) - \(\theta_2\)</p> <p>Taking tan on both sides,</p> <p>tan\(\theta\) = tan(\(\theta_1\) - \(\theta_2\))</p> <p>or, tan\(\theta\) = \(\frac {tan\theta_1 - tan\theta_2}{1 + tan\theta_1\theta_2}\)</p> <p>or, tan\(\theta\) = \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>&there4; \(\theta\) = tan<sup>-1</sup>(\(\frac {m_1 - m_2}{1 + m_1m_2}\)) <sub>Ans</sub></p>

Q39:

Find the angle between two lines whose equation are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.


Type: Long Difficulty: Easy

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Answer: <p>Given equations are:</p> <p>a<sub>1</sub>x + b<sub>1</sub>y + c<sub>1</sub> = 0............................(1)</p> <p>a<sub>2</sub>x + b<sub>2</sub>y + c<sub>2</sub> = 0............................(2)</p> <p>Let: m<sub>1</sub> and m<sub>2</sub> represents the slope of the lines (1) and (2) respectively.</p> <p>Slope of (1), m<sub>1</sub>= -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {a_1}{b_1}\)</p> <p>Slope of (2), m<sub>2</sub>= -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {a_2}{b_2}\)</p> <p>Let: \(\theta\) be the angle between the two given lines. Then,</p> <p>tan\(\theta\) =&plusmn; (\(\frac {m_1 - m_2}{1 + m_1m_2})\)</p> <p>or, tan\(\theta\) =&plusmn; (\(\frac {(\frac {-a_1}{b_1}) - (\frac {-a_2}{b_2})}{1 + (\frac {-a_1}{b_1})(\frac {-a_2}{b_2})})\)</p> <p>or, tan\(\theta\) =&plusmn; (\(\frac {\frac {-a_1b_2 + a_2b_1}{b_1b_2}}{\frac {b_1b_2 + a_1a_2}{b_1b_2}})\)</p> <p>or, tan\(\theta\) =&plusmn; (\(\frac {a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2})\)</p> <p>&there4; \(\theta\) = tan<sup>-1</sup>&plusmn; (\(\frac {a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2})\) <sub>Ans</sub></p>

Q40:

Find the equation of the sides of an equilateral triangle whose vertex is (-1, 2) and base is y = 0.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 150px;"><img src="/uploads/h2.jpg" alt="b" width="150" height="150"><figcaption><br></figcaption></figure><p>Let: ABC is an equilateral triangle with base BC on x-axis.</p> <p>i.e. y = 0, \(\angle\)CBA = 60&deg; and \(\angle\)XCA = 120&deg;</p> <p>Slope of line AB = m<sub>1</sub> = tan\(\theta\) = tan 60&deg; = \(\sqrt 3\)</p> <p>Slope of line CA = m<sub>2</sub> = tan\(\theta\) = tan 120&deg; = - \(\sqrt 3\)</p> <p>The eq<sup>n</sup> of AB is:</p> <p>y - y<sub>1</sub> = m(x - x<sub>1</sub>)</p> <p>or, y - 2 = \(\sqrt 3\)(x + 1)</p> <p>or, y - 2 = \(\sqrt 3\)x + \(\sqrt 3\)</p> <p>&there4;\(\sqrt 3\)x - y + 2 + \(\sqrt 3\) = 0</p> <p>The eq<sup>n</sup> of the sides CA is:</p> <p>y - y<sub>1</sub> = m(x - x<sub>1</sub>)</p> <p>or, y - 2 = -\(\sqrt 3\) (x + 1)</p> <p>or, y - 2 = -\(\sqrt 3\)x - \(\sqrt 3\)</p> <p>&there4;\(\sqrt 3\)x + y - 2 + \(\sqrt 3\) = 0</p> <p>Hence, the required lines are:\(\sqrt 3\)x - y + 2 + \(\sqrt 3\) = 0 and\(\sqrt 3\)x + y - 2 + \(\sqrt 3\) = 0 <sub>Ans</sub></p>

Q41:

Find the equation of the straight line passing through the point of intersection of the lines x + 2y = 3 and 2x - 3y = 20 and perpendicular to the line 2x - 3y + 5 = 0.


