Question 1: Point P divides the line segment joining the points A (8, 0) \ and \ B (16, -8) in the ratio 3 : 5 . Find the co-ordinates of point P . Also, find the equation of the line through P and parallel to 3x + 5y = 7 .

Answer:

Given P divides A (8, 0) \ and \ B (16, -8) in the ratio 3 : 5

Ratio: m_1:m_2 = 3 : 5

Let the coordinates of the point P \ be \ (x, y) . Therefore

x = \frac{3 \times 16+5 \times 8}{3+5} = \frac{88}{8}  = 11 

y =  \frac{3 \times (-8)+5 \times 0}{3+5}   = \frac{-24}{8}   =  -3

Therefore P = (11, -3)

Equation of the line given : 3x+5y=7 \Rightarrow slope = m = \frac{-3}{5}

Therefore the required equation is

y - y_1 = m(x-x_1)

y -(-3) = \frac{-3}{5} (x-11) or 5y +3x=18

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Question 2: The line segment joining the points A (3, -4) \ and \ B (-2, 1) is divided in the ratio 1 : 3 at point P in it. Find the co-ordinates of P . Also, find the equation of the line through P and perpendicular to the line 5x - 3y = 4 .

Answer:

Given P divides A (3,-4) \ and \ B (-2,1) in the ratio 1:3

Ratio: m_1:m_2 = 1:3

Let the coordinates of the point P \ be \ (x, y) . Therefore

x = \frac{1 \times (-2)+3 \times 3}{1+3} = \frac{7}{4}   

y = \frac{1 \times 1+3 \times (-4)}{1+3}  = \frac{-11}{4} 

Therefore P = ( \frac{7}{4} , \frac{-11}{4} )

Equation of the line given : 5x - 3y = 4 \Rightarrow slope = m_1 = \frac{5}{3}

Therefore the slope of the line perpendicular to the above line = \frac{-3}{5}

Therefore the required equation is

y - y_1 = m(x-x_1)

y - \frac{-11}{4} = \frac{-3}{5} (x- \frac{7}{4} )   or  10y +6x+17 = 0

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Question 3: A line 5x + 3y + 15 = 0 meets y-axis at point P . Find the co-ordinates of point P . Find the equation of a line through P and perpendicular to x - 3y + 4 = 0 .

Answer:

At y-axis,  \ x = 0 \Rightarrow y = -5

Therefore the coordinate of P = (0, -5)

x - 3y + 4 = 0 \Rightarrow slope = \frac{1}{3}

Therefore the slope of a like perpendicular to this line = -3

Hence the line passing through (0, -5) with a slope of -3 is

y-(-5) = -3(x-0) \Rightarrow y+3x+5 = 0 

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Question 4: Find the value of k for which the lines kx - 5y + 4 = 0 and 5x - 2y + 5 = 0 are perpendicular to each other. [2003]

Answer:

Slope of  kx - 5y + 4 = 0   \Rightarrow m_1 = \frac{k}{5} 

Slope of 5x - 2y + 5 = 0   \Rightarrow m_2 = \frac{5}{2} 

Since the two lines are perpendicular, m_1. m_2 = -1 

\Rightarrow \frac{k}{5} . \frac{5}{2} = -1  

\Rightarrow k = -2 

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Question 5: A straight line passes through the points P (-1, 4) \ and \ Q (5, -2) . It intersects the co-ordinate axes at points A \ and \ B . M is the mid-point of the line segment AB . Find:

The equation of line

The co-ordinates of A \ and \ B

The co-ordinates of M [2003]

Answer:

P (-1, 4) \ and \ Q (5, -2)

The equation of the line:

y - 4 = \frac{-2-4}{5-(-1)} (x+1)

\Rightarrow y-4 = -1 (x+1)

\Rightarrow y+x=3

The x-intercept A = (3,0) and the y-intercept B = (0,3)

The coordinate of M = ( \frac{3+0}{2} , \frac{0+3}{2} ) = ( \frac{3}{2} , \frac{3}{2} )

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Question 6: (1, 5) \ and \ (-3, -1) are the co-ordinates of vertices A \ and \  C respectively of rhombus ABCD . Find the equations of the diagonals AC \ and \  BD .

