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Perpendicular to the line 2x + 3y = 12, passing through the point (7,3)

User Joshua Belden
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1 Answer

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8 votes

Answer:

2y-3x+15 = 0

Explanation:

let the equation of the perpendicular line: y=mx+c.

Given equation : 2x+3y=12

=> y=-2/3 x+4

so the sope of the equation is m1= -2/3.

as the two lines are mutually perpendicular,

so, m1*m2=-1

=> (-2/3)*m2= -1

=> m2= 3/2

the paerpendicular line : y=(3/2) x+c.

this line pass through (7,3)

so, 3 = (3/2)*7+c

=> c = -15/2

so the perpendicular line y = (3/2)*x -15/2

=> 2y-3x+15 = 0

User Brian McFarland
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2.9k points