Question

Find the equation of straight line which passes through the point (2,-3) and the point of intersection of the lines x+y +4 = 0 and 3x-y-8=0

• A. x-2y-7 = 0
• B. 2x-y-7 = 0
• C. x-2y+7 = 0
• D. 2x-y+7 = 0

Answer 1

The correct option is B

2x-y-7 = 0

Any line through the intersection of the lines x + y + 4 = 0 and 3x - y - 8 = 0

Has the equation (x + y + 4) + $\lambda$ (3x - y - 8) = 0-------------(1)

This equation also passes through the point (2,-3)

So, this point should satisfy the equation.

(2+3+4) + $\lambda$ (6+3-8) = 0

3 + $\lambda$ = 0

$\lambda$ = -3

Putting the value of $\lambda$ in equation 1 required equation is

(x + y + 4) -3(3x-y-8) = 0

x + y + 4 - 9x + 3y + 24 = 0

-8x + 4y + 28 = 0

-2x + y + 7 = 0

2x - y - 7 = 0

Alternative method

We can find the intersection between x+y+4 = 0 and 3x - y - 8 = 0

Adding both the equation 4x - 4 = 0

x = 1

Substituting x in equation (2)

1 + y + 4 = 0

y = -5

so, equation line which passes through (1,-5) and (2,-3)

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

2x - y - 7 = 0