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Find the equations of the line that passes trough point (3,4)and is perpendicular to the given line. 3x+2y=4

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The point is given (3,4) and the line is given 3x+2y=4 which is perpendicular to the equation of line determined.

Then slope of the equation line is


m=\frac{-1}{slope\text{ of perpendicular line}}

The slope of perpendicular line is


y=-(3)/(2)x+2

slope is -3/2.

Then the slope od equation of line is


m=-(1)/(-(3)/(2))=(2)/(3)

Then the equation of line obtained is


y=mx+c
4=(2)/(3)*3+c
4-2=c\text{.}
c=2

The equation of line formed is


y=(2)/(3)x+2

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