Subjective Questions
Q1:
A point which divides a line joining points (4 , 3) and (5 , 6) in the ratio 2 : 3.
Type: Long
Difficulty: Easy
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Answer: <p>Let , the given points be <br>(x<sub>1</sub> , y<sub>1</sub>) = (4 , 3) amd (x<sub>2</sub> , y<sub>2</sub>) = (5 ,6)<br>ratio m<sub>1</sub>:m<sub>2</sub> = 2 : 3 or m<sub>1</sub> = 2 and m<sub>2</sub> = 3<br>x = \(\frac{m1x2+m2x1}{m1+m2}\) and y = \(\frac{m1y2 + m2y1}{m1 + m2}\)<br><br>\(\therefore\) x = \(\frac{2.5+3.4}{2+3}\) and y = \(\frac{2.6 + 3.3}{2 + 3}\)<br><br>= \(\frac{10 + 12}{5}\) = \(\frac{22}{5}\) and y = \(\frac{12+9}{5}\) = \(\frac{21}{5}\)<br>\(\therefore\) (x , y) = \(\frac{22}{5}\) , \(\frac{21}{5}\) Ans.</p>
Q2:
Find the co-ordinates of :
A points which divides a line joining points (4,3) and (5,6) in the ratio 2:3.
Type: Long
Difficulty: Easy
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Answer: <p>Let the fiven points be</p> <p>(x<sub>1</sub> y<sub>1</sub>)=(4,3) and (x<sub>2</sub> y<sub>2</sub>)=(5,6)</p> <p>ratio m<sub>1</sub>:m<sub>2</sub>=2:3 or m<sub>1</sub>=2 and m<sub>2</sub>=3</p> <p>Using section formula</p> <p>x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\) </p> <p>x=\(\frac{2.5+3.4}{2+3}\) and y=\(\frac{2.6+3.3}{2+3}\) </p> <p>=\(\frac{10+12}{5}\) =\(\frac{22}{5}\) and y=\(\frac{12+9}{5}\) =\(\frac{21}{5}\) </p> <p>∴(x,y)=(\(\frac{22}{5}\),\(\frac{21}{5}\)).Ans. </p>
Q3:
Find the co-ordinates of :
A points which divides a line joining points (-3,4) and (-8,7) in the ratio 3:4.
Type: Long
Difficulty: Easy
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Answer: <p>Let the fiven points be</p> <p>(x<sub>1</sub> y<sub>1</sub>)=(-3,-4) and (x<sub>2</sub> y<sub>2</sub>)=(-8,7)</p> <p>ratio m<sub>1</sub>:m<sub>2</sub>=3:4 or m<sub>1</sub>=3 and m<sub>2</sub>=4</p> <p>Using section formula</p> <p>x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)</p> <p>x=\(\frac{3×-8+4×-3}{3+4}\) ,y=\(\frac{3×7+4×-4}{3+4}\)</p> <p>=\(\frac{-24-12}{7}\) , y=\(\frac{21-16}{7}\)</p> <p>∴(x,y)=(\(\frac{-36}{7}\),\(\frac{5}{7}\)).Ans.</p>
Q4:
Find the co-ordinates of :
A points which divides a line joining points (2,-3) and (-6,5) in the ratio 4:5.
Type: Long
Difficulty: Easy
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Answer: <p>Let the fiven points be</p> <p>(x<sub>1</sub> y<sub>1</sub>)=(2,-3) and (x<sub>2</sub> y<sub>2</sub>)=(-6,5)</p> <p>ratio m<sub>1</sub>:m<sub>2</sub>=4:5</p> <p>Using section formula</p> <p>x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)</p> <p>x=\(\frac{4 ×-6+5×2}{4+5}\) ,y=\(\frac{4 × 5 +5 ×-3}{4+5}\)</p> <p>=\(\frac{-24+10}{9}\) , y=\(\frac{20-15}{9}\)</p> <p>=\(\frac{-14}{9}\), \(\frac{5}{9}\)</p> <p>∴(x,y)=(\(\frac{-14}{9}\),\(\frac{5}{9}\)).Ans.</p>
Q5:
Find the co-ordinates of :
A points which divides a line joining points (-3,-4) and (-8,7) in the ratio 3:(-4).