Type: Long Difficulty: Easy

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Answer: <p>Given equations are:</p> <p>x + 2y = 3...........................(1)</p> <p>2x - 3y = 20......................(2)</p> <p>Eq<sup>n</sup> (1) is multipliedby 2 and subtract with eq<sup>n</sup> (2)</p> <table width="111"><tbody><tr><td>2x</td> <td>+</td> <td>4y</td> <td>=</td> <td>6</td> </tr><tr><td>2x</td> <td>-</td> <td>3y</td> <td>=</td> <td>20</td> </tr><tr><td>-</td> <td>+</td> <td></td> <td></td> <td>-</td> </tr><tr><td></td> <td></td> <td>7y</td> <td>=</td> <td>-14</td> </tr></tbody></table><p>y = -\(\frac {14}7\)</p> <p>&there4; y = -2</p> <p>Putting the value of y in eq<sup>n</sup> (1)</p> <p>x + 2y = 3</p> <p>or, x + 2&times; (-2) = 3</p> <p>or, x - 4 = 3</p> <p>or, x = 3 + 4</p> <p>&there4; x = 7</p> <p>The point of intersection is: (7, -2)</p> <p>The given eq<sup>n</sup> is:</p> <p>2x - 3y + 5 = 0.............................(3)</p> <p>The eq<sup>n</sup> (3) changes in perpendicular form:</p> <p>-3x - 2y + k = 0..........................(4)</p> <p>The point (7, -2) passes through eq<sup>n</sup> (4)</p> <p>-3&times; 7 - 2&times; (-2) + k = 0</p> <p>or, -21 + 4 + k = 0</p> <p>or, -17 + k = 0</p> <p>&there4; k = 17</p> <p>Putting the value of k in eq<sup>n</sup> (4)</p> <p>-3x - 2y + 17 = 0</p> <p>&there4; 3x + 2y - 17 = 0 <sub>Ans</sub></p>

Q42:

Find the equation of the straight line passing through the point that divides the join of (-3, 4) and (7, 1) in the ratio 3 : 2 and perpendicular to the line.


Type: Long Difficulty: Easy

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Answer: <p>Given points are: (-3, -4) and (7, 1)</p> <p>Eq<sup>n</sup> of two points is:</p> <p>y - y<sub>1</sub> = \(\frac {y_2 - y_1}{x_2 - x_1}\)(x - x<sub>1</sub>)</p> <p>or, y + 4 = \(\frac {1 + 4}{7 + 3}\)(x + 3)</p> <p>or, y +4 = \(\frac 5{10}\)(x + 3)</p> <p>or, y +4 = \(\frac 12\)(x + 3)</p> <p>or, 2y + 8 = x + 3</p> <p>or, x + 3 - 2y - 8 = 0</p> <p>&there4; x - 2y - 5 = 0.............................(1)</p> <p>The points (-3, -4) and (7, 1) divides by a point P(x, y) in the ratio 3 : 2.</p> <p>Using section formula,</p> <p>P(x, y) = (\(\frac {m_1x_2 + m_2x_1}{m_1 + m_2}\), \(\frac {m_1y_2 + m_2y_1}{m_1 + m_2})\)</p> <p>or,P(x, y) = (\(\frac {3 &times; 7 + 2 &times; (-3)}{3 + 2}\), \(\frac {3 &times; 1 + 2 &times; (-4)}{3 + 2}\))</p> <p>or,P(x, y) = (\(\frac {21 - 6}5\), \(\frac {3 - 8}5\))</p> <p>or,P(x, y) = (\(\frac {15}5\), \(\frac {-5}5\))</p> <p>&there4; P(x, y) = (3, -1)</p> <p>The eq<sup>n</sup> (1) changes in perpendicular form is:</p> <p>-2x - y + k = 0....................................(2)</p> <p>The point (3, -1) passes through eq<sup>n</sup> (2)</p> <p>-2&times; 3 - (-1) + k = 0</p> <p>or, -6 + 1 + k = 0</p> <p>or, -5 + k = 0</p> <p>&there4; k = 5</p> <p>Putting the value of k in eq<sup>n</sup> (2)</p> <p>-2x - y + 5 = 0</p> <p>&there4; 2x + y - 5 = 0 <sub>Ans</sub></p>

