Alcohol
Alcohol is defined as an organic compound containing the hydroxyl group (-OH) attached to a saturated carbon atom or hydrocarbon radical. This note has brief introduction about alcohol and also the uses and properties of different types of alcohol.
Summary
Alcohol is defined as an organic compound containing the hydroxyl group (-OH) attached to a saturated carbon atom or hydrocarbon radical. This note has brief introduction about alcohol and also the uses and properties of different types of alcohol.
Things to Remember
- Alcohol is defined as an organic compound containing the hydroxyl group (-OH) attached to a saturated carbon atom or a hydrocarbon radical.
- Types of Alcohol: methyl alcohol and ethyl alcohol
- Methyl alcohol is colorless liquid boiling at 650C
- Ethyl alcohol is colorless liquid but it has taste.
- Ethyl alcohol is soluble in water.
- The boiling point of ethyl alcohol is 780C and freezing point of ethyl alcohol is -1170C.
MCQs
No MCQs found.
Subjective Questions
Q1:
Find the equation of the lines represented by:
3x2 - 5xy - 2y2 - x + 2y = 0
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>3x<sup>2</sup> - 5xy - 2y<sup>2</sup> - x + 2y = 0</p> <p>or, 3x<sup>2</sup> - 6xy + xy - 2y<sup>2</sup> - x + 2y = 0</p> <p>or, 3x(x - 2y) + y(x - 2y) - 1(x - 2y) = 0</p> <p>or, (x - 2y) (3x + y - 1) = 0</p> <p>Either: x - 2y = 0</p> <p>Or: 3x + y - 1 = 0</p> <p>∴ The required equations are: x - 2y = 0 and 3x + y - 1 = 0<sub>Ans</sub></p>
Q2:
Find the separate equation of the line given by equation:
x2 - 3xy + 2y2 = 0
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>x<sup>2</sup> - 3xy + 2y<sup>2</sup> = 0</p> <p>or, x<sup>2</sup> - 2xy - xy + 2y<sup>2</sup> = 0</p> <p>or, x(x - 2y) - y(x - 2y) = 0</p> <p>or, (x - 2y) (x - y) = 0</p> <p>Either: x - 2y = 0</p> <p>Or: x - y = 0</p> <p>∴ The required seperate equations are:x - 2y = 0 andx - y = 0 <sub>Ans</sub></p>
Q3:
Show that the straight lines represented by the equation 6x2 + 11xy - 6y2 = 0 are perpendicular to each other.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>The given equation is:6x<sup>2</sup> + 11xy - 6y<sup>2</sup> = 0..................(1)</p> <p>The homogeneous equation of second is: ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0.................(2)</p> <p>Comparing (1) and (2)</p> <p>a = 6</p> <p>2h = 11 i.e. h = \(\frac {11}2\)</p> <p>b = -6</p> <p>Now.</p> <p>a + b = 0</p> <p>or, 6 - 6 = 0</p> <p>∴ 0 = 0</p> <p>Hence, a + b = 0 is satisfied by the given equation so the equation6x<sup>2</sup> + 11xy - 6y<sup>2</sup> = 0 are perpendicular to each other. <sub>Proved</sub></p> <p></p>
Q4:
Prove that the straight lines represented by the equation x2 - 4xy + 4y2 = 0 are coincident to each other.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given equation is:x<sup>2</sup> - 4xy + 4y<sup>2</sup> = 0......................(1)</p> <p>The homogenous equation of second degree is:</p> <p>ax<sup>2</sup>+ 2hxy + by<sup>2</sup> = 0...........................(2)</p> <p>Comparing (1) and (2)</p> <p>a = 1</p> <p>2h = -4 i.e. h = \(\frac {-4}2\) = -2</p> <p>b = 4</p> <p>Now,</p> <p>h<sup>2</sup> = ab</p> <p>or, (-2)<sup>2</sup> = 1× 4</p> <p>∴ 4 = 4</p> <p>Hence, h<sup>2</sup> = ab is satisfied by the given equation so the equationx<sup>2</sup> - 4xy + 4y<sup>2</sup> = 0 are coincident to each other. <sub>Proved</sub></p>
Q5:
Write down the condition for the lines represented by the equation ax2 + 2hxy + by2 = 0 to be coincident and perpendicular to each other.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0</p> <p>When: h<sup>2</sup> = ab then straight lines are coincident each other.</p> <p>When: a + b = 0 then straight lines are perpendicular each other.</p>
Q6:
Find the homogeneous equation of the second degree from the pair of lines:
x = 2y and 2x = y
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Given eq<sup>n</sup> of line are:</p> <p>x = 2y</p> <p>i.e. x - 2y = 0..........................(1)</p> <p>2x = y</p> <p>i.e. 2x - y = 0..........................(2)</p> <p>Combined eq<sup>n</sup> of (1) and (2) is:</p> <p>(x - 2y) (2x - y) = 0</p> <p>or, 2x<sup>2</sup>- xy - 4xy + 2y<sup>2</sup> = 0</p> <p>or, 2x<sup>2</sup> - 5xy + 2y<sup>2</sup> = 0</p> <p>∴ The required eq<sup>n</sup> is: 2x<sup>2</sup> - 5xy + 2y<sup>2</sup> = 0 <sub>Ans</sub></p>
Q7:
