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\[\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\].