perpendicular from the point P4,4 to the straight lines 3x+4y+5=0 and y=mx+7  meet at Q and R, respectively. If the area of triangle PQR is maximum, then the value of 3 is

Since PQ is of fixed length, Area of ΔPQR=12PQRPsinθThis will be maximum if sinθ=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, θ becomes 900 i,e., the line becomes perpendicular to 3x+4y+5=0. Then the slope of line is 43.Also, slope of PR'=4−74−0=−34=Slope of given lineThat is the in new position of the line, R coincides with R'.