<s>[INST] Given known and and unknown line equation and maximising area of triangle condition. 
Q) If from a point $P\equiv(4,4)$ 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=__ $ ?

First I drew the known line $-3x+4y+5=0$ and the unknown line $y=mx+7$ then I fixed the perpendicular distance from $P(4,4)$ as a fixed value $b$ .
I also know that area of a triangle $\dfrac{1}2 \times base \times height$ so to maximise this area we need $base=height$
I took $base$ as my assumed constant $b$ 
Then I found out $b=9/5$
Then I substituted in distance formula and I have no idea how to proceed further 
$\frac{9}{5}=\frac{-4m+4+c}{\sqrt{m^{2}+1}}$
 [/INST] First, I found the equation of lines which form the triangle. 


*

*$y-4=\frac{-1}{m}(x-4)$ which is the line perpendicular to $y=mx+7$ and passing through $P(4,4)$

*$y-4=\frac{4}{3}(x-4)$ represents the line perpendicular to $3x+4y+5=0$ and passing through $P(4,4)$
Next, I need information about vertices of the triangle. One of the vertex can be obtained by solving $y-4=\frac{-1}{m}(x-4)$ and $y=mx+7$ which is
$$h=\frac{4-3m}{1+m^2}$$
and $$k=\frac{4m-3m^2}{1+m^2}+7$$
Now, area of triangle is given by $\frac{1}{2}ab\sin\theta$. It easy to observe that one of the side is fixed. (The side formed by given point and fixed line). Let that side be $a$. Now, all that remains is to maximize $b\sin\theta$ which represents the altitude of the triangle. 
To get altitude of the triangle, I need to find the perpendicular distance from $(h,k)$ to side $a$. 
Side $a$ is given by the line $y-4=\frac{4}{3}(x-4)$ which can be written as $$3y-4x+4=0$$
the perpendicular distance formula, $$d(m)=\frac{3k-4h+4}{\sqrt{4^2+3^2}}$$
The graph of $d(m)$ looks like this:


So, when the lines are perpendicular, you get maximum area of triangle.
  That is, $m=\frac{4}{3}$

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