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  • Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Q : 10         Find the equation of the line passing through  (-3,5)  and perpendicular to the line through the points  (2,5)  and (-3,6).

Answers (1)

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It is given that the line passing through  (-3,5)  and perpendicular to the line through the points  (2,5)  and (-3,6)
Let the slope of the line passing through the point (-3,5) is m and
Slope of line  passing through points (2,5) and (-3,6)
m' = \frac{6-5}{-3-2}= \frac{1}{-5}
Now this line is perpendicular to line passing through point (-3,5)
Therefore,
m= -\frac{1}{m'} = -\frac{1}{\frac{1}{-5}}= 5

Now, equation of line passing through point (x_1,y_1) and with slope m is
(y-y_1)= m(x-x_1)
equation of line passing through point (-3 , 5) and with slope  5 is
(y-5)= 5(x-(-3))\\ \\ (y-5)=5(x+3)\\ y-5=5x+15\\ 5x-y+20=0
Therefore, equation of line  is 5x-y+20=0 

Posted by

Gautam harsolia

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