Question

Let $A=\left( 3,-4 \right)$ and $B=\left( 1,2 \right)$ and $P=\left( 2k-1,2k+1 \right)$ is a variable point such that $PA+PB$ is the minimum. Then $k$ is

Answer 1

Hint: For these kinds of questions, we need to make use of differentiation. We need to make the first rule of differentiation. First of all, we use the distance formula from geometry and find the distance between $PA,PB$. After doing so, we add them. We will get an equation in terms of $k$. We will differentiate it with respect to $k$. Upon doing so, we equate it to $0$ by using the first rule of differentiation. After we get, we might get one value of $k$ or two values of $k$. If we get two, we should use the second rule of differentiation too.

Complete step-by-step answer:
First rule of differentiation let's equate the first derivative of a function to $0$.
Now let us calculate the distance of $PA,PB$ using the formula $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ .
$\begin{align}
  & \Rightarrow PA=\sqrt{{{\left( 2k-1-3 \right)}^{2}}+{{\left( 2k+1+4 \right)}^{2}}} \\
 & \Rightarrow PA=\sqrt{{{\left( 2k-4 \right)}^{2}}+{{\left( 2k+5 \right)}^{2}}} \\
\end{align}$
Let us square on both sides. Upon doing so, we get the following :
\[\begin{align}
  & \Rightarrow PA=\sqrt{{{\left( 2k-1-3 \right)}^{2}}+{{\left( 2k+1+4 \right)}^{2}}} \\
 & \Rightarrow {{\left( PA \right)}^{2}}={{\left( 2k-4 \right)}^{2}}+{{\left( 2k+5 \right)}^{2}} \\
 & \Rightarrow {{\left( PA \right)}^{2}}=4{{k}^{2}}+16-16k+4{{k}^{2}}+25+20k \\
 & \Rightarrow {{\left( PA \right)}^{2}}=8{{k}^{2}}+4k+41 \\
\end{align}\]
$\begin{align}
  & \Rightarrow PB=\sqrt{{{\left( 2k-1-1 \right)}^{2}}+{{\left( 2k+1-2 \right)}^{2}}} \\
 & \Rightarrow PB=\sqrt{{{\left( 2k-2 \right)}^{2}}+{{\left( 2k-1 \right)}^{2}}} \\
\end{align}$
Let us square on both sides. Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow PB=\sqrt{{{\left( 2k-1-1 \right)}^{2}}+{{\left( 2k+1-2 \right)}^{2}}} \\
 & \Rightarrow {{\left( PB \right)}^{2}}={{\left( 2k-2 \right)}^{2}}+{{\left( 2k-1 \right)}^{2}} \\
 & \Rightarrow {{\left( PB \right)}^{2}}=4{{k}^{2}}+4-8k+4{{k}^{2}}+1-4k \\
 & \Rightarrow {{\left( PB \right)}^{2}}=8{{k}^{2}}-12k+5 \\
\end{align}$
Let us ${{\left( PA \right)}^{2}}+{{\left( PB \right)}^{2}}$ .
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow {{\left( PA \right)}^{2}}+{{\left( PB \right)}^{2}}=8{{k}^{2}}+4k+41+8{{k}^{2}}-12k+5 \\
 & \Rightarrow {{\left( PA \right)}^{2}}+{{\left( PB \right)}^{2}}=16{{k}^{2}}-8k+45 \\
\end{align}$
We got an equation in $k$. Let us differentiate it with respect to $k$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow \frac{d\left( {{\left( PA \right)}^{2}}+{{\left( PB \right)}^{2}} \right)}{dk}=\frac{d\left( 16{{k}^{2}}-8k+45 \right)}{dk} \\
 & \Rightarrow \frac{1}{2}\frac{d\left( PA \right)}{dk}+\frac{1}{2}\frac{d\left( PB \right)}{dk}=32k-8 \\
 & \Rightarrow \frac{d\left( PA \right)}{dk}+\frac{d\left( PB \right)}{dk}=2\left( 32k-8 \right)=0 \\
 & \Rightarrow 32k-8=0 \\
 & \Rightarrow k=\frac{8}{32}=\frac{1}{4} \\
\end{align}$
Since , we only got one value of $k$ , we can say it for sure that at this particular value of $k$ , the sum of the distances $PA+PB$ is the minimum.
$\therefore $ Value of $k$ for which $PA+PB$ is the minimum is $\frac{1}{4}$.

Note: We should remember the formulae of derivatives of all the functions. We should also remember all the rules of differentiation since these alone can help us solve problems. We should be able to identify that a particular question is from calculus or from another chapter. A lot of practice is required so as to identify the topic asked in the question. We should be careful while solving since there is a lot of scope for calculation errors.