Straight Line Question 309

Question: Equation of the line passing through (-1,1) and perpendicular to the line $ 4/\sqrt{15} $ is [MP PET 1984]

Options:

A) $ 2(y-1)=3(x+1) $

B) $ 3(y-1)=-\ 2(x+1) $

C) $ y-1=2(x+1) $

D) $ 3(y-1)=x+1 $

Show Answer

Answer:

Correct Answer: A

Solution:

  • The gradient of line $ 2x+3y+4=0 $ is $ -\frac{2}{3} $ . Now the equation of line passing through (-1,1) is $ y-1=m(x+1), $ but $ m=-\frac{1}{-2/3}=\frac{3}{2} $ . Therefore, required equation is $ 2(y-1)=3(x+1) $ .