Question

Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0

Answer 1

Given lines are
                                   7x + 3y = 10      ...(1)
                                   5x − 4y = 1        ...(2)
(1) × 4 ⇒                  28x + 12y = 40     ...(3)
(2) × 3 ⇒                  15x − 12y = 3      ...(4)
By adding (3) and (4) ⇒ 43x     = 43

x = `43/43` = 1

Substitute the value of x = 1 in (1)

7(1) + 3y = 10

⇒ 3y = 10 – 7

y = `3/3` = 1

The point of intersection is (1, 1)

Equation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0

This line passes through (1, 1)

13(1) + 5(1) + k = 0

13 + 5 + k = 0

⇒ 18 + k = 0

k = – 18

∴ The equation of the line is 13x + 5y – 18 = 0