Find a pair of eqn of straight lines represented by y2 = x2 and hence prove that they are perpendicular to each other.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given eq<sup>n</sup> is: y<sup>2</sup> = x<sup>2</sup></p> <p>or, x<sup>2</sup> - y<sup>2</sup> = 0</p> <p>or, (x - y) (x + y) = 0</p> <p>either: x + y = 0</p> <p>Or, x - y = 0</p> <p>Slope of eq<sup>n</sup> x + y = 0 is: m<sub>1</sub> = -\(\frac 11\) = -1</p> <p>Slope of eq<sup>n</sup> x -y = 0 is: m<sub>2</sub> = -\(\frac 1{-1}\) = 1</p> <p>m<sub>1</sub>× m<sub>2</sub> = -1× 1 = -1</p> <p>Hence, they are perpendicular to each other. <sub>Proved</sub></p>
Q8:
Prove that the angles between the lines represented by 6x2 - 5xy - 6y2 = 0 is at right angle.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given eq<sup>n</sup> is:</p> <p>6x<sup>2</sup> - 5xy - 6y<sup>2</sup> = 0...............................(1)</p> <p>The eq<sup>n</sup>of homogenous is:</p> <p>ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0..........................(2)</p> <p>Comparing eq<sup>n</sup> (1) and (2)</p> <p>a = 6</p> <p>b = -6</p> <p>Now,</p> <p>a + b = 0</p> <p>or, 6 - 6 = 0</p> <p>∴ 0 = 0</p> <p>Hence, the angle between two lines is 90°. <sub>Proved</sub></p>
Q9:
Find the angle between a pair of straight lines represented by the equation x2 - 2xy sec\(\alpha\) + y2 = 0.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given eq<sup>n</sup> is:</p> <p>x<sup>2</sup> - 2xy sec\(\alpha\) + y<sup>2</sup> = 0...........................(1)</p> <p>The homogenous eq<sup>n</sup> is:</p> <p>ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0..............................(2)</p> <p>Comparing eq<sup>n</sup> (1) and (2)</p> <p>a = 1</p> <p>h = - sec\(\alpha\)</p> <p>b = 1</p> <p>Now,</p> <p>tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or, tan\(\theta\) =± \(\frac {2\sqrt {(-sec\alpha)^2 - 1 × 1}}{1 + 1}\)</p> <p>or, tan\(\theta\) =± \(\frac {2\sqrt {sec^2\alpha - 1}}2\)</p> <p>or, tan\(\theta\) =± \(\sqrt {sec^2\alpha - 1}\)</p> <p>or, tan\(\theta\) =± \(\sqrt {tan^2\alpha}\)</p> <p>or, tan\(\theta\) =± tan\(\alpha\)</p> <p>∴ \(\theta\) = \(\alpha\) <sub>Ans</sub></p>
Q10:
Find a single equations representing the line pairs x cos\(\alpha\) +y sin\(\alpha\) = 0 and x sin\(\alpha\) + y cos\(\alpha\) = 0.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given eq<sup>n</sup> are:</p> <p>x cos\(\alpha\) + y sin\(\alpha\) = 0...................(1)</p> <p>x sin\(\alpha\) + y cos\(\alpha\) = 0...................(2)</p> <p>Combined eq<sup>n</sup> of (1) and (2)</p> <p>(x cos\(\alpha\) + y sin\(\alpha\)) (x sin\(\alpha\) + y cos\(\alpha\)) = 0</p> <p>or, x<sup>2</sup> cos\(\alpha\).sin\(\alpha\) +xy cos<sup>2</sup>\(\alpha\) + xy sin<sup>2</sup>\(\alpha\) + y<sup>2</sup> sin\(\alpha\).cos\(\alpha\) = 0</p> <p>or, x<sup>2</sup> cos\(\alpha\).sin\(\alpha\) + y<sup>2</sup> sin\(\alpha\).cos\(\alpha\) + xy (cos<sup>2</sup>\(\alpha\) + sin<sup>2</sup>\(\alpha\)) = 0</p> <p>or, sin\(\alpha\).cos\(\alpha\) (x<sup>2</sup> + y<sup>2</sup>) + xy× 1 = 0</p> <p>∴(x<sup>2</sup> + y<sup>2</sup>) sin\(\alpha\).cos\(\alpha\) + xy = 0 <sub>Ans</sub></p>
Q11:
Find the obtuse angle between a pair of lines represented by the equation:
x2 + 4xy + y2 = 0
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Given equation is:</p> <p>x<sup>2</sup> + 4xy + y<sup>2</sup> = 0.............................(1)</p> <p>The homogeneous equation of second degree is:</p> <p>ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0.....................(2)</p> <p>Comparing (1) and (2)</p> <p>a = 1</p> <p>h =2</p> <p>b = 1</p> <p>Now,</p> <p>tan\(\theta\) = ± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or, tan\(\theta\) = ± \(\frac {2\sqrt {2^2 - 1 × 1}}{1 + 1}\)</p> <p>or, tan\(\theta\) = ± \(\frac {2\sqrt {4 - 1}}2\)</p> <p>or, tan\(\theta\) = ± \(\sqrt 3\)</p> <p>∴ obtuse angle (\(\theta\) = tan<sup>-1</sup> (-\(\sqrt 3\))</p> <p>∴ \(\theta\) = (90 + 30)° = 120°<sub>Ans</sub></p>
Q12:
Find the angle between two straight lines represented by:
3x2 + 2y2 - 5xy = 0
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given equation is:</p> <p>3x<sup>2</sup> + 2y<sup>2</sup> - 5xy = 0</p> <p>The homogeneous equation of second degree is:</p> <p>ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0.....................(2)</p> <p>Comparing (1) and (2)</p> <p>a = 1</p> <p>h =2</p> <p>b = 1</p> <p>Now,</p> <p>tan\(\theta\) = \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or, tan\(\theta\) = \(\frac {2\sqrt {(\frac {-5}2)^2 - 3 × 2}}{3 + 2}\)</p> <p>or, tan\(\theta\) = \(\frac {2\sqrt {\frac {25}4 - \frac 61}}5\)</p> <p>or, tan\(\theta\) = \(\frac {2\sqrt {\frac {25 - 24}4}}5\)</p> <p>or, tan\(\theta\) = \(\frac {2 × \frac 12}{\frac 51}\)</p> <p>or, tan\(\theta\) = \(\frac 15\)</p> <p>or, \(\theta\) = tan<sup>-1</sup>(\(\frac 15\))</p> <p>∴ \(\theta\) = 11.31° <sub>Ans</sub></p>
