Question

A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms a right angle triangle OPQ, where O is the origin. If the area of the triangle is least, then the slope of the line PQ is?

Answer 1

Hint: Assume the equation of line as $y=mx+c$, where $m$ is the slope and $c$ is the intercept on y-axis. Satisfy the point (1, 2) in the equation of the line and find the point where it will meet the coordinate axes. Find the area of the triangle in terms of $m$ and then differentiate the area with respect to $m$. Equate the derivative with zero and find the value of $m$. Substitute the values of $m$ in the equation to find the case in which area is minimum.

Complete step-by-step solution -
Let us assume that the equation of the line is $y=mx+c$. Then, substituting the point (1, 2) in the equation of line, we get,
$\begin{align}
  & 2=m\times 1+c \\
 & m+c=2 \\
 & \Rightarrow c=2-m.....................(i) \\
\end{align}$
Therefore, the equation of line can be written as $y=mx+2-m$. Let us assume that this line cuts the x-axis and y-axis at P and Q respectively.

Substituting: $y=0$,
$x=\dfrac{m-2}{m}$
Therefore, point P is $\left( \dfrac{m-2}{m},0 \right)$.
Substituting: $x=0$,
\[\] $y=2-m$
Therefore, point Q is $\left( 0,2-m \right)$.
Now, area of the triangle OPQ
\[\begin{align}
  & =\dfrac{1}{2}\times \text{base}\times \text{height} \\
 & \text{=}\dfrac{1}{2}\times OP\times OQ \\
 & =\dfrac{1}{2}\times \dfrac{(m-2)}{m}\times (2-m) \\
 & =\dfrac{-1}{2}\times \dfrac{{{(m-2)}^{2}}}{m} \\
\end{align}\]
Expanding the square term and then separating the terms, we get,
$\begin{align}
  & A=\dfrac{-1}{2}\times \left( \dfrac{{{m}^{2}}+4-4m}{m} \right) \\
 & =\dfrac{-1}{2}\times \left( m+\dfrac{4}{m}-4 \right) \\
\end{align}$
Differentiating the area with respect to $m$, we get,
$\begin{align}
  & \dfrac{dA}{dm}=\dfrac{-1}{2}\times \left( 1-\dfrac{4}{{{m}^{2}}}-0 \right) \\
 & \dfrac{dA}{dm}=\dfrac{-1}{2}\times \left( 1-\dfrac{4}{{{m}^{2}}} \right) \\
\end{align}$
To find the minimum value, we have to substitute this derivative with zero. Therefore,
$\begin{align}
  & \dfrac{-1}{2}\times \left( 1-\dfrac{4}{{{m}^{2}}} \right)=0 \\
 & 1-\dfrac{4}{{{m}^{2}}}=0 \\
 & 1=\dfrac{4}{{{m}^{2}}} \\
 & {{m}^{2}}=4 \\
 & \Rightarrow m=\pm 2 \\
\end{align}$
So, we got two values of $m$. One is maxima and the other is minima. To determine the minima, let us substitute the values of $m$ in the area.
Case (i): Substituting $m=2$, we get,
\[\begin{align}
  & A=\dfrac{-1}{2}\times \dfrac{{{(m-2)}^{2}}}{m} \\
 & =\dfrac{-1}{2}\times \dfrac{{{(2-2)}^{2}}}{2} \\
 & =0 \\
\end{align}\]
Case (ii): Substituting $m=-2$, we get,
$\begin{align}
  & A=\dfrac{-1}{2}\times \dfrac{{{(m-2)}^{2}}}{m} \\
 & =\dfrac{-1}{2}\times \dfrac{{{(-2-2)}^{2}}}{-2} \\
 & =4 \\
\end{align}$
Clearly, we can see that area is zero in the 1st case but we know that area of the triangle cannot be zero. Therefore, the slope of the line is -2.

Note: After finding the 1st derivative, one more alternate method can be applied to find the point of minima. Differentiate the area 2nd time and substitute the value of $m$, that was obtained in the 1st derivative, in the 2nd derivative. If the value of this 2nd derivative is negative then the substituted value of $m$ is the point of maxima and if the value of this 2nd derivative is positive then the substituted value of $m$ is the point of minima.