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If (-2,6) is the image of the point (4,2) with respect to line L=0, then L=0 a) 6x-4y-7=0 ​b) 2x+3y-5=0 c) 3x-2y+5=0 ​d) 3x-2y+10=0

If (-2,6) is the image of the point (4,2) with respect to line L=0, then L=0
a) 6x-4y-7=0
​b) 2x+3y-5=0
c) 3x-2y+5=0
​d) 3x-2y+10=0

Grade:8

3 Answers

CHANDRA KIRAN
35 Points
6 years ago
Given,
(-2,6) is the image of the point (4,2) w.r.t. L=0.
So, L=0 is the perpendicular bisector of (-2,6) , (4,2).
Therefore, L=0 is (-2-4)x+(6-2)y = [(-2)2+62]-{42+22]/2.
So, (-6x)+4y = 10
3x-2y+5 = 0.[C].
CHANDRA KIRAN
35 Points
6 years ago
If you feel that the above meathod is lengthy or complicated you can directly go from options by checking each option.
Anuj zanwar
13 Points
2 years ago
For above condition the line must pass trought midpoint of two given points and there is only one which passes through the midpoint of two given points.

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