Question

Find the equation of the lines through the point of intersection of the lines x-3y+1 = 0 and 2x+5y-9 = 0 and whose distance from the origin is $\sqrt{5}$.

• A. x-2y+5 = 0
• B. 2x+y+5 = 0
• C.
• D. 2x+y-5 = 0

Answer 1

The correct option is D

2x+y-5 = 0

Equation of family of straight line is

(x - 3y + 1) + $\lambda$(2x + 5y - 9) = 0

(1 + 2$\lambda$)x + (-3 + 5$\lambda$)y + (1 - 9$\lambda$) = 0-------------(1)

Distance of this equation from origin is $\sqrt{5}$

$\frac{1-9\lambda }{\sqrt{(1+2\lambda {)}^2+(-3+5\lambda {)}^2}}=\sqrt{5}$

Squaring on both sides

$\frac{(1-9\lambda {)}^2}{\sqrt{(1+2\lambda {)}^2+(-3+5\lambda {)}^2}}=5$

1 + 81${\lambda }^2$ - 18$\lambda$ = 5 [1 + 4${\lambda }^2$ + 4$\lambda$ + 9 + 25${\lambda }^2$ - 30$\lambda$]

1 + 81${\lambda }^2$ - 18$\lambda$ = 50 + 145${\lambda }^2$ - 130$\lambda$

64${\lambda }^2$ - 112$\lambda$ + 4y = 0

$(8\lambda {)}^2-2\times 8\lambda \times 7+7^2=0$

$(8\lambda -7{)}^2=0$

$\lambda =\frac{7}{8}$

Substituting the value of $\lambda$ in equation (1) we get,

$(1+2\times \frac{7}{8})x+(-3+5\times \frac{7}{8})y+(1-9\times \frac{7}{8})=0$

22x + 11y - 55 = 0

2x + y - 5 = 0

Required equation is 2x + y - 6 = 0