Q13:
Find the obtuse angle between the lines represented by the equation 12x2 - 23xy + 5y2 = 0.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>The given equation is:</p> <p>12x<sup>2</sup> - 23xy + 5y<sup>2</sup> = 0.....................(1)</p> <p>The homogeneous equation of second degree is:</p> <p>ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0.....................(2)</p> <p>Comparing (1) and (2)</p> <p>a =12</p> <p>h = -\(\frac {23}2\)</p> <p>b = 5</p> <p>Now,</p> <p>tan\(\theta\) = ±\(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or,tan\(\theta\) = ± \(\frac {2\sqrt {(\frac {-23}2)^2 - 12 × 5}}{12 + 5}\)</p> <p>or,tan\(\theta\) = ± \(\frac {2\sqrt {\frac {529}4 - 60}}{17}\)</p> <p>or,tan\(\theta\) = ± \(\frac {2\sqrt {\frac {529 - 240}4}}{17}\)</p> <p>or,tan\(\theta\) = ± \(\frac {2\sqrt {289}}2\)× \(\frac 1{17}\)</p> <p>or,tan\(\theta\) = ± \(\frac {17}{17}\)</p> <p>∴tan\(\theta\) = ± 1</p> <p>Taking +ve sign,</p> <p>tan\(\theta\) = 1</p> <p>or, tan\(\theta\) = tan 45°</p> <p>∴ \(\theta\) = 45°</p> <p>Taking -ve sign,</p> <p>tan\(\theta\) = -1</p> <p>or, tan\(\theta\) = tan (180 - 35)</p> <p>or, tan\(\theta\) = tan 135°</p> <p>∴ \(\theta\) = 135°</p> <p>Hence, the obtuse angle between the pair of equation is: 135°. <sub>Ans</sub></p> <p></p>
Q14:
Find the acute angle between the pair of lines through origin represented by homogeneous equation of the second degree 2x2 - 5xy + 2y2 = 0.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>The homogenous equation of second degree is:</p> <p>ax<sup>2</sup> + 2hxy +by<sup>2</sup> = 0................................(1)</p> <p>The given equation is:</p> <p>2x<sup>2</sup> - 5xy + 2y<sup>2</sup> = 0...................................(2)</p> <p>Comparing (1) and (2)</p> <p>a = 2</p> <p>2h = -5 i.e. h = -\(\frac 52\)</p> <p>b = 2</p> <p>If \(\theta\) be the angle between pair of lines:</p> <p>tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or, tan\(\theta\) =± \(\frac {2\sqrt {(\frac {-5}2)^2 - 2 × 2}}{2 + 2}\)</p> <p>or, tan\(\theta\) =± \(\frac {2\sqrt {\frac {25}4 - \frac 41}}4\)</p> <p>or, tan\(\theta\) =± \(\frac {\sqrt {\frac {25 - 16}4}}2\)</p> <p>or, tan\(\theta\) =± \(\sqrt {\frac 94}\)× \(\frac 12\)</p> <p>or, tan\(\theta\) =± \(\frac 32\)× \(\frac 12\)</p> <p>or, tan\(\theta\) =± \(\frac 34\)</p> <p>∴ \(\theta\) = tan<sup>-1</sup>(± \(\frac 34\))</p> <p>Hence, the acute angle (\(\theta\)) =tan<sup>-1</sup>(± \(\frac 34\)) <sub>Ans</sub></p>
Q15:
Find the separate equation of the lines contained by the equation 6x2 + 5xy - 3x - 2y - 6y2 = 0 and prove that two lines are perpendicular each other.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given equation is:</p> <p>6x<sup>2</sup> + 5xy - 3x - 2y - 6y<sup>2</sup> = 0</p> <p>or, 6x<sup>2</sup> + 9xy - 4xy - 6y<sup>2</sup> - 3x - 2y = 0</p> <p>or, 3x (2x + 3y) - 2y (2x + 3y) - 1 (2x + 3y) = 0</p> <p>or, (2x + 3y) (3x - 2y - 1) = 0</p> <p>Either: 2x + 3y = 0..........................(1)</p> <p>Or: 3x - 2y - 1 = 0............................(2)</p> <p>Slope of equation (1), m<sub>1</sub> = -\(\frac 23\)</p> <p>Slope of equation (2), m<sub>2</sub> = \(\frac 32\)</p> <p>Again,</p> <p>m<sub>1</sub>× m<sub>2</sub> = \(\frac {-2}3\)× \(\frac 32\) = -1</p> <p>The product of two slopes = -1</p> <p>Hence, these equations are perpendicular each other. <sub>Hence, Proved</sub></p>
Q16:
To find the equation of a pair of straight lines through origin.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Let: the two equation of the lines through origin be:</p> <p>a<sub>1</sub>x + b<sub>1</sub>y = 0..............................(1)</p> <p>a<sub>2</sub>x + b<sub>2</sub>y = 0..............................(2)</p> <p>Combined equation of (1) and (2) is:</p> <p>(a<sub>1</sub>x + b<sub>1</sub>y) (a<sub>2</sub>x + b<sub>2</sub>y) = 0</p> <p>or, a<sub>1</sub>a<sub>2</sub>x<sup>2</sup> + a<sub>1</sub>b<sub>2</sub>xy + a<sub>2</sub>b<sub>1</sub>xy + b<sub>1</sub>b<sub>2</sub>y<sup>2</sup> = 0</p> <p>or,a<sub>1</sub>a<sub>2</sub>x<sup>2</sup> + (a<sub>1</sub>b<sub>2</sub>+ a<sub>2</sub>b<sub>1</sub>)<sub></sub>xy + b<sub>1</sub>b<sub>2</sub>y<sup>2</sup> = 0</p> <p>Let:</p> <p>a<sub>1</sub>a<sub>2</sub> = a</p> <p>b<sub>1</sub>b<sub>2</sub> = b</p> <p>(a<sub>1</sub>b<sub>2</sub>+ a<sub>2</sub>b<sub>1</sub>) = 2h</p> <p>Now,</p> <p>ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0</p> <p>Thus,ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0 is homogenous equation of second degree. <sub>Ans</sub></p>