Q43:

If (2, 3) and (-6, 5) are the ends points of the diagonal of a square. Find the equation of the other diagonal.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 150px;"><img src="/uploads/i5.jpg" alt="f" width="150" height="150"><figcaption><br></figcaption></figure><p>Here,</p> <p>The diagonals of the square bisect each other.</p> <p>Mid-point of AC = (\(\frac {x_1 + x_2}2\), \(\frac {y_1 + y_2}2\)) = (\(\frac {2 - 6}2\), \(\frac {3 + 5}2\)) = (\(\frac {-4}2\), \(\frac 82\)) = (-2, 4)</p> <p>The eq<sup>n</sup> of the points (2, 3) and (-6, 5) is:</p> <p>y - y<sub>1</sub> = \(\frac {y_2 - y_1}{x_2 - x_1}\)(x - x<sub>1</sub>)</p> <p>or, y - 3 = \(\frac {5 - 3}{-6 - 2}\)(x - 2)</p> <p>or, y - 3 = \(\frac 2{-8}\)(x - 2)</p> <p>or, y - 3 = \(\frac 1{-4}\)(x - 2)</p> <p>or, - 4y + 12 = x - 2</p> <p>or, x - 2 + 4y - 12 = 0</p> <p>&there4; x + 4y - 14 = 0.............................(1)</p> <p>Since, the diagonals of the square are perpendicular to each other.</p> <p>The eq<sup>n</sup> (1) change in perpendicular form:</p> <p>4x - y + k = 0.....................................(2)</p> <p>The point (-2, 4) passes through eq<sup>n</sup> (2)</p> <p>4&times; (-2) - 4 + k = 0</p> <p>or, -8 - 4 + k = 0</p> <p>or, - 12 + k = 0</p> <p>&there4; k = 12</p> <p>Putting the value of k in eq<sup>n</sup> (2)</p> <p>4x - y + 12 = 0</p> <p>&there4; The required equation is:4x - y + 12 = 0 <sub>Ans</sub></p>

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Environmental Effects of Population Growth

Environmental Effects of Population Growth

The increasing population results into increasing demands, needs and services. Their needs are unlimited but country has limited resources to fulfill their needs and desire. The growing population needs more fuel, timber, food, cloth, and more facilities of education and health. The people started to exploit the natural resources which results in the environment degradation. Thus society, culture, natural resources and economic development are affected by the growing population.

It is really important to maintain a sustainable balance between population growth and available resources. If the population growth is in proportion to the means and resources of a country, it does not have a negative effect on its social- economic condition. When socio- economic condition is improved, it helps to control population growth. So there must be a proper and balance between the population growth and economic growth rate.

In this respect, various consequences of population growth are as follows:

Environmental effects

The environment of any area is affected by the composition of population and growth rate of population. If there is any imbalance between population growth and the environment, the environment gets degraded and people have to face many difficulties to meet their needs and demands. Let's analyze how various elements are affected by population growth:

Forest and vegetation

Forest and vegetation are the most important aspects of human life. Various species of vegetation play an important role increasing a balance in the environment, they help to keep environment clean, fresh, green and beautiful. They protect the soil from landslides and soil erosion. The needs and demands of people are increasing along with the growth of population. It leads to degradation of forest and vegetation.

