find the equation of the straight line perpendicular to the 2x-3y=0 and passing through the point of intersection of the line x+3y-1=0 and x-2y+4=0
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Answer:
Given Lines
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
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