Q17:
To prove that the homogenous equations of second degree ax2 + 2hxy + by2 = 0 always represents a pair of straight lines through origin.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>The given equation is: ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0 if a≠ 0, then:</p> <p>The given equation is multiplied by 'a' on both sides:</p> <p>a<sup>2</sup>x<sup>2</sup> + 2ahxy + aby<sup>2</sup> = 0</p> <p>or (ax)<sup>2</sup> + 2 (ax) (hy) + (hy)<sup>2</sup> - h<sup>2</sup>y<sup>2</sup> + aby<sup>2</sup> = 0</p> <p>or, (ax + hy)<sup>2</sup> - (h<sup>2</sup> - ab) y<sup>2</sup> = 0</p> <p>or, (ax + hy)<sup>2</sup> - {(\(\sqrt {(h^2 - ab)y})}<sup>2</sup> = 0</p> <p>or, (ax + hy + \(\sqrt {(h^2 - ab)y}\)) (ax + hy - \(\sqrt {(h^2 - ab)y}\)) = 0</p> <p>Either:(ax + hy + \(\sqrt {(h^2 - ab)y}\)) = 0.................(1)</p> <p>Or:(ax + hy - \(\sqrt {(h^2 - ab)y}\)) = 0..........................(2)</p> <p>Equation (1) and (2) are satisfied (0, 0) so both straight lines will passes through origin.</p> <p>Again,</p> <p>If a = 0</p> <p>The equation ax<sup>2</sup>+ 2hxy + by<sup>2</sup> = 0 will be</p> <p>2hxy + by<sup>2</sup> = 0</p> <p>or, y(2hx + by) = 0</p> <p>Either: y = 0.......................(3)</p> <p>Or: 2hx + by = 0..............(4)</p> <p>Equation (3) and (4) are satisfied by the co-ordinates (0, 0).</p> <p>Hence, the second degree homogeneous equation: ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0 always represents a pair of straight lines through the origin. <sub>Proved</sub></p>
Q18:
Find the angles between the pair of lines represented by the equation ax2 + 2hxy + by2 = 0.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given equationax<sup>2</sup> + 2hxy + by<sup>2</sup>= 0 represents two straight lines that passes through the origin.</p> <p>Let: y = m<sub>1</sub>x and y = m<sub>2</sub>x are the two lines.</p> <p>y -m<sub>1</sub>x = 0............................(1)</p> <p>y - m<sub>2</sub>x = 0............................(2)</p> <p>Product of the equation (1) and (2) is:</p> <p>(y - m<sub>1</sub>x) (y - m<sub>2</sub>x) = 0</p> <p>or, m<sub>1</sub>m<sub>2</sub>x<sup>2</sup> - m<sub>1</sub>xy - m<sub>2</sub>xy + y<sup>2</sup> = 0</p> <p>or, m<sub>1</sub>m<sub>2</sub>x<sup>2</sup> - (m<sub>1 </sub>+m<sub>2</sub>)xy + y<sup>2</sup> = 0...................(3)</p> <p>Given equationax<sup>2</sup> + 2hxy + by<sup>2</sup>= 0</p> <p>\(\frac ab\)x<sup>2</sup> + \(\frac {2h}b\) xy + y<sup>2</sup> = 0...................(4)</p> <p>Comparing equation (3) and (4), we get:</p> <p>m<sub>1</sub> + m<sub>2</sub> = -\(\frac {2h}b\) and m<sub>1</sub>m<sub>2</sub> = \(\frac ab\)</p> <p>Let \(\theta\) be the angle between the lines:</p> <p>tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, tan\(\theta\) =± \(\frac {\sqrt {(m_1 + m_2)^2 - 4m_1m_2}}{1 + m_1m_2}\) [\(\because\) a - b = \(\sqrt {(a + b)^2 - 4ab}\)]</p> <p>or, tan\(\theta\) =± \(\frac {\sqrt {\frac {4h^2}{b^2} - \frac {4a}b}}{1 + \frac ab}\)</p> <p>or, tan\(\theta\) =± \(\frac {\sqrt {\frac {4h^2 - 4ab}{b^2}}}{\frac {b + a}b}\)</p> <p>or, tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)× \(\frac bb\)</p> <p>or, tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>∴\(\theta\) = tan<sup>-1</sup>(± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))<sub> Ans</sub></p>
Q19:
Find the single equation passing through the point (1, 1) and parallel to the lines represented by the equation x2 - 5xy + 4y2 = 0.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>x<sup>2</sup> - 5xy + 4y<sup>2</sup> = 0</p> <p>or, x<sup>2</sup> - xy - 4xy + 4y<sup>2</sup> = 0</p> <p>or, x(x - y) - 4y(x - y) = 0</p> <p>or, (x - y) (x - 4y) = 0</p> <p>Either: x - y = 0.................(1)</p> <p>Or: x - 4y = 0......................(2)</p> <p>The eq<sup>n</sup> (1) changes into parallel form is:</p> <p>x - y + k<sub>1</sub> = 0......................(3)</p> <p>The point (1, 1) passes through eq<sup>n</sup> (1)</p> <p>1 - 1 + k<sub>1</sub> = 0</p> <p>∴ k<sub>1</sub> = 0</p> <p>Putting the value of k<sub>1</sub> in eq<sup>n</sup> (3)</p> <p>x - y + 0 = 0</p> <p>x - y = 0............................(4)</p> <p>The eq<sup>n</sup>(2) changes in parallel form is:</p> <p>x - 4y + k<sub>2</sub> = 0..................(5)</p> <p>The point (1, 1) passes through eq<sup>n</sup> (5)</p> <p>1 - 4× 1 + k<sub>2</sub> = 0</p> <p>or, -3 + k<sub>2</sub> = 0</p> <p>∴ k<sub>2</sub> = 3</p> <p>Putting the value of k<sub>2</sub> in eq<sup>n</sup> (5)</p> <p>x - 4y + 3 = 0.......................(6)</p> <p>The eq<sup>n</sup> of pair of line is:</p> <p>(x - y) (x - 4y + 3) = 0</p> <p>or, x<sup>2</sup> - 4xy + 3x - xy - 4y<sup>2</sup> - 3y = 0</p> <p>∴x<sup>2</sup> - 5xy - 4y<sup>2</sup> + 3x - 3y = 0<sub>Ans</sub></p>
