Question

Find the equation of lines passing through intersection of lines x+y+4 = 0 and 3x-y-8 = 0 and equally inclined to axis..

• A. x-y-6 = 0
• B. x+y+4 = 0
• C. x+y+6 = 0
• D. 3x-y-8 = 0

Answer 1

Answer: (A), (B)

x+y+4 = 0

Equation of straight line passing through intersection of (x+y+4) and 3x - y - 8 = 0 is

(x+y+4) + $\lambda$(3x-y-8) = 0

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

Let m be the slope of the line

Then m =$\frac{-(1+3\lambda )}{1-\lambda }=\frac{1+3\lambda }{\lambda -1}$

As lines are equally inclined with axis, therefore

m = $\tan {45}^{\circ }$ or m = $\tan {135}^{\circ }$

$m=\pm 1$

$\frac{3\lambda +1}{\lambda -1}=1or\frac{3\lambda +1}{\lambda -1}=-1$

3$\lambda$ + 1 = $\lambda$ - 1 3$\lambda$ + 1 = 1 - $\lambda$

2$\lambda$ = -2 4$\lambda$ = 0

$\lambda$ = -1 $\lambda$ = 0

Putting the value of $\lambda$ in equation 2

We get,

(x+y+4) - 1(3x-y-8) = 0

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

-2x + 2y + 12 = 0

x - y - 6 = 0

and

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

x + y + 4 = 0

x - y - 6 = 0 and x + y + 4 = 0 are the equations of the required lines.