Type: Long
Difficulty: Easy
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Answer: <p>Let the given points be</p> <p>(x<sub>1</sub> y<sub>1</sub>)=(-3,-4) and (x<sub>2</sub> y<sub>2</sub>)=(-8,7)</p> <p>ratio m<sub>1</sub>:m<sub>2</sub>==3:(-4) or m<sub>1</sub>=3 and m<sub>2</sub>=-4</p> <p>Using section formula</p> <p>x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)</p> <p>x=\(\frac{3×-8+(-4)×-3}{3+(-4)}\) ,y=\(\frac{3×7+(-4)×-4}{3+(-4)}\)</p> <p>=\(\frac{-24+12}{3-4}\) , y=\(\frac{21+16}{3-4}\)</p> <p>=\(\frac{-12}{-1}\), =\(\frac{37}{-1}\)</p> <p>= 12 , =-37</p> <p>∴(x,y)=(12,-37).Ans.</p>
Q6:
Find the co-ordinates of a point which divides the line joining the points (1,-2) and (4,7) in the following ratio:
- internally in the ratio1:2
- externally in the ratio 2:3
Type: Long
Difficulty: Easy
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Answer: <ul><li>Let the given points be</li> </ul><p>(x<sub>1</sub> y<sub>1</sub>)=(1,-2) and (x<sub>2</sub> y<sub>2</sub>)=(4,7)</p> <p>Internal ratio m<sub>1</sub>:m<sub>2</sub>=1:2</p> <p>Using formula,</p> <p>x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)</p> <p>or, x=\(\frac{1 × 4 + 2×1 }{1+2}\) ,y=\(\frac{1 ×7+2×-2}{1+2}\)</p> <p>=\(\frac{4+2}{3}\) , =\(\frac{7-4}{3}\)</p> <p>=\(\frac{6}{3}\), =\(\frac{3}{3}\)</p> <p>= 2 , =1</p> <p>∴(x,y)=The required point=(2,1).Ans.</p> <ul><li>Let the given points be <p>(x<sub>1</sub> y<sub>1</sub>)=(1,-2) and (x<sub>2</sub> y<sub>2</sub>)=(4,7)</p> <p>Let (x,y) divides externally in the ratio m<sub>1</sub>:m<sub>2</sub>=2:3 or m<sub>1</sub>=2 and m<sub>2</sub>=3</p> <p>Using formula,</p> <p>x=\(\frac{m_1x_2-m_2x_1}{m_1-m_2}\) and y=\(\frac{m_1y_2-m_2y_1}{m_1-m_2}\)</p> <p>or, x=\(\frac{2 × 4 - 3 ×1 }{2-3}\) ,y=\(\frac{2 ×7-3×-2}{2-3}\)</p> <p>=\(\frac{8-3}{-1}\) , =\(\frac{14+6}{-1}\)</p> <p>=\(\frac{5}{-1}\), =\(\frac{20}{-1}\)</p> <p>= -5 , =-20</p> <p>∴The required point=(-5,-20).Ans.</li> </ul>
Q7:
What are the cco-ordinates of the points which divide the line joining the points (4,-5) and (6,3) internally and externally in the ratio 2:5?
Type: Long
Difficulty: Easy
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Answer: <p>Here, Let the given points be (x<sub>1</sub>,y<sub>1</sub>)= (4,-5) and (x<sub>2</sub>,y<sub>2</sub>) = (6,3) and m<sub>1</sub>:m<sub>2</sub>=2:5</p> <p>Let (x,y) divides the given points internally then</p> <p>Using formula,</p> <p>x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\) and y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)</p> <p>x=\(\frac{2×6+5×4}{2+5}\) ,y=\(\frac{2×3+5×-5}{2+5}\)</p> <p>=\(\frac{12+20}{7}\) , y=\(\frac{6-25}{7}\)</p> <p>=\(\frac{32}{7}\), =\(\frac{-19}{7}\)</p> <p>∴The required point =(\(\frac{32}{7}\),\(\frac{-19}{7}\)).Ans.</p> <p>If (x<sup>1</sup>,y<sup>1</sup>) divides externally then using formula</p> <p>x'=\(\frac{m_1x_2-m_2x_1}{m_1-m_2}\) and y'=\(\frac{m_1y_2-m_2y_1}{m_1-m_2}\)</p> <p></p> <p>x'=\(\frac{2×6-5×4}{2-5}\) ,y'=\(\frac{2×3-5 ×-5)}{2-5}\)</p> <p>=\(\frac{12-20}{-3}\) , y=\(\frac{6+25}{-3}\)</p> <p>=\(\frac{-8}{-3}\), =\(\frac{31}{-3}\)</p> <p>∴The required point=\(\frac{8}{3}\), =\(\frac{-31}{3}\) Ans.</p> <p></p> <p></p> <p></p> <p></p> <p></p>
Q8:
Find the co-ordinates of the midpoint of the line joining the points A(-3,-6) and B(1,-2).
