Question

passing through(-1, 2), one is parallel to x+3y-1=0 and the other is perpendicular to 2x-3y-1=0​

Answer 1

Given:

Passing through(-1, 2), one is parallel to x+3y-1=0 and the other is perpendicular to 2x-3y-1=0​

To find:

What is the joint equation of lines passing through (-1,2), one is parallel to x+3y-1=0 and the other is perpendicular to 2x-3y-1=0?

Solution:

From given, we have,

Equation of the line parallel to x+3y-1 = 0 is given by, x+3y+p =0.

If it passes through the point (-1,2) then -1+6+p = 0 gives p=-5.  

Therefore, the equation of the parallel line is x+3y-5=0 ....(i).

Equation of the line perpendicular to 2x-3y-1=0 is given by, 3x+2y+q=0.  

If it passes through the point (-1,2) then -3+4+q=0 gives q=-1.  

Therefore, the equation of the perpendicular line is 3x+2y-1=0  ..…(ii).

Hence the combined equation of (i) and (ii) is given by,

(x+3y-5)(3x+2y-1) = 0  

3x^2 + 11xy + 6y^2 – 16x - 15y + 5=0

Hence the required joint equation.