Q20:
Find the single equation of pair of straight lines passing through the origin and perpendicular to the line pairs represented by x2 - xy - 2y2 = 0.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given equation is:</p> <p>x<sup>2</sup> - xy - 2y<sup>2</sup> = 0</p> <p>or, x<sup>2</sup> - 2xy + xy - 2y<sup>2</sup> = 0</p> <p>or, x (x - 2y) + y (x - 2y) = 0</p> <p>or, (x - 2y) (x + y) = 0</p> <p>Either: x - 2y = 0.........................(1)</p> <p>Or: x + y = 0.................................(2)</p> <p>The eq<sup>n</sup> (1) changes in perpendicular form is:</p> <p>-2x - y + k<sub>1</sub> = 0</p> <p>2x + y - k<sub>1</sub> = 0..........................(3)</p> <p>The eq<sup>n</sup> (3) passes through origin (0, 0):</p> <p>2× 0 + 0 - k<sub>1</sub> = 0</p> <p>∴ k<sub>1</sub> = 0</p> <p>Putting the value of k<sub>1</sub> in eq<sup>n</sup> (3)</p> <p>2x + y - 0 = 0</p> <p>2x + y = 0........................(4)</p> <p>The eq<sup>n</sup> (2) change in perpendicular form is:</p> <p>x - y + k<sub>2</sub> = 0........................(5)</p> <p>The eq<sup>n</sup> (5) passes through origin (0, 0)</p> <p>0 - 0 + k<sub>2</sub> = 0</p> <p>∴ k<sub>2</sub> = 0</p> <p>Putting the value of k<sub>2</sub> in eq<sup>n</sup> (5)</p> <p>x - y + 0 = 0</p> <p>x - y = 0...........................(6)</p> <p>The equation of the pairs of lines is:</p> <p>(2x + y) (x - y) = 0</p> <p>or, 2x<sup>2</sup> - 2xy + xy - y<sup>2</sup> = 0</p> <p>∴2x<sup>2</sup> - xy - y<sup>2</sup> = 0<sub>Ans</sub></p>
Q21:
Find the single equation of the pair of straight lines passing through the point (2, 3) and perpendicular to the line 3x2 - 8xy + 5y2 = 0.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Given eq<sup>n</sup> of line is:</p> <p>3x<sup>2</sup> - 8xy + 5y<sup>2</sup><sup></sup>= 0</p> <p>or, 3x<sup>2</sup> - 3xy - 5xy + 5y<sup>2</sup><sup></sup>= 0</p> <p>or, 3x (x - y) - 5y (x - y) = 0</p> <p>or, (x - y) (3x - 5y) = 0</p> <p>Either: x - y = 0.......................(1)</p> <p>Or: 3x - 5y = 0.........................(2)</p> <p>The eq<sup>n</sup> (1) changes in perpendicular form is:</p> <p>x + y + k<sub>1</sub> = 0...........................(3)</p> <p>The point (2, 3) passes through eq<sup>n</sup> (3)</p> <p>2 + 3 + k<sub>1</sub> = 0</p> <p>or, 5 + k<sub>1</sub> = 0</p> <p>∴ k<sub>1</sub> = -5</p> <p>Putting the value of k<sub>1</sub> in eq<sup>n</sup> (3)</p> <p>x + y - 5 = 0..........................(4)</p> <p>The eq<sup>n</sup> (2) change in perpendicular form is:</p> <p>5x + 3y + k<sub>2</sub> = 0...................(5)</p> <p>The point (2, 3) passes through eq<sup>n</sup> (5)</p> <p>5× 2 + 3× 3 + k<sub>2</sub> = 0</p> <p>or, 10 + 9 + k<sub>2</sub> = 0</p> <p>or, 19 + k<sub>2</sub> = 0</p> <p>∴ k<sub>2</sub> = - 19</p> <p>Putting the value of k<sub>2</sub> in eq<sup>n</sup> (5)</p> <p>5x + 3y - 19 = 0......................(6)</p> <p>The eq<sup>n</sup> of pairs of lines is:</p> <p>(x + y - 5) (5x + 3y - 19) = 0</p> <p>or, 5x<sup>2</sup> + 3xy - 19x + 5xy + 3y<sup>2</sup> - 19y - 25x - 15y + 95 = 0</p> <p>∴ 5x<sup>2</sup> + 8xy + 3y<sup>2</sup> - 44x - 34y + 95 = 0<sub>Ans</sub></p>
Q22:
Find the equation of two lines which passes through the point (3, -1) and perpendicular to the line pair x2 - xy - 2y2 = 0.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>x<sup>2</sup> - xy - 2y<sup>2</sup> = 0</p> <p>or, x<sup>2</sup> - 2xy + xy - 2y<sup>2</sup> = 0</p> <p>or, x(x - 2y) + y(x - 2y) = 0</p> <p>or, (x - 2y) (x + y) = 0</p> <p>Either: x - 2y = 0..................(1)</p> <p>Or: x + y = 0...........................(2)</p> <p>The eq<sup>n</sup> (1) change in perpendicular form is:</p> <p>2x + y + k<sub>1</sub> = 0.......................