ISC – Maths – XII – Important Q.A.

Straight Lines [2 – dimensions]

Q.1. Find the equation of the line passing through the intersection of the straight lines 2x + 3y = 1 and 3x + 4y = 6 and perpendicular to the line 5x – 2y = 7.

Solution : –

Let the equation of line through the intersection of

2x + 3y = 1  ————— (1) and

3x + 4y = 6 —————- (2)  be

(2x + 3y – 1) + k (3x + 4y – 6) = 0  ————-  (3)

Slope of (3)  =  {– (2 + 3k)}/(3 + 4k)

Slope of the line  5x – 2y = 7  ——————– (4)  is 5/2

As, (3) and (4) are perpendicular to each other , hence

{– (2 + 3k)}/(3 + 4k) × 5/2 = – 1

Or,       10 + 15 k = 6 + 8 k

Or,                  7k  =  – 4   =>  k = – 4/7

Putting k = – 4/7 in (3) we get

(2x + 3y – 1) – 4/7(3x + 4y – 6) = 0

Or,              7(2x + 3y – 1) – 4(3x +4y – 6) = 0

Or,                         2x + 5y + 17 = 0 .   [Ans.]

Q.2. A straight line is parallel to the x-axis and passes through the intersection of the line x + 2y + 1 = 0 and y = x + 7. Find the equation of the straight line.

Solution : –

Equation of line through the intersection of lines x + 2y + 1 =  0 and x – y + 7 = 0 is

x + 2y + 1 + k (x – y +7) = 0   —————— (1)

Or,                        x(1 + k) + y (2 – k) + 1 + 7k = 0

This line is parallel to x-axis , coefficient of x is zero.

Therefore,             1 + k = 0 => k = –1.

Putting value of k in (1) we get  x + 2y +1 – 1 (x – y + 7) = 0

Or,                                              x + 2y + 1 – x + y – 7 = 0

Or,                                                    3y – 6 = 0 => y = 2.   [Ans.]

Q.3. A straight line  y = mx passes through the intersection of the straight lines x + 2y – 1 = 0 and 2x – y + 3 = 0. Find the value of m.

Solution : –

The given lines are , x + 2y – 1 = 0  and  2x – y + 3 = 0 .

Solving these we get , x = –1 and y = 1.

Therefore, the point of intersection is (–1 , 1).

The line y = mx passes through (–1 , 1) , then m = –1.  [Ans.]

Q.4. Find the equation of the straight line through origin and passing through the intersection of lines x – 2y + 3 = 0 and 3x + 5y + 6 = 0.

Solution : –

Let the line through the intersection of lines x – 2y + 3 = 0 and 3x + 5y + 6 = 0 be :

(x – 2y + 3) + k (3x + 5y + 6) = 0      ——————– (1)

This passes through (0, 0)

Therefore,         3 + 6 k = 0 => k = – 1/2.

Putting the value of k in (1), we get

(x – 2y + 3) – 1/2 (3x + 5y + 6) = 0

Or,             2(x – 2y + 3) – (3x + 5y + 6) = 0

Or,                   2x – 4y + 6 – 3x – 5y – 6 = 0

Or,                                              – x – 9y = 0

Or,                                                 x + 9y = 0 . [Ans.]

Q.5. Find the equation of the straight line passing through the intersection of the lines 2x + 3y = 1 and 3x + 2y = 2 and making equal intercepts on the axes.

Solution : –

Let the equation of lines through the point of intersection of 2x + 3y = 1 and 3x + 2y = 2 be :

(2x + 3y – 1) + k (3x + 2y – 2) = 0 ——————- (1)

Or,                    2x + 3y – 1 + 3kx + 2ky – 2k = 0

Or,                               (2 + 3k) x + (3 + 2k) y = (1 + 2k)

Or,             {(2 + 3k)/(1 + 2k)} x + {(3 + 2k)/(1 + 2k)}= 1

Or,                     x/{(1 + 2k)/(2 + 3k)} + y/{(1 + 2k)/(3 + 2k)} = 1

As per question,

(1 + 2k)/(2 + 3k) = (1 + 2k)/(3 + 2k)

Or,                                2 + 3k = 3 + 2k => k = 1

Putting the value of k in (1), we get

(2x + 3y – 1) + 1.(3x + 2y – 2) = 0

Or,                                                   5x + 5y – 3 = 0  [Ans.]

Q.6. Find the equation of the straight line passing through the intersection of the straight lines x + y + 9 = 0 and y – x – 4 = 0 and perpendicular to the straight line 2x + 3y + 5 = 0.

Solution : –

Let equation of lines passing through the point of intersection of x + y + 9 = 0 and y – x – 4 = 0 be :

(x + y + 9) + k (– x + y – 4) = 0  —————– (1)

Or,                       x + y + 9 – k x + k y – 4k = 0

Or,                        (1 + k) y = (k – 1) x + (4k – 9)

Or,                                   y = {(k – 1)/(k + 1)} + (4k – 9)/(k + 1)

Thus slope of (1)                             m₁ =  (k – 1)/(k + 1)

2x + 3y + 5 = 0

Or,                                                y =  – (2/3) x – 5/3     ————- (2)

And slope of (2)                  m₂ =  – 2/3

Line (1) is perpendicular to line (2)

Therefore,                m₁×m₂ = – 1

Or,                             {(k – 1)/(k + 1)}(– 2/3) = – 1

Or,                              3k + 3 = 2k – 2    =>     k = – 5

Putting the value of k in (1), we get

(x + y + 9) – 5 (– x + y – 4) = 0

Or,                                                      6x – 4y + 29 = 0 [Ans.]

Q.7. The lines x – 2y + 6 = 0 and 2x – y – 10 = 0 intersect at the point A. Find the equation of the line making an angle 45º with the positive direction of the x-axis and passing through the point A.

Solution : –

The lines are : x – 2y + 6 = 0  and 2x – y – 10 = 0

Solving the two equations we get, A (26/3, 22/3).

Slope of the line is m = tan 45º = 1.

Equation of line is y – 22/3 = 1 (x – 26/3)

Or,               3y – 22 = 3x – 26

Or,               3x – 3y – 4 = 0. [Ans.]