Question

Consider the points $A(0,1)$ and $B(2,0)$ and $P$ be a point on the line $4x + 3y + 9 = 0$. The coordinates of $P$ such that $|PA - PB|$ is maximum are
A. $(\dfrac{{12}}{5},\dfrac{{ - 1}}{5})$
B. $(\dfrac{{ - 84}}{5},\dfrac{{13}}{5})$
C. $(\dfrac{{ - 24}}{5},\dfrac{{17}}{5})$
D. $(0, - 3)$

Answer 1

Hint: The given question shall be solved with the help of diagrams. Also, any expression containing mod (| |) actually expresses the distance. So here $|PA - PB|$ expresses distance.

Complete step by step solution:
The given two points are $A(0,1)$ and $B(2,0)$. Since $P$ is a point which lies on the line $4x + 3y + 9 = 0$, so$PA$ and $PB$ can be shown as given below:

(Source: Geogebra Application)

As $P$ is a point on the line $4x + 3y + 9 = 0$, we can show this line as given below:

From the figure, it is evident that $|PA - PB|$ is actually the distance between point $A$ and point $B$. Let the coordinates of $P$ be $(x,y)$. So when we move $P$ upwards the line, we stop at a point where $P(x,y)$, $A(0,1)$ and $B(2,0)$ becomes collinear. This means that $P$, $A$and $B$ lies on the same line as shown below:

Hence, this is where $P(x,y)$ must be located on the line $4x + 3y + 9 = 0$ so as to maximize $|PA - PB|$. Now, to find the coordinates of $P$, we will solve the equation for both the lines, the first one is on which $P$ lies, i.e. $4x + 3y + 9 = 0$ and the second one is on which $A$ and $B$ lies.
But for the second line, we will have to find the equation of that line on which $A(0,1)$ and $B(2,0)$ lies. So we will use the two-point method, such that
$ \Rightarrow ({x_1},{y_1}) = (0,1)$
$ \Rightarrow {x_1} = 0$ and $ \Rightarrow {y_1} = 1$
And, $ \Rightarrow ({x_2},{y_2}) = (2,0)$
$ \Rightarrow {x_2} = 2$ and $ \Rightarrow {y_2} = 0$
On substituting these values in the equation for the two-point method below, we will get
\[ \Rightarrow (y - {y_1}) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1})\]
\[ \Rightarrow (y - 1) = \dfrac{{0 - 1}}{{2 - 0}}(x - 0)\]
On simplifying, we will get
\[ \Rightarrow (y - 1) = \dfrac{{ - 1}}{2}x\]
On multiplying both sides by $2$, we will get
\[ \Rightarrow 2(y - 1) = - x\]
\[ \Rightarrow 2y - 2 = - x\]
On rearranging variables and constants on different sides, we will get
\[ \Rightarrow 2y + x = 2\]
\[ \Rightarrow x + 2y - 2 = 0\]
Hence, the equation of the line on which $A$ and $B$lies is \[x + 2y - 2 = 0\].
Now we will solve $4x + 3y + 9 = 0$ and \[x + 2y - 2 = 0\], so as to obtain the coordinates of point $P$. For this we will find the value of $x$ from \[x + 2y - 2 = 0\], such that
$ \Rightarrow x = 2 - 2y$
We will now substitute this value of $x$ into the equation $4x + 3y + 9 = 0$, such that
$ \Rightarrow 4(2 - 2y) + 3y + 9 = 0$
$ \Rightarrow 8 - 8y + 3y + 9 = 0$
On arranging the variables and constants on the left hand side and the right hand side respectively, we will get
$ \Rightarrow 3y - 8y = - 9 - 8$
$ \Rightarrow - 5y = - 17$
Dividing both sides by $ - 5$, we get
$ \Rightarrow y = \dfrac{{17}}{5}$
Now, on putting this value of $y$ in the equation \[x + 2y - 2 = 0\], we will get
\[ \Rightarrow 2 \times \dfrac{{17}}{5} + x - 2 = 0\]
\[ \Rightarrow \dfrac{{34}}{5} + x = 2\]
On subtracting $\dfrac{{34}}{5}$ from both the sides, we will get
\[ \Rightarrow x = 2 - \dfrac{{34}}{5}\]
\[ \Rightarrow x = \dfrac{{10 - 34}}{5}\]
Further solving, we will get
\[ \Rightarrow x = \dfrac{{ - 24}}{5}\]
Hence the coordinates of the point $P$ for which $|PA - PB|$ is maximum is $(\dfrac{{ - 24}}{5},\dfrac{{17}}{5})$.

Hence, the correct answer is option C.

Note:
It will seem appropriate at first to apply the distance formula here so as to calculate $|PA - PB|$ but doing so will not lead us to any answer. Therefore in such questions, we must avoid using the distance formula and try to solve the question with the help of graphs.