Question

find theequayion of the straight line perpendicular to the line 2x+3y=0 and passing throgh the point intersection of the lines x+3y-1=0and x-3y-4=0​

Answer 1

Step-by-step explanation:

x+3y−1=0---(1)

x−2y+4=0---(2)

On solving eq (1) and (2) we get

5y=5⇒y=1

x+3−1=0⇒x=−2

Intersection point of line eq (1) and (2) is (−2,1)

Equation of line perpendicular to 3x+2y=0 is

2x−3y=λ----(3)

Above equation passing through (−2,1)

−4−3=λ⇒λ=−7

From eq (3)

2x−3y+7=0