Answer:

Midpoint M = ( \frac{-3+1}{2} , \frac{-1+5}{2} ) = (-1, 2)

Slope of AC = \frac{-1-5}{-3-1} = \frac{3}{2}

Equation of  AC :

y - 5 =  \frac{3}{2} (x-1) \Rightarrow 3x-2y+7=0

Slope of BD = - \frac{2}{3}

Therefore equation of BD :

y - 2 = - \frac{2}{3} (x+1) \Rightarrow 3y+2x=4

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Question 7: Show that A (3, 2), B (6, -2) \ and \ C (2, -5) can be vertices of a square. Find the co-ordinates of its fourth vertex D , if ABCD is a square. Without using the co-ordinates of vertex D , find the equation of side AD of the square and the equation of diagonal BD .

Answer:

Mid point of AC = ( \frac{3+2}{2} , \frac{-5+2}{2} ) = ( \frac{5}{2} , \frac{-3}{2} )

Let the coordinate of D be (x, y)

In a square, the diagonals bisect each other. Therefore

\frac{x+6}{2} = \frac{5}{2} \Rightarrow x = -1

\frac{y-2}{2} = \frac{-3}{2} \Rightarrow y = -1

Hence D is (-1, -1)

Slope  of AB = \frac{-2-2}{6-3} = \frac{-4}{3}

Since AB \perp AD , slope of AD = \frac{3}{4}

Hence the equation of AD :

y - 2 = \frac{3}{4} (x-3) \Rightarrow 4y-3x+1 = 0

Slope of BD = \frac{-2+1}{6+1} = - \frac{1}{7}

Hence the equation of BD :

y +2 = - \frac{1}{7} (x-6) \Rightarrow 7y+x+8=0

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Question 8: A line through origin meets the line x = 3y + 2 at right angles at point X . find the co-ordinates of point X .

Answer:

Given x = 3y + 2 … … … … (i)

Slope of line x = 3y + 2 is \frac{1}{3}

Slope of perpendicular = -3

The equation of a line passing through (0,0) and having slope 3 is

y - 0 = -3 (x-0) \Rightarrow y = -3x   … … … … (i)

Solving equations (i) and (ii)

x= 3 (-3x) + 2 \Rightarrow x = \frac{1}{5}

Hence y = - \frac{3}{5}

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Question 9: A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.

Answer:

Let y-intercept be (0, y) and x-intercept be (x, 0)

Given (3,2) is the mid point of (0,y) and (x,0) . Therefore:

y-intercept = (0, 4)

x-intercept = (6, 0)

Slope of line = \frac{0-4}{6-0} = - \frac{2}{3}

Equation of line:

y-0 = - \frac{2}{3} (x-6) \Rightarrow 3y+2x=12

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Question 10: Find the equation of the line passing through the point of intersection of 7x + 6y = 71 \ and \ 5x - 8y = -23 ; and perpendicular to the line 4x -2y = 1 .

Answer:

Solve equations

7x + 6y = 71     … … … … (i)

5x - 8y = -23    … … … … (ii)

Multiply (i) by 4 and (ii) by 3 and then add the equations, we get

43x = 215 \Rightarrow x = 5 

Substituting x = 5  in (i) we get

7(5)+6y=71 \Rightarrow y = 6 

Therefore the intercept is (5, 6) 

Sloe of line 4x -2y = 1 is 2 

Therefore the slope of perpendicular = - \frac{1}{2} 

Hence the equation of the perpendicular:

y - 6 = - \frac{1}{2} (x-5) 

2y -12 = -x+5 

2y+x=17 

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Question 11: Find the equation of the line which is perpendicular to the line \frac{x}{a} - \frac{y}{b} =1 at the point where this line meets y-axis .

