Question

Find the slope of the straight line which is perpendicular to the straight line joining the points $\left( { - 2,6} \right){\text{ and }}\left( {4,8} \right)$ ?
$\left( a \right){\text{ }}\dfrac{1}{3} \\\ \left( b \right){\text{ 3}} \\\ \left( c \right){\text{ - 3}} \\\ \left( d \right){\text{ - }}\dfrac{1}{3} \\\$

Answer 1

Hint- Use the relation between the slopes of two lines which are perpendicular to each other which is ${\text{slop}}{{\text{e}}_1} \times {\text{slop}}{{\text{e}}_2} = - 1$.

It’s given that we have to find the slope of a line which is perpendicular to a straight line joining the points$\left( { - 2,6} \right){\text{ and }}\left( {4,8} \right)$.
Now the slope of any line passing through $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right){\text{ is }}\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$…………………… (1)
Thus using the equation 1 we have slope of line passing through $\left( { - 2,6} \right){\text{ and }}\left( {4,8} \right)$is
${{\text{m}}_1} = \dfrac{{8 - 6}}{{4 - \left( { - 2} \right)}} = \dfrac{2}{6}{\text{ = }}\dfrac{1}{3}$
Now if two lines are perpendicular then their slope are related using the equation ${\text{slop}}{{\text{e}}_1} \times {\text{slop}}{{\text{e}}_2} = - 1$
Let us suppose the required slope is ${{\text{m}}_2}$so
${{\text{m}}_1} \times {m_2} = - 1$
Putting value of ${{\text{m}}_1}$we get
$\dfrac{1}{3} \times {m_2} = - 1$
${m_2} = - 3$
Hence option (c) is the right answer.

Note-Whenever two lines are perpendicular to each other than theirs slope are related as ${{\text{m}}_1} \times {m_2} = - 1$
We can easily find the slope of any line using 2 of its passing points via the concept of$\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.