From a point $P=(3,4)$, perpendiculars PQ and PR are drawn to line $3x+4y-7=0$ and a variable line $y-1=m(x-7)$ respectively

Question

From a point $P=(3,4)$, perpendiculars PQ and PR are drawn to line $3x+4y-7=0$ and a variable line $y-1=m(x-7)$ respectively, find the maximum area of $\triangle PQR$.

On EduRev website, they have taken the point R to be $(7,1)$. Why? Can't the foot of the perpendicular be some other point?

On toppr website, they are taking general PR in terms of m. And then they say for maximum area, $PR=5$. For this, they are comparing the slope of PQ. That's more confusing.

Can anyone explain this? Thanks.

Answer 1

The reason why the perpendicular cannot be some other point is that the variable line, which can be rewritten as $$y = m \cdot (x-7) + 1$$, where m is the slope of the line has a fixed at a point $(7,1)$. As you can see from the coordinates of $x$ and $y$. From the elementary slope formula it follows...$y_{2} - y_{1} = a \cdot (x_{2} - x_{1})$ where $x_{1} = 7$ and $y_{1} =1$ here.

Considering this, If we take a look at the graphs of these functions... . We can calculate and see that the second answer cannot be true with both perpendiculars $PQ$ and $PR$ drawn to lines respectively. Finally $m$ must be $0.75$. The slope equation evaluates then to $0.75x - \frac {25}{3}$

Answer 2

This is how I understand the problem setup:

Point $P(3,4)$ is given. point $Q$ is the foot of the perpendicular from $P$ to $3x+4y−7=0$. That means that the slope of $PQ$ is $4/3$. Since $PQ$ goes through $P$, the equation of that line is $$y=\frac43 x$$ Then point $Q$ is at the intersection of the two lines, so $$\begin{cases}y_Q=\frac43x_Q\\3x_Q+4y_Q-7=0\end{cases}$$ From here $$x_Q=\frac{21}{25}\\y_Q=\frac{28}{25}$$ The second line goes through point $A(7,1)$, and has the equation $$y=mx-7m+1$$ The slope of this line is $m$, so the slope of perpendicular to the line is $-1/m$. $R$ is the foot of the perpendicular from $P$. Then the equation of $PR$ line is $$y-y_P=-\frac 1m(x-x_P)\\y=-\frac1m x+\frac3m+4$$ Then $R$ is at the intersection, so $$\begin{cases}y_R=-\frac1m x_R+\frac3m+4\\y_R=mx_R-7m+1\end{cases}$$ I will let you calculate $x_R$ and $y_R$, both are some functions of $m$. Then, the area of $\triangle PQR$ is given by the half of length of $PQ$ (which is a constant) multiplied by the distance ($h$) from $R$ to the $PQ$ line. This is given by $$h=\frac{|-4x_R+3y_R|}{\sqrt{4^2+3^2}}$$ Find the maximum of $h$, then find the desired area.