(3)</p> <p>The point (3, -1) passes through eq<sup>n</sup> (3)</p> <p>2× 3 - 1 + k<sub>1</sub> = 0</p> <p>or, 6 - 1 + k<sub>1</sub> = 0</p> <p>∴ k<sub>1</sub>= -5</p> <p>Putting the value of k<sub>1</sub> in eq<sup>n</sup> (3)</p> <p>2x + y - 5 = 0..........................(4)</p> <p>The eq<sup>n</sup> (2) change in perpendicular form is:</p> <p>x - y + k<sub>2</sub> = 0.........................(5)</p> <p>The point (3, -1) passes througheq<sup>n</sup> (5)</p> <p>3 + 1 + k<sub>2</sub> = 0</p> <p>∴ k<sub>2</sub> = -4</p> <p>Putting the value of k<sub>2</sub> in eq<sup>n</sup> (5)</p> <p>x - y - 4 = 0.........................(6)</p> <p>The eq<sup>n</sup> of pair of lines is:</p> <p>(2x + y - 5) (x - y - 4) = 0</p> <p>or, 2x<sup>2</sup> - 2xy - 8x + xy - y<sup>2</sup> - 4y - 5x + 5y + 20 = 0</p> <p>∴ 2x<sup>2</sup> - xy- y<sup>2</sup> - 13x + y + 20 = 0<sub>Ans</sub></p>
Q23:
Find the equation of the pair of lines represented by the equation 2x2 + 5xy + 3y2 = 0. Also find the angle between them.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>2x<sup>2</sup> + 5xy + 3y<sup>2</sup> = 0</p> <p>or, 2x<sup>2</sup> + 3xy + 2xy + 3y<sup>2</sup> = 0</p> <p>or, x(2x + 3y) + y(2x + 3y) = 0</p> <p>or, (2x + 3y) (x + y) = 0</p> <p>Either: 2x + 3y = 0</p> <p>Or: x + y = 0</p> <p>Given eq<sup>n</sup> is: 2x<sup>2</sup> + 5xy + 3y<sup>2</sup> = 0......................(1)</p> <p>Homogenous eq<sup>n</sup> is: ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0...................(2)</p> <p>Comparing eq<sup>n</sup> (1) and (2)</p> <p>a = 2</p> <p>h = \(\frac 52\)</p> <p>b = 3</p> <p>We know that:</p> <p>tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or,tan\(\theta\) =± \(\frac {2\sqrt {(\frac 52)^2 - 2 × 3}}{2 + 3}\)</p> <p>or, tan\(\theta\) =± \(\frac {2\sqrt {\frac {25}4 - 6}}5\)</p> <p>or,tan\(\theta\) =± \(\frac {2\sqrt {\frac {25 - 24}4}}5\)</p> <p>or,tan\(\theta\) =± \(\frac {2\sqrt {\frac 14}}5\)</p> <p>or,tan\(\theta\) =± \(\frac {2 × \frac 12}5\)</p> <p>or,tan\(\theta\) =± \(\frac 15\)</p> <p>Taking +ve sign,</p> <p>\(\theta\) = tan<sup>-1</sup> (\(\frac 15\)) = 11.31°</p> <p>Taking -vesign,</p> <p>\(\theta\) = (180° - 11.31°) = 168.69°</p> <p>∴ Required eq<sup>n</sup> are: 2x + 3y = 0 and x + y = 0 and angle between them are: 11.31° and 168.69°. <sub>Ans</sub></p>
Q24:
If \(\alpha\) be the angle made by the straight lines represented by the equation x2 + 2xy sec\(\theta\) + y2 = 0. Prove that: \(\alpha\) = \(\theta\). Find the pairs of equations of the lines.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Given eq<sup>n</sup> is:x<sup>2</sup> + 2xy sec\(\theta\) + y<sup>2</sup> = 0........................(1)</p> <p>Homogenous eq<sup>n</sup> is: ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0..........(2)</p> <p>Comparingeq<sup>n</sup>(1) and (2)</p> <p>a = 1</p> <p>h = sec\(\theta\)</p> <p>b = 1</p> <p>If \(\alpha\) be the angle between pairs of lines then,</p> <p>tan\(\alpha\) = \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or, tan\(\alpha\) = \(\frac {2\sqrt {sec^2\theta - 1 × 1}}{1 + 1}\)</p> <p>or, tan\(\alpha\) = \(\frac {2\sqrt {sec^2\theta - 1}}2\)</p> <p>or, tan\(\alpha\) = \(\sqrt {tan^2\theta}\)</p> <p>or, tan\(\alpha\) = tan\(\theta\)</p> <p>∴\(\alpha\) = \(\theta\) <sub>Proved</sub></p> <p>Again,</p> <p>The given eq<sup>n</sup> is:</p> <p>y<sup>2</sup> + 2xy sec\(\theta\) + x<sup>2</sup> = 0..........................(3)</p> <p>The quadratic eq<sup>n</sup> is:</p> <p>ax<sup>2</sup> + bx + c = 0.............................(4)</p> <p>Comparing (3) and (4)</p> <p>a = 1</p> <p>b = 2x sec\(\theta\)</p> <p>c = x<sup>2</sup></p> <p>We know,</p> <p>x = \(\frac {-b ± \sqrt {b^2 - 4ac}}{2a}\)</p> <p>or, y = \(\frac {-(2x sec\theta) ± \sqrt {(2x sec\theta)^2 - 4 × 1 × x^2}}{2 × 1}\)</p> <p>or, y = \(\frac {-2x sec\theta ± \sqrt {4x^2 sec^2\theta - 4x^2}}2\)</p> <p>or, y = \(\frac {-2x sec\theta ± \sqrt {4x^2 (sec^2\theta - 1)}}2\)</p> <p>or, y = \(\frac {-2x sec\theta ± 2x\sqrt {tan^2\theta}}2\)</p> <p>or, y = \(\frac {-2x sec\theta ± 2x tan\theta}2\)</p> <p>or, y = \(\frac {2(-x sec\theta ± xtan\theta)}2\)</p> <p>or, y = - xsec\(\theta\)± x tan\(\theta\)</p> <p>Taking +ve sign:</p> <p>y = -xsec\(\theta\) + xtan\(\theta\)</p> <p>Taking -ve sign:</p> <p>y = -xsec\(\theta\) - xtan\(\theta\)</p> <p>∴ The required eq<sup>n</sup> are:y = -xsec\(\theta\) + xtan\(\theta\) andy = -xsec\(\theta\) - xtan\(\theta\) <sub>Ans</sub></p>
Q25:
Prove that the angle between the lines represented by the equation x2 - 2xy cosec\(\theta\) + y2 = 0 is (\(\frac p2\) - \(\theta\)).
