Q.
In △ABC if A(0,0) , internal angle bisector through vertex B is x+2y−5=0 and perpendicular bisector of the side AC is 3x−y−10=0 , then the absolute value of slope of line BC is
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a
12
b
2
c
-32
d
4
answer is C.
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Detailed Solution
Point C will be image of A in 3x−y−10=0x−03=y−0−1=−2−1010C=x,y=6,−2 Image of A in x+2y−5=0 lie on BCx−01=y−02=−2−55 (∵Image formula h−x1a=k−y1b=−2ax1+by1+ca2+b2x,y=2,4∴ Slope of BC,mBC=6−4=−32
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