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>Let the given points be (x<sub>1</sub>,y<sub>1</sub>) =A(-3,-6) and (x<sub>2</sub>,y<sub>2</sub>) = B(1,-2) If (x,y) be the mid point, then using</p> <p> x=\(\frac{x_1+x_2}{2}\) and y=\(\frac{y_1+y_2}{2}\)</p> <p>∴ x =\(\frac{-3+1}{2}\) and y=\(\frac{-6-2}{2}\)</p> <p> =\(\frac{-2}{2}\), =\(\frac{-8}{2}\)</p> <p>∴x=-1, =-4</p> <p>∴(x,y)=(-1,-4).Ans.</p> <p></p>
Q9:
Find the distance of the following points.
a. (2 , 3) and (4 , 3)
Type: Short
Difficulty: Easy
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Answer: <p>Here , given P(x1 , y1) = (2 , 3) and Q(x2 , y2) = (4 , 3)<br>Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)<br>Distance , PQ = \(\sqrt{(4-2)^{2} + (3-3)^{2}}\) = \(\sqrt{2^{2} + 0^{2}}\) = \(\sqrt{2^{2}}\) = 2.<br>PQ = 2 units. Ans.</p>
Q10:
If the co-ordinates of the midpoint of a line joining the points M(1,4) and N(x,y) is (-2,2),what are the value of x and y?
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>Let the mid point of M(1,4) and N(x,y) be (-2,2)</p> <p> -2=\(\frac{1+x}{2}\) and 2=\(\frac{4+y}{2}\)</p> <p>or, -4=1+x, 4=4+y</p> <p>or, -4=-1=x, 4-4=y</p> <p>or, -5=x, 0=y</p> <p>∴(x,y)=(-5,0).Ans.</p>
Q11:
Find the distance between the given pairs of points.
(-1 , 3) and (5 , 1)
Type: Short
Difficulty: Easy
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Answer: <p>Here given , P(x1, ,y1) = (-1 , 3) and Q(x2 , y2) = (5 , 1)<br>Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)<br>Distance PQ = \(\sqrt{(5 -(-1)^{2} + (1 -3)^{2}}\) = \(\sqrt{6^{2} + 2^{2}}\) = \(\sqrt{36 + 4}\)<br>\(\therefore\) PQ = \(\sqrt{40}\) = \(\sqrt{4 \times 10}\) = 2\(\sqrt{10}\) Units. Ans.</p>
Q12:
If one end and the midpoint of a line are (4,4) and (-3,2) respectively, find the co-ordinates of the other end point of the line.
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>Let one end (x<sub>1</sub>y<sub>1</sub>)=(4,4) they let other end =(x<sub>2</sub>y<sub>2</sub>)</p> <p>Midpoint (x,y)=(-2,2)</p> <p>Using formula,</p> <p>x=\(\frac{x_1+x_2}{2}\) and y=\(\frac{y_1+y_2}{2}\)</p> <p></p> <p>-2=\(\frac{4+x_2}{2}\), 2=\(\frac{y_1+y_2}{2}\)</p> <p>or, -4=4+x<sub>2</sub> 4=4+y<sub>2</sub></p> <p>or, -4-4=x<sub>2</sub> 0=y<sub>2</sub></p> <p>or, -8=x<sub>2</sub>, 0=y<sub>2</sub></p> <p>∴(x<sub>2</sub>y<sub>2</sub>)=(-8,0).Ans.</p>
Q13:
Find the distance between the given pairs of points.
