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

\\

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

\\

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

\\

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

\\

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)

\\

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

\\

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

\\

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

\\

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{}{2}, 2)

Therefore we have A(x, 0), O(\frac{}{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

\\

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

\\

Advertisements