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Given equation is:x<sup>2</sup> - 2xy cosec\(\theta\) + y<sup>2</sup> = 0.....................(1)</p> <p>Homogenous equation is: ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0......................(2)</p> <p>Comparing eq<sup>n</sup> (1) and (2)</p> <p>a = 1</p> <p>h = -cosec\(\theta\)</p> <p>b = 1</p> <p>If \(\alpha\) be the angle between pair of lines then:</p> <p>tan\(\alpha\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or, tan\(\alpha\) =± \(\frac {2\sqrt {(-cosec\theta)^2 - 1 × 1}}{1 + 1}\)</p> <p>or, tan\(\alpha\) =± \(\frac {2\sqrt {(-cosec\theta)^2 - 1}}2\)</p> <p>or, tan\(\alpha\) =± \(\frac {2\sqrt {cosec^2\theta - 1}}2\)</p> <p>or, tan\(\alpha\) =± \(\sqrt {cot^2\theta}\)</p> <p>∴ tan\(\alpha\) =± cot\(\theta\)</p> <p>Taking +ve sign,</p> <p>tan\(\alpha\) = cot\(\theta\)</p> <p>tan\(\alpha\) = tan(\(\frac p2\) - \(\theta\))</p> <p>∴\(\alpha\) = (\(\frac p2\) - \(\theta\))<sub>Hence, Proved</sub></p>
Q26:
Find the equations of two lines represented by the equation 2x2 + 7xy + 3y2 = 0. Also find the angle between them.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given equation is:</p> <p>2x<sup>2</sup> + 7xy + 3y<sup>2</sup> = 0</p> <p>or, 2x<sup>2</sup> + 6xy + xy + 3y<sup>2</sup> = 0</p> <p>or, 2x(x + 3y) + y(x + 3y) = 0</p> <p>or, (x + 3y)(2x + y) = 0</p> <p>The two equation represented by2x<sup>2</sup> + 7xy + 3y<sup>2</sup> = 0 are:</p> <p>2x + y = 0......................(1)</p> <p>x + 3y = 0......................(2)</p> <p>Now,</p> <p>Slope of equation (1), m<sub>1</sub> = -\(\frac {x-coefficient}{y-coefficient}\) = -2</p> <p>Slope of equation (2),m<sub>2</sub> =-\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 13\)</p> <p>If the angle between the lines be \(\theta\) then:</p> <p>tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)</p> <p>or, tan\(\theta\) =± \(\frac {-2 + \frac 13}{1 + (-2) (-\frac 13)}\)</p> <p>or, tan\(\theta\) = ± \(\frac {\frac {-6 + 1}3}{\frac {3 + 2}3}\)</p> <p>or, tan\(\theta\) = ± \(\frac {-5}3\)× \(\frac 35\)</p> <p>or, tan\(\theta\) = ± 1</p> <p>Taking +ve sign,</p> <p>tan\(\theta\) = tan 45°</p> <p>∴ \(\theta\) = 45°</p> <p>Taking -ve sign,</p> <p>tan\(\theta\) = tan (180 - 45)° = tan 135°</p> <p>∴ \(\theta\) = 135°</p> <p>∴ Required angles (\(\theta\)) = 45° and 135°<sub>Ans</sub></p>
Q27:
Find the equation of a pair of lines represented by the equation 2x2 + 3xy - 2y2 = 0. Also find the angle between them.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>The given equation is:</p> <p>2x<sup>2</sup> + 3xy - 2y<sup>2</sup> = 0</p> <p>or, 2x<sup>2</sup> + 4xy - xy - 2y<sup>2</sup> = 0</p> <p>or, 2x(x + 2y) - y(x + 2y) = 0</p> <p>or, (x + 2y) (2x - y) = 0</p> <p>∴ Equation are: x + 2y = 0 and 2x - y = 0</p> <p>Again,</p> <p>2x<sup>2</sup> + 3xy - 2y<sup>2</sup> = 0............................(1)</p> <p>The homogenous equation of second degree is:</p> <p>ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0........................(2)</p> <p>Comparingequation (1) and (2)</p> <p>a = 2</p> <p>h = \(\frac 32\)</p> <p>b = -2</p> <p>tan\(\theta\) = \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or, tan\(\theta\) = \(\frac {2\sqrt {(\frac 32)^2 - 2 × (-2)}}{2 + (-2)}\)</p> <p>or, tan\(\theta\) = \(\frac {2\sqrt {\frac 94 + \frac 41}}{2 - 2}\)</p> <p>or, tan\(\theta\) = \(\frac {2\sqrt {\frac {9 + 16}4}}0\)</p> <p>or, tan\(\theta\) =∞</p> <p>∴ \(\theta\) = tan<sup>-1</sup>∞ = 90°</p> <p>∴ The required equations are: x + 2y = 0 and 2x - y = 0 and angle between the pairs of straight lines is 90°.<sub>Ans</sub></p> <p></p>
Q28:
Find the obtuse angle between the pair of lines represented by the equation 2x2 - 3xy + y2 = 0. Also find the pair of lines.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>Given equation is:</p> <p></p> <p>2x<sup>2</sup>- 3xy + y<sup>2</sup> = 0......................(1)</p> <p>The homogenous equation of second degree is:</p> <p>ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0..................(2)</p> <p>Comparing equation (1) and (2)</p> <p>a = 2</p> <p>h = -\(\frac 32\)</p> <p>b =1</p> <p>Now,</p> <p>tan\(\theta\) = ± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or, tan\(\theta\) = ± \(\frac {2\sqrt {(\frac {-3}2)^2 - 2 × 1}}{2 + 1}\)</p> <p>or, tan\(\theta\) = ± \(\frac {2\sqrt {\frac 94 - \frac 21}}3\)</p> <p>or, tan\(\theta\) = ± \(\frac {2\sqrt {\frac {9 - 8}4}}3\)</p> <p>or, tan\(\theta\) = ± \(\frac {2 × \frac 12}3\)</p> <p>or, tan\(\theta\) = ± \(\frac 13\)</p> <p>For the obtuse angle,</p> <p>tan\(\theta\) = -\(\frac 13\)</p> <p>\(\theta\) = tan<sup>-1</sup> (\(\frac 13\))</p> <p>Now,</p> <p>2x<sup>2</sup> - 3xy + y<sup>2</sup> = 0</p> <p>or, 2x<sup>2</sup> - 2xy - xy + y<sup>2</sup> = 0</p> <p>or, 2x(x - y) - y(x - y) = 0</p> <p>or, (x - y) (2x - y) = 0</p> <p>∴ The required pairs of lines are: x - y = 0 and 2x - y = 0. <sub>Ans</sub></p>
