Two lines L1: 2y - 3x - 6 = 0 and L2: 3y + x - 20 = 0 interest at a point A.

Question

(a) find the coordinates of A.

(b) A third line L3 is perpendicular to L2 at point A. Find the equation of L3 in the form y = mx + c.

(c) Another line L4 is parallel to L1 and passes through (-1,3). Find the x and y intercepts of L4

Answer 1

I would use substitution.
from the 2nd equation: y = 20-x
sub that into the first:
3y + x - 20 = 0
3(20-x) + x = 20
60 - 3x + x = 20
-2x = -40
x = 20
from there find y to get point A

slope of 3y + x - 20 = 0 is -1
so the slope of L# has to be +1
and its equation is y = x + b
sub in the values of point A to find b

since L3 is parallel to 2y - 3x - 6 = 0 it must look like
2y - 3x - c = 0 , that is, it will only differ by the constant
plut in (-1,3) to find the value of c

I will assume you know how to find the x and y intercepts.

Answer 2

I misread the 2nd equation, (should put on glasses)

here is the correct version:
I would use substitution.
from the 2nd equation: x = 20-3y
sub that into the first:
2y - 3x - 6 = 0
2y - 3(20-3y) = 6
2y - 60 + 9y = 6
11y = 66
y = 6
from there find x to get point A