Q.

If from P(4,4)  perpendiculars to the straight lines 3x+4y+5=0  and  y=mx+7  meet at Q and R and area of triangle PQR is maximum, then  m is equal to

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a

4/3

b

−3/4

c

−1

d

1

answer is A.

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Detailed Solution

since PQ is of fixed lengtharea of ΔPQR=12|PQ||RP|sinθThis will be maximum if sinθ=1&RP  is maximum.Since line y=mx+7  rotates about(0,7) , if PR'  is perpendicular to the line than PR’ is maximum value of PR.∴m=−(4−04−7)=43.
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