Many forest and vegetation are being cleared by the people for fodder, grass and their settlement. Thus, deforestation has become an acute problem in Nepal. Due to deforestation many species of wildlife, medicinal herbs and other important forest products are in acute shortage. The more people, the more is the demand of firewood. This trend in Nepal has resulted into rapid deforestation and loss of natural resources and means.

Aquatic lives and terrestrial animals

The growth of population has affected the marine life adversely. The water resources such as underground water, river, well, pond, tank, sea and ocean, etc. have begun to dry and the existing ones have been polluted. The survival of fish and amphibians has disappearance. It has badly affected man, other creatures and plants. Thus, the growth of population has made the condition of aquatic creatures very difficult. The terrestrial part refers to the earth crust. It composed of soil, sand, dust particles and stone. Thousands of creatures are living population has creature environmental pollution. It has created noise pollution. Various kinds of insects, birds, creatures and animals are vanishing because their habitat has been destroyed by man.

Air

Living organisms cannot survive without fresh air. Fresh air saves the life. Polluted air causes diseases. Oxygen content of the atmosphere decreases by deforestation. Contents of carbon dioxide and other toxic gases increase in the atmosphere.

Unmanaged urbanization and industrialization produces smoke, dust particles and other harmful wastes. Insecticides and other toxic chemicals mix inn air and pollute the fresh air of environment. Polluted air can produce different kinds of diseases like skin diseases, eye irritation, lung diseases, etc. Vegetation is also affected by the smoke and dust particle.

Water

The source of water is adversely affected by the growth of population. The quality water is being degraded due to growing human activities in the environment. Sometimes, soil is deposited in the sources of water. It causes water pollution. Dumping of garbage and other waste, discharge of excreta materials into river make the water polluted. People become the victim of various types of diseases by drinking polluted water. Water is also polluted by excessive use of chemical fertilizers and insecticides. Chemical fertilizers insecticides remain in soil for a long period of time and slowly move down and pollute the underground water.

Land

Land is the most important natural resource and sources of varieties of food. Human activities are responsible for land pollution and the loss of productivity of the land. Use of fertilizers, insecticides, etc. kill the natural fertility of the soil and soil finally losses its quality. The pressure on land increases with the increasing population. The size of land cannot increase along with the growth of population. There is a high possibility of imbalance between the population and the available land.

Floods

Floods are usually called natural calamities. But human activities have contributed to the sharp rise in floods. Cultivation of land, deforestation, overgrazing and mining has removed water absorbing vegetation and soil. Urbanization also increases flooding. Heavy rainfall occurs during the monsoon period. This down-power runs off from the mountain so rapidly that most of it cannot be controlled and used. The massive runoff water causes periodic flooding. During, the monsoon period, the speed and level of water is increased in streams, rivulet and rivers.

Traditional method of cultivation and deforestation in the hilly region has caused soil erosion and landslide. These activities have also reduced the water holding capacity of soil which has favored the flooding. Flood transport top soil from the naked mountains and deposits in the plain areas. Each year millions of the flood is landslides, soil erosion, loss fertile land, killing of the people and property damages, sometimes, villages completely get submerged under the flooded water.

Noise pollution

Human settlement has become very dense many places of Nepal due to the rapid growth in population. People conduct a lot of activities in the growth of urbanization and establishment of much unpleasant or unwanted sound that causes discomfort. Noise increases along with the growth of urbanization and establishment of many industries in the town areas. The sound produces by increasing number of buses, trucks, motorcycle, trains, airplanes and heavy use of radio, television, etc is polluting environment and causing sound pollution. In Nepal, the level of noise is too high which may cause loss of hearing power as well as various other socio-psychological disorders. The measurement of sound is expressed in decibel. The sound with more than eighty db is considered very harmful which degrades the whole environment.

Lesson

Causes and Effects of Population Change

Subject

Enviroment Population and Health

Grade

Grade 9

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