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KEAM 2011 Math Paper

Question : 7 of 120

Marks: +1, -0

Solution:

The equation of the line passing through the points (-2,3) and (6,-5) is

(y-3)=\;\frac{-5-3}{6+2}(x+2)

\Rightarrow\;\;(y-3)=-1(x+2)

\Rightarrow\;\;y-3=-x-2

\Rightarrow\;\;y=-x+1

...(i)

Now, the slope of perpendicular bisector of this line is

=\;\frac{-1}{(-1)}=1

and the perpendicular bisector passing through the mid point of this line is (2,-1)

Then, equation of perpendicular bisector is

(y+1)=1(x-2)

\Rightarrow\;\;y+1=x-2

\Rightarrow\;\;x-y=3

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