Answer:

Slope of line \frac{x}{a} - \frac{y}{b} =1 is \frac{b}{a}

Therefore slope of line perpendicular to given line = - \frac{a}{b}

y-intercept = b

Therefore the equation of line passing through (0,b) and having slope of - \frac{a}{b}  is:

y - b =  - \frac{a}{b} (x-0)

by-b^2 = -ax

ax+by=b^2

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Question 12: O (0, 0), A (3, 5)  \ and \ B (-5, -3) are the vertices of a triangle OAB . Find:

(i) The equation of the median of triangle OAB through vertex O

(ii) The equation of altitude of triangle OAB through vertex B

Answer:

Mid point of AB =  ( \frac{-5+3}{2} , \frac{-3+5}{2} ) = (-1, 1)

Therefore the equation of median of  \triangle OAB through O is

(i) y - 0 = \frac{1-0}{-1-0} (x-0) \Rightarrow y = -x  \  or \ y+x = 0

(ii) Slope of OA = \frac{5-0}{3-0} = \frac{5}{3}

Slope of line perpendicular to OA  = - \frac{3}{5}

Therefore the equation of altitude of \triangle OAB through B 

y - (-3) = - \frac{3}{5} (x-(-5))

y+3 = - \frac{3}{5} (x+5)

5y+15 = -3x-15

5y+3x+30 = 0

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Question 13: Determine whether the line through points (-2, 3) \ and \ (4, 1) is perpendicular to the line 3x = y + 1 . Does line 3x = y +1 bisect the line segment joining the two given points?

Answer:

Slope of line passing through (-2,3)  and (4,1) = \frac{1-3}{4-(-2)} = - \frac{1}{3} 

Slope of 3x = y + 1 is 3 

Slope of perpendicular = - \frac{1}{3} 

Therefore line passing through (-2, 3)  and (4, 1)  is perpendicular to 3x = y + 1

Mid point of (-2, 3)  and (4, 1) = ( \frac{4-2}{2} , \frac{1+3}{2} ) = (1,2) 

Substituting (1, 2)  in 3x=y+1   we get that it satisfies the equation. Therefore 3x=y+1  bisects the line joining (-2, 3)  and (4, 1) 

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Question 14: Given a straight line x \cos 30^o+y \sin 30^o=2 . Determine the equation of the other line which is parallel to its and passes through (4, 3) .

Answer:

Given x \cos 30^o + y \sin 30^o = 2 

Slope of this line = - \frac{ \cos 30^o}{ \sin 30^o} = -\sqrt{3} 

Equation of line with slope  -\sqrt{3}  and passing through (4, 3)  is

y-3 =  -\sqrt{3} (x-4) 

y+\sqrt{3} x= 4 \sqrt{3}+3 

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Question 15: Find the value of k such that the line (k-2)x+(k+3)y-5=0 is:

(i) Perpendicular to the line 2x - y + 7 = 0     (ii) Parallel to it.

Answer:

Given (k-2)x+(k+3)y-5=0

Slope of this line = - \frac{k-2}{k+3}

Slope of line 2x-y+7 = 0 is 2

Slope of line perpendicular to this line = - \frac{1}{2}

(i) If perpendicular

- \frac{1}{2} = - \frac{k-2}{k+3}

k+3 = 2k-4

k= 7

(ii) If parallel

2 = - \frac{k-2}{k+3}

2k+6 = -k +2 

3k = -4 \Rightarrow k = - \frac{4}{3}

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Question 16: The vertices of a triangle ABC are A (0, 5), B (-1, -2) \ and \  C (11, 7) . Write down the equation of BC . Find:

(i) The equation of the line through A and perpendicular to BC .

(ii) The co-ordinates of the point P , where the perpendicular through A , as obtained in (i.), meets BC .

Answer:

(i) Slope of BC = \frac{7-(-2)}{11-(-1)} = \frac{9}{4} = \frac{3}{4}

Slope of line perpendicualr to BC = - \frac{4}{3}

Therefore equation of line passing through A (0,5) with slope - \frac{4}{3} is:

y-5 = - \frac{4}{3} (x-0)

3y-15 = -4x

3y+4x=15   … … … … (i)

(ii) Equation of BC

y-7 = \frac{3}{4} (x-11)

4y - 28 = 3x - 33

4y-3x= -5    … … … … (ii)

Solving (i) and (ii)  we get x = 3 and y = 1 .

Therefore P is (3, 1)

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l5Question 17: From the given figure, find:

(i) The co-ordinates of A, B, \ and \  C .

