Question

The inclination of a straight line passing through the point $\left( -3,6 \right)$ and the midpoint of the line joining the point $\left( 4,-5 \right)$ and $\left( -2,9 \right)$ is
(a) $\dfrac{\pi }{4}$
(b) \[\dfrac{\pi }{6}\]
(c) \[\dfrac{\pi }{3}\]
(d) \[\dfrac{3\pi }{4}\]

Answer 1

Hint:The formula of midpoint which is given by $x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$ and use of the formula for inclination is given by $\tan \theta =\dfrac{{{y}_{4}}-y}{{{x}_{4}}-x}$ also the angle here is taken in anti clockwise direction. So the angle is positive here. Also the point $\left( {{x}_{4}},{{y}_{4}} \right)$ is the point through which the line passes.

Complete step-by-step answer:

The inclination of a straight line is nothing but an angle, denoted by $\theta ={{\tan }^{-1}}\left( \dfrac{{{y}_{4}}-y}{{{x}_{4}}-x} \right)$ and this is represented in the following diagram.

Clearly the angle of incarnation has been taken from an anti clockwise direction. This angle is calculated in an anti-clockwise direction that the line makes with x-axes. If in case the inclination is negative then this only means that the angle is taken in clockwise direction.
The formula for finding inclination is basically the slope. That is $\tan \theta =m$ and $m$ is the slope of the line. Clearly it is given that the line is passing through the point $\left( -3,6 \right)$ and so this is going to be our first point that we will substitute in the formula of slope.
The other point is given as a midpoint of the line joining the point $\left( 4,-5 \right)$ and $\left( -2,9 \right)$ so now we use the formula of midpoint which is given by $x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$
By using midpoint formula,
$\begin{align}
  & \Rightarrow x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2} \\
 & \Rightarrow x=\dfrac{4+\left( -2 \right)}{2} \\
 & \Rightarrow x=\dfrac{2}{2} \\
 & \Rightarrow x=1 \\
\end{align}$
Therefore the value of x is,
And the value of y is given by,
$\begin{align}
  & \Rightarrow y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \\
 & \Rightarrow y=\dfrac{-5+9}{2} \\
 & \Rightarrow y=\dfrac{4}{2} \\
 & \Rightarrow y=2
\end{align}$
So the value of mid point is $\left( x,y \right)=\left( 1,2 \right)$ and now we have both points through which the line passes, and these are $\left( -3,6 \right)\,,\left( 1,2 \right)$ so using the formula of inclination we have $\tan \theta =\dfrac{{{y}_{4}}-y}{{{x}_{4}}-x}$ this is also equal to m. Here $\left( {{x}_{4}},{{y}_{4}} \right)=\left( -3,6 \right)$ and $\left( x,y \right)=\left( 1,2 \right)$ therefore we get $\tan \theta =\dfrac{{{y}_{4}}-y}{{{x}_{4}}-x}$ and after substituting the value we lead to $\tan \theta =\dfrac{6-2}{-3-1}$
$\begin{align}
  & \Rightarrow \dfrac{6-2}{-3-1}=\dfrac{4}{-4} \\
 & \Rightarrow \tan \theta =-1
\end{align}$
 Since the value of $\tan \left( \dfrac{3\pi }{4} \right)=-1$ then this results into $\tan \theta =\tan \left( \dfrac{3\pi }{4} \right)$ thus, the value $\theta =\dfrac{3\pi }{4}$
Hence the correct option is (d).

Note: If at first the angle is not a right angled triangle that is the value of inclination is not exactly 90 degree then put $\theta ={{\tan }^{-1}}\left( \dfrac{{{y}_{4}}-y}{{{x}_{4}}-x} \right)$ where the point $\left( {{x}_{4}},{{y}_{4}} \right)$ is the point through which the line passes. This will result in the right answer. Here the answer is $\theta =\dfrac{3\pi }{4}$ but for further understanding the general value of $\theta $ is given by $\theta =\dfrac{3\pi }{4}+k\pi$, where $k=1,2,3,4,...$