Question

Find the slope of the line which passes through the points (3,-2) and (3, 4).

Answer 1

Hint: The slope of the line is also known as gradient of a line is a number that gives the direction and steepness of the line. If we have two coordinates \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] then slope of a line passing through these points is as follows:
\[slope=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\]

Complete step-by-step answer:
We have been given the points (3, -2) and (3, 4) through which the line passes. Now we know that the slope of a line passing through the points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is given by as follows:
\[slope=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\]
So we have \[{{x}_{1}}=3,{{y}_{1}}=-2,{{x}_{2}}=3,{{y}_{2}}=4\]
\[\Rightarrow slope=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)=\dfrac{4-(-2)}{3-3}=\dfrac{6}{0}=\infty \] (infinity)
Since we know that if the slope is infinity means the line is perpendicular to the x-axis i.e. it makes an angle of \[{{90}^{\circ }}\] with the x-axis.
Therefore, the required slope of the line is \[\infty \] (infinity).

Note: Remember that if the slope of a line is equal to zero then it is parallel to the x-axis and if the slope tends to infinity then it is perpendicular to x-axis. Also, you can remember that if the x-coordinates of the two points through which line passes are same then it must be perpendicular to the x-axis and if y-coordinates of the two points through which line passes are same then it must be parallel to the x-axis.