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

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The correct Answer is:
4

Since PQ is of fixed length,
Area of `Delta PQR=(1)/(2)|PQ||RP|sintheta`
This will be maximum if `sintheta=1` and RP is maximum .
Since `y=mx+7` passes through `R'(0,7)` it must be rotated about (0,7) such that in new position, `theta` becomes `90^@` i.e., the line becomes perpendicular to `3x+4y+5=0`, then the slope of line `(4)/(3)`.
That is of `PR'=(4-7)/(4-0)=-(3)/(4)` = slope of given line. That is the in new position of the line. R coincides with R'.
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