(ii) The equation of the line through A and parallel to BC [2004]

Answer:

Slope of BC = \frac{0-2}{3-(-1)} = \frac{-2}{4} = - \frac{1}{2}

The equation of line parallel to BC and passing through A(2,3)

y-3 = - \frac{1}{2} (x-2)

2y-6 = -x+2

2y+x=8

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Question 18: P (3, 4), Q (7, -2) \ and \ R (-2, -1) are the vertices of triangle PQR . Write down the equation of the median of the triangle through R [2005]

Answer:

Mid point of PQ = ( \frac{3+7}{2} , \frac{4-2}{2} ) = (5,1)

Therefore equation passing through (5,1) and P(-2,-1) is

y-1 = \frac{-1-1}{-2-5} (x-5)

y-1 = \frac{2}{7} (x-5)

7y-7 = 2x-10

7y-2x+3=0

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Question 19: A ((8, -6), B (-4, 2) \ and \ C (0, -10) are vertices of a triangle ABC . If P is the mid-point of AC , use co-ordinate geometry to show that PQ is parallel to BC . Give a special name to quadrilateral PBCQ .

Answer:

Coordinates of P = ( \frac{-4+8}{2} , \frac{2-6}{2} ) = (2, -2)

Coordinates of Q = ( \frac{8+0}{2} , \frac{-6-10}{2} ) = (4,-8)

Slope of PQ =  \frac{-8-(-2)}{4-2} = \frac{-6}{2} = -3

Slope of BC =  \frac{-10-2}{0-(-4)} = \frac{-12}{4} = -3

Therefore PB \parallel BC .

PQBC is a trapezoid.

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Question 20: A line AB meets the x-axis at point A and y-axis at point B . The point P (-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2 . Find:

(i) The co-ordinates of A \ and \ B .

(ii) Equation of the line through P and perpendicular to AB .

Answer:

AP:PB = 1:2

(i) Let A (0,y) and B(x,0)

Therefore -4 =  \frac{1 \times x+2 \times 0}{3}  \Rightarrow x = -6

Similarly,  -2 =  \frac{1 \times 0+2 \times y}{3}  \Rightarrow y= -3

Therefore B(-6, 0) and A (0, -3)

(ii)  Slope of  AB =  \frac{-3-0}{0-(-6)} = \frac{-3}{6} = \frac{-1}{2}

Slope of line perpendicular to AB  = 2

Therefore the equation of line passing through P(-4, -2) with slope 2 :

y - (-2) = 2(x-(-4))

y+2 = 2(x+4)

y = 2x+6

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Question 21: A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from the positive side of y-axis . Find the equation of the line. [1992]

Answer:

x-intercept  = (-2, 0)

y-intercept = (0, 3)

Equation of line

y-3 = \frac{3-0}{0-(-2)} (x-0)

y-3 = \frac{3}{2} x

2y -6 = 3x

2y = 3x+6

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Question 22: Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units. [2002]

Answer:

x-intercept = (4, 0)

Equation of line passing through (2,3) and (4, 0)

y - 0 = \frac{0-3}{4-2} (x-2)

y = - \frac{3}{2} (x-4)

2y = -3x+12

2y + 3x = 12

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Question 23: The given figure (not drawn to scale) shows two straight lines AB \ and \ CD . If equation of the line AB is y =x+1 and equation of CD is y = \sqrt{3}x-1 . Write down the inclination of lines AB \ and \ CD ; also, find the angle between AB \ and \ CD [1989]

Answer:l2

AB: y = x+1

CD: y = \sqrt{3}x - 1

Slope of AB = 1

tan \alpha_1 = 1 \Rightarrow \alpha_1 = 45^o

Slope of CD = \sqrt{3}

tan \alpha_2 = 1 \Rightarrow \alpha_2 = 60^o

Therefore 45^o+(180^o-60^o) + \theta = 180^o

\Rightarrow \theta = 15^o

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Question 24: Write down the equation of the line whose gradian is \frac{2}{3} and which passes through P , where P divides the line segment joining A (-2, 6) \ and \ B (3, -4) in the ratio 2 : 3 [2001]

Answer:

Given  P divides the line segment joining A (-2, 6) \ and \ B (3, -4) in the ratio 2 : 3

Let the coordinates of P = (x, y)

Therefore

x = \frac{2 \times 3 + 3 \times (-2)}{2+3} = \frac{6-6}{5} = 0

y = \frac{2 \times (-4) + 3 \times 6}{2+3} = \frac{-8+18}{5} = 2

\therefore P(0,2)

Equation of a line passing through P(0,2) with slope \frac{3}{2}

y-2 = \frac{3}{2} (x-0)

2y - 4 = 3x \Rightarrow 2y = 3x+4

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Question 25: The ordinate of a point lying on the line joining points (6, 4) \ and \ (7, -5) \ is \ -23 . Find the co-ordinates of that point.

