Suppose that the points $\left( h,k \right),\left( 1,2 \right)\text{ }and\text{ }\left( -3,4 \right)$ lie on the line ${{L}_{1}}$ . If a line ${{L}_{2}}$ passing through the points $\left( h,k \right)\text{ }and\text{ }\left( 4,3 \right)$ is perpendicular to ${{L}_{1}}$ , then $\dfrac{k}{h}$ equals :
(a) $3$
(b) $-\dfrac{1}{7}$
(c) $\dfrac{1}{3}$
(d) $0$

Hint: So here in this question we need to Find the slope of ${{L}_{1}}$ and ${{L}_{2}}$ . Consider ${{m}_{1}}$ and ${{m}_{2}}$ as slope of ${{L}_{2}}$ . after finding the slopes we have to find the values of the coordinates $\left( h,k \right)$ and then we have to find the values of h and k for the value $\dfrac{k}{h}$ then we will get the final answer

Complete step by step answer:

So let us understand this by drawing the lines ${{L}_{1}}$ and ${{L}_{2}}$ .



As both ${{L}_{1}}$ and ${{L}_{2}}$ passes through $\left( h,k \right)$ therefore $\left( h,k \right)$ is intersecting point of ${{L}_{1}}$ and ${{L}_{2}}$ . Also, it is given in the question that ${{L}_{1}}$ and ${{L}_{2}}$ are perpendicular, so the product slopes of ${{L}_{1}}$ and ${{L}_{2}}$ is equal to $-1$ .
 $\therefore {{m}_{1}}.\text{ }{{m}_{2}}=-1$
Now , we can see that the line ${{L}_{1}}$ is passing through the points $\left( h,k \right),\left( 1,2 \right)\text{ }and\text{ }\left( -3,4 \right)$ . The line ${{L}_{2}}$ is passing through the points $\left( h,k \right)\text{ }and\text{ }\left( 4,3 \right)$ . We will use these points to find the slope of the line ${{L}_{1}}$ and line ${{L}_{2}}$ .
If a line ${{l}_{1}}$ is passing through points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ then
 \[Slope\text{ }of\text{ }{{l}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}.................\left( 1 \right)\]
Let us consider that ${{m}_{1}}$ is slope of ${{L}_{1}}$ and ${{m}_{2}}$ is slope of ${{m}_{2}}$ . Then we can write
${{m}_{1}}=\dfrac{4-2}{-3-1}=\dfrac{2}{-4}=-\dfrac{1}{2}....................\left( 2 \right)$
 As we have seen above that if two lines are perpendicular then , the product of slopes of both the lines will be $-1$ .
 $\therefore {{m}_{1}}{{m}_{2}}=-1$
From $\left( 2 \right)$ we have
 $\begin{align}
  & {{m}_{1}}=-\dfrac{1}{2} \\
 & \therefore -\dfrac{1}{2}.\text{ }{{m}_{2}}=-1 \\
 & \Rightarrow {{m}_{2}}=2 \\
\end{align}$
Now , we have ${{m}_{1}}=-\dfrac{1}{2}$ and ${{m}_{2}}=2$ .
As we have the value of ${{m}_{1}}\text{ }and\text{ }{{m}_{2}}$ so we will use this to find the equation of line ${{L}_{1}}\text{ }and\text{ }{{L}_{2}}$ .
Using $\left( 1 \right)$ we can write
 $\begin{align}
  & \dfrac{2-k}{1-h}=-\dfrac{1}{2} \\
 & \Rightarrow 2\left( 2-k \right)=-1\left( 1-h \right) \\
 & \Rightarrow 4-2k=-1+h \\
 & \Rightarrow h+2k=5............\left( 3 \right) \\
 & and\text{ }\dfrac{3-k}{4-h}=2 \\
 & \Rightarrow 3-k=2\left( 4-h \right) \\
 & \Rightarrow 3-k=8-2h \\
 & \Rightarrow 2h-k=8-3 \\
 & \Rightarrow 2h-k=5.............\left( 4 \right) \\
\end{align}$
Let us solve equation $\left( 3 \right)$ and $\left( 4 \right)$ .
From equation $\left( 4 \right)$ , we have
  $\begin{align}
  & 2h-k=5 \\
 & \Rightarrow k=2h-5..........\left( 5 \right) \\
\end{align}$
Let us put the value of $k$ in equation $\left( 3 \right)$ .
Then we have
 $\begin{align}
  & h+2(2h-5)=5 \\
 & \Rightarrow h+4h-10=5 \\
 & \Rightarrow 5h-10=5 \\
 & \Rightarrow 5h=5+10 \\
 & \Rightarrow 5h=15 \\
 & \Rightarrow h=\dfrac{15}{5} \\
 & \Rightarrow h=3 \\
\end{align}$
Then from $\left( 5 \right)$
 $k=2\times 3-5=6-5=1$
Thus $\dfrac{k}{h}=\dfrac{1}{3}$
Hence the correct option is (c).

Note:
The students must remember the formula to find the slope of a line. Also, the student must remember the condition of the slope when the lines are perpendicular. Here in this type of questions the student must plot the coordinates properly then we will get the idea of the question properly and also do not forget the formula of finding the slope if you don’t know the formula then it is very difficult to solve the question $\left( {{y}_{2}}-{{y}_{1}} \right)=m\left( {{x}_{2}}-{{x}_{1}} \right)$ remember this formula.