(1 , -2) and (-2 , 2)
Type: Short
Difficulty: Easy
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Answer: <p>Here given , P(x1 , y1) = (1 , -2) and Q(x2, , y2) = (-2 , 2)<br>Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)<br>Distance PQ = \(\sqrt{(-1 -3)^{2} + [-1 - (-1)]^{2}}\)<br> = \(\sqrt{(-4)^{2} + (-1 = 1)^{2}}\)<br>= \(\sqrt{4^{2} + 0^{2}}\) = \(\therefore\) PQ = \(\sqrt{4^{2}}\) = 4 units. Ans.</p>
Q14:
Find the distance between the given pairs of points.
(-6 , 7) and (-1 , -5)
Type: Short
Difficulty: Easy
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Answer: <p>Here given , P(x1, y1) = (-6 , 7) and Q(x2 , y2) = (-1 , -5)<br>Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)<br>Distance PQ = \(\sqrt{[-1 -(-6)]^{2} + (-5 -7)^{2}}\)<br>= \(\sqrt{-1 ^{2} + 6^{2} + (-12)^{2}}\) = \(\sqrt{5 ^{2} + 144}\)<br>= \(\sqrt{25 + 144}\) = \(\sqrt{169}\)<br>\(\therefore\) PQ = 13 Units Ans.</p>
Q15:
In what ratio is the line joining the points (2,-4) and (-3,6) is divided by
x-axis?
Type: Long
Difficulty: Easy
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Answer: <p>Here given points,A(2,-4) and B(-3,6)</p> <p>Let the points on x-axis be P(x,0) which divides AB in the ratio k:1 then,</p> <p>Using formula y=\(\frac{m_1y_2+m_2y_1}{m_1+m_2}\)</p> <p>0=\(\frac{k×6+1×-4}{k+1}\)</p> <p>or, 0×(k+1)=6k-4</p> <p>or, 6k-4=0</p> <p>or, 6k=4, or k=\(\frac{4}{6}\)=\(\frac{2}{3}\)</p> <p>∴This shows that x-axis divides the line AB internally in the ratio 2:3.Ans.</p>
Q16:
Find the distance bewteen the given pairs of points.
(4 , 3) and (-2 , 2)
Type: Short
Difficulty: Easy
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Answer: <p>Here given , P(x1 , y1) = (4 , 3) and Q(x2 , y2) = (3 , -6)<br>Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1) ^{2}}\)<br>Distance PQ = \(\sqrt{(-2 , -4)^{2} + (2 - 3 )^{2}}\) = \(\sqrt{(-6)^{2} + (-1)^{2}}\)<br>= \(\sqrt{36 + 1}\) = \(\sqrt{37}\) Units. Ans.</p>
Q17:
Find the distance from the given pairs of points.
(-2 , 6) and (3 , -6)
Type: Short
Difficulty: Easy
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Answer: <p>Here , given P(x1 , y1) = (-2 , 6) and Q(x2 , y2) = (3 , -6)<br>Using formula , PQ = \(\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}\)<br>Distance PQ = \(\sqrt{[ 3 - (-2)^{2} + (-6 -6)^{2}}\)<br>= \(\sqrt{(3 + 2)^{2} + (-12)^{2}}\) = \(\sqrt{5 ^{2} + 144}\)<br>= \(\sqrt{25 + 144}\) = \(\sqrt{169}\) = 13 Units. Ans.</p>
Q18:
Find the points whose x-coordinates is 3 and which is on the line joining the points P(7,-3) and Q(-2,-5).
