Question

# Slope of the straight line which is perpendicular to the straight line joining the points $\left( { - 2,6} \right)$ and $\left( {4,8} \right)$ is equal to:A. $\dfrac{1}{3}$B. $3$C. $- 3$D. $- \dfrac{1}{3}$

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Hint: In this question you first find out the slope of the given two points. Then use the condition of perpendicularity to find slope of the straight line which is perpendicular to the straight line joining by the given points. So, use this concept to reach the solution of the problem.

Let the given points be $\left( {{x_1},{y_1}} \right) = \left( { - 2,6} \right)$ and $\left( {{x_2},{y_2}} \right) = \left( {4,8} \right)$
We know that the slope of the straight lines joining by the two points $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$is given by slope $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
So, slope of the given points is ${m_1} = \dfrac{{8 - 6}}{{4 - \left( { - 2} \right)}} = \dfrac{2}{{4 + 2}} = \dfrac{2}{6} = \dfrac{1}{3}$
If two lines of slopes ${m_1}{\text{ and }}{m_2}$ are perpendicular, then the condition for perpendicularity is ${m_1}{m_2} = - 1$.
Let the slope of required line is ${m_2}$
By using the condition of perpendicularity we have
$\Rightarrow {m_1}{m_2} = - 1 \\ \Rightarrow \dfrac{1}{3}({m_2}) = - 1 \\ \Rightarrow {m_2} = - 1 \times 3 \\ \therefore {m_2} = - 3 \\$
Therefore, Slope of the straight line which is perpendicular to the straight line joining the points $\left( { - 2,6} \right)$ and $\left( {4,8} \right)$ is equal to $- 3$
Thus, the correct option is C. $- 3$

Note: In the given problem we need not to find the complete straight-line equation formed by joining the points $\left( { - 2,6} \right)$ and $\left( {4,8} \right)$. Since we have to find only the slope, it is enough to find the slope of the straight line joining by the given points.