Answer:

Let the ordinate of a point lying on the line joining points (6, 4) \ and \ (7, -5) \ is \ -23 be (x, -23)

Equation of line passing through (6,4) and (7, -5) 

y - 4 = \frac{-5-4}{7-6} (x-6)

y-4 = -9(x-6)

y+9x=58

Therefore if y = -23 , then

-23+9x=58 \Rightarrow x = 9

Therefore the point is (9, -23)

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Question 26: Point A \ and \ B have co-ordinates (7, 3) \ and \ (1, 9) respectively. Find:

(i) The slope of AB

(ii) The equation of perpendicular bisector of the line segment AB

(iii) The value of p \ of \ (-2, p) lies on it [2008]

Answer:

(i) Slope of AB = \frac{9-(-3)}{1-7} = \frac{12}{-6} = -2

(ii) Mid point of AB = ( \frac{7+1}{2} , \frac{-3+9}{2} ) = (4,3)

Therefore equation of line passing through (4,3) and slope \frac{1}{2} is

y - 3 = \frac{1}{2} (x-4)

2y -6 = x-4

2y= x+2

(iii) p \ of \ (-2, p) lies on it

Therefore 2(p) = (-2)+2 \Rightarrow p = 0

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Question 27: A \ and \ B are two points on the x-axis \ and \ y-axis respectively. P (2, -3) is the mid-point of AB . Find the

l3(i) Co-ordinates of A \ and \ B

(ii) Slope of line AB

(iii) Equation of line AB [2010]

Answer:

Let A(x,0) and B(0, y)

P(2, -3) is the mid point

(i) Therefore  2 = \frac{0+x}{2} \Rightarrow x = 4

-3 = \frac{y+0}{2} \Rightarrow y = -6

Hence A(4,0) and B(0, -6)

(ii) Slope of AB = \frac{-6-0}{0-4} = \frac{-6}{-4} = \frac{3}{2}

(iii) Equation of AB

y = 0 = \frac{3}{2} (x-4)

2y = 3x-12

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Question 28: The equation of a line is 3x + 4y - 7 = 0 . Find:

(i) Slope of the line.

(ii) The equation of a line perpendicular to the given line and passing through the intersection of the lines x -y + 2 = 0 and 3x + y- 10 = 0 [2010]

Answer:

(i) Slope = - \frac{3}{4}

(ii) Slope of perpendicular = \frac{4}{3}

For point of intersection solve x -y + 2 = 0 and 3x + y- 10 = 0

y = 4 and x = 2 

Therefore intersection = (2, 4)

Therefore equation of line

y-4 = \frac{4}{3} (x-2)

3y-12 = 4x-8

3y = 4x+4

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Question 29: ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4) . Find:  (i) Co-ordinates of A     (ii) Equation of diagonal BD [2011]

Answer:

(i) Mid point of BD = ( \frac{5+2}{2} , \frac{-4+8}{2} )= ( \frac{7}{2} , 2)

Therefore we have A(x, 0), O( \frac{7}{2} , 2) and C(4, 7)

O is the mid point of AC as well  (diagonals of a parallelogram bisect each other)

Hence \frac{x+4}{2} = \frac{7}{2} \Rightarrow x = 3

and \frac{y+7}{2} = 2 \Rightarrow y = -3

Hence A ( 3, -3)

(ii) Equation of BD

y - 8 = \frac{-4-8}{2-5} (x-5)

y-8 = 4(x-5)

y - 8 = 4x - 20

y + 12 = 4x

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Question 30: Given equation of line L_1  is y = 4 .

(i) Write the slope of line L_2  is L_2  is the bisector of angle O

l4(ii) Write the co-ordinates of point P .

(iii) Find the equation of L_2

Answer:

L_1 : y = 4

(i) \angle \alpha = 45^o

Therefore slope m = \tan 45^o = 1

(ii) Therefore P (4, 4)

(iii) Equation of line L_2

y - 0 = 1 (x-0) \Rightarrow y = x