Q29:
Find separate equation of the lines contained by the equation 6x2 + 5xy - 3x + 2y - 6y2 = 0 and show that the lines are perpendicular to each other.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>6x<sup>2</sup> + 5xy - 3x + 2y - 6y<sup>2</sup> = 0</p> <p>or, 6x<sup>2</sup> + 5xy - 6y<sup>2</sup> - 3x + 2y = 0</p> <p>or, 6x<sup>2</sup> + 9xy - 4xy - 6y<sup>2</sup> - 3x + 2y = 0</p> <p>or, 3x(2x + 3y) - 2y(2x + 3y) - 1(3x - 2y) = 0</p> <p>or, (2x + 3y - 1) (3x - 2y) = 0</p> <p>Either: 2x + 3y - 1 = 0...........(1)</p> <p>Or: 3x - 2y = 0.........................(2)</p> <p>Slope of equation (1), m<sub>1</sub> = \(\frac {-3}{-2}\) = \(\frac 32\)</p> <p>Slope of equation (2),m<sub>2</sub> = -\(\frac 23\)</p> <p>∴ m<sub>1</sub>× m<sub>2</sub> = \(\frac 32\)× \(\frac {-2}3\) = -1</p> <p>Hence, the product of two slopes = -1 so these equations are perpendicular to each other. <sub>Proved</sub></p>
Q30:
Find the equation of the pair of lines represented by the equation x2 - 2xy cosec\(\theta\) + y2 = 0. Also find the angle between them.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>The given equation is:</p> <p>x<sup>2</sup> - 2xy cosec\(\theta\) + y<sup>2</sup> = 0</p> <p>or, (y)<sup>2</sup> - (2x cosec\(\theta\)) y + (x)<sup>2</sup> = 0..................(1)</p> <p>The quadratic equation is:</p> <p>ax<sup>2</sup> + bx + c = 0......................(2)</p> <p>Comparing (1) and (2)</p> <p>a = 1</p> <p>b = -2x cosec\(\theta\)</p> <p>c = x<sup>2</sup></p> <p>x = \(\frac {-b ± \sqrt {b^2 - 4ac}}{2a}\)</p> <p>y = -(-2x cosec\(\theta\))± \(\frac {\sqrt {(-2x cosec\theta)^2 - 4(1) (x^2)}}{2× 1}\)</p> <p>y = \(\frac {2x cosec\theta ± \sqrt {4x^2 cosec^2\theta - 4x^2}}2\)</p> <p>y = \(\frac {2x cosec\theta ± 2x \sqrt {cosec^2\theta - 1}}2\)</p> <p>y = \(\frac {2(x cosec\theta ± x cot\theta)}2\)</p> <p>y = x cosec\(\theta\)± x cot\(\theta\)</p> <p>The pair of lines are:</p> <p>y - x (cosec\(\theta\) - cot\(\theta\)) = 0 and</p> <p>y - x (cosec\(\theta\) + cot\(\theta\)) = 0</p> <p>Comparing x<sup>2</sup> -2xy cosec\(\theta\) + y<sup>2</sup> = 0 and ax<sup>2</sup> + 2hxy + by<sup>2</sup> = 0</p> <p>a = 1</p> <p>h = - cosec\(\theta\)</p> <p>b = 1</p> <p>We know that:</p> <p>tanA =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)</p> <p>or, tanA = ± \(\frac {2\sqrt {(-cosec\theta)^2 - 1 × 1}}{1 + 1}\)</p> <p>or, tanA =± \(\frac {2\sqrt {cosec^2\theta - 1}}2\)</p> <p>or, tanA =± cot\(\theta\)</p> <p>∴ A = tan<sup>-1</sup> (± cot\(\theta\))</p> <p>∴ The required equations are: y - x (cosec\(\theta\) - cot\(\theta\)) = 0 and y - x (cosec\(\theta\) + cot\(\theta\)) = 0 and angle (A) = tan<sup>-1</sup> (± cot\(\theta\)). <sub>Ans</sub></p>
Videos
Pair Of Straight Lines Example -1 l Maths Geometry
Pair Of Straight Lines Example - 2 l Maths Geometry
Pair Of Straight Lines Example - 6 l Maths Geometry
Alcohol
Alcohol is defined as an organic compound containing the hydroxyl group (-OH) attached to a saturated carbon atom or a hydrocarbon radical.
Methyl alcohol
Methyl alcohol is the 1st member of monohybrid alcohol. It is prepared in the industrial scale from methane obtained from natural gas.
-
Physical Properties
- It is a colorless liquid boiling at 65°C and melting at -98°C
- It is miscible with water.
- It has a sharp odor and burning taste.
-
Use
- It is used for dry cleaning
- It is used as an excellent solvent for fats, oils, paints, etc.
- It is used as antifreeze for automobile radiation.
- It is used to make perfumes, paints, and synthetic fabrics.
Ethyl alcohol
- Molecular formula: C2H5OH
- Structural formula:
-
Occurrence:
It is the earliest representative of this class. It is simply called alcohol. It is also called ethanol. It is an important alcohol. -
Physical properties:
- It is a colorless liquid but has a taste.
- It is soluble in water.
- Its boiling point is 78°C and freezing point is -117°C.
-
Uses:
- It is used as an alcoholic beverage.
- It is used as a solvent for drugs, oils, perfumes, etc.
- It is used as a preservative for biological specimens.
- It is used as a fluid in the thermometer.
- It is used in the manufacture of polyethene.
Lesson
Hydrocarbon And Their Derivatives
Subject
Science
Grade
Grade 10
Recent Notes
No recent notes.
Related Notes
No related notes.