The image of the point (−8,12) with respect to the line mirror 4x+7y+13=0 is
A
(−16,2)
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B
(16,−2)
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C
(−16,−2)
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D
(−2,−16)
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Solution
The correct option is C(−16,−2) Let the image of the point P(−8,12) in the line mirror AB be Q(α,β) and mid point of PQ be M.
Then, coordinates of M are (α−82,β+122).
Since, M lies on 4x+7y+13=0, ∴2α−16+7β+842+13=0⇒4α+7β+78=0⋯(1)
Since, PQ is perpendicular to AB,
therefore, Slope of AB× Slope of PQ=−1 ⇒(−47)×(β−12α+8)=−1 ⇒7α−4β+104=0⋯(2)
On solving equation (1) and (2), we get α=−16,β=−2
Hence, the image of (−8,12) in the line mirror 4x+7y+13=0 is (−16,−2)
Alternate Solution :
Using the image formula x+84=y−127=−2(4⋅(−8)+7⋅12+13)42+72⇒x=−16;y=−2
So image will be (−16,−2)