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  • Find equation of the line parallel to the line 3x-4y+2=0 and passing through the point (-2, 3).

Q : 7      Find equation of the line parallel to the line  3x-4y+2=0  and passing through  the point (-2,3). 

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It is given that line is parallel to line  3x-4y+2=0 which implies that the slopes of both the lines are equal
we can rewrite it as
y = \frac{3x}{4}+\frac{1}{2}
The slope of line 3x-4y+2=0  =  \frac{3}{4}
Now, the equation of the line passing through the point (-2,3) and with slope \frac{3}{4} is
(y-3)=\frac{3}{4}(x-(-2))
4(y-3)=3(x+2)
4y-12=3x+6
3x-4y+18= 0
Therefore, the equation of the line is  3x-4y+18= 0

Posted by

Gautam harsolia

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