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