Type: Long
Difficulty: Easy
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Answer: <p>Here, Let (3,y) be a point in the line PQ with co-ordinates P(7,-3) and Q(-2,-5) and as it divided PQ in the ratio k:1 then,</p> <p>Using formula, x=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\)</p> <p>3=\(\frac{k×-2+1×7}{k+1}\) or, 3k +3=-2k+7</p> <p>or, 3k +2k=7-3,or, 5k = 4 ∴k=\(\frac{4}{5}\)</p> <p>∴Ratio m<sub>1</sub>:m<sub>2</sub>=4:5</p> <p>Again using y =\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\)</p> <p>or, y=\(\frac{4 ×-5+5 ×-3}{4+5}\) =\(\frac{-20-15}{9}\)</p> <p>=\(\frac{-35}{9}\)</p> <p>∴ Requierd point (x,y)=(3,\(\frac{-35}{9}\)).Ans.</p> <p></p>
Q19:
If A(0 , 0) , B(3 , -4) , C(-3 , 4) , D(-2 , 2) and E (10 , - 3) are five points , find the distance between the following point.
1. A and B
Type: Short
Difficulty: Easy
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Answer: <p>Here, given A(0 , 0) and B(3 , -4)<br>\(\therefore\) AB = \(\sqrt{(3 - 0)^{2} + (4 - 0)^{2}}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 units.</p>
Q20:
If A(0 , 0) , B(3 , -4) , C(-3 , 4) , D(-2 , 2) and E (10 , - 3) are five points , find the distance between the following point.
1. B and C
Type: Short
Difficulty: Easy
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Answer: <p>Here given , A(0 , 0) amd C(-3 , 4)<br>\(\therefore\) AC = \(\sqrt{(-3 -0)^{2} + (4 - 0)^{2}}\)<br>= \(\sqrt{(-3)^{2} + (4)^{2}}\)<br>= \(\sqrt{9 + 16}\)<br>= \(\sqrt{25}\) = 5 units. Ans.</p>
Q21:
If A(0 , 0) , B(3 , -4) , C(-3 , 4) , D(-2 , 2) and E (10 , - 3) are five points , find the distance between the following point.
1. A and C
Type: Short
Difficulty: Easy
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Answer: <p>Here given , A(0 , 0) and C (-3 , 4)<br>\(\therefore\) AC = \(\sqrt{(-3 - 0)^{2}+(4 - 0)^{2} }\)<br>= \(\sqrt{(-3)^{2} + (4)^{2}}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 units. Ans.</p>
Q22:
In what ratio is the line joining the points (2,-4) and (-3,6) is divided by
y-axis
Type: Long
Difficulty: Easy
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Answer: <p>Let the point Q(0,y) on y-axis divides AB in the ratio k:1,then,</p> <p>Using formula y=\(\frac{m_1x_2+m_2x_1}{m_1+m_2}\)</p> <p>0=\(\frac{k×-3+1×2}{1+2}\)</p> <p>or, 0=3k+2</p> <p>or, 3k=2∴k=\(\frac{2}{3}\)</p> <p>∴y-axis divides internally in the ratio 2:3.Ans.</p>
Q23:
Prove that a triangle with the vertices (2 , 4) (6 , 4) and (6 , 7) is a right angled triangle.
Type: Long
Difficulty: Easy
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Answer: <p>Let , ABC be a triangle whose vertices are A(2 , 4) B(6 , 4) , C(6 , 7)<br>Using distance formula ,<br>AB = \(\sqrt{(6 - 2)^{2} + (4 - 4)^{2}}\) = \(\sqrt{4^{2}}\) = 4<br><br>BC = \(\sqrt{(6 - 6)^{2} + (7 - 4)^{2}}\) = \(\sqrt{3^{2}}\) = 3<br><br>CA= \(\sqrt{(2 -6)^{2} + (4 - 7)^{2}}\) = \(\sqrt{16 + 9}\) = \(\sqrt{25}\) = 5<br><br>Again here ,<br>AC \(^2\) = AB\(^2\) + BC\(^2\)<br>or , 5\(^2\) = 4\(^2\) + 3\(^2\)<br>\(\therefore\) 25 = 25<br>Since , h\(^2\) = p + b\(^2\) is satisfied so by pythagoras theorem \(\triangle\) ABC is a right angled triangle.</p>
Q24: Find the distance between the points A(-2,1) and B(4,3).
Type: Short
Difficulty: Easy
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Answer: Here,A(-2,1) and B(4,3) be any two points.
Using distance formula,we get
(AB)2 =(x2-x1)2+(y2-y1)2
=(-2-4)2 + (1-3)2
=(-6)2 +(-2)2
=36+4 =40
