Question

Find the slope of the lines :
$\left(i\right)$ Passing through the points $(3,-2)\text{and}(-1,4)$
$\left(ii\right)$ Passing through the points $(3,-2)\text{and}(7,-2)$
$\left(iii\right)$ Passing through the points $(3,-2)\text{and}(3,4)$
$\left(iv\right)$ Making inclination of ${60}^{\circ }$ with the positive direction of $x-$axis.

Answer 1

$\left(i\right)$ Slope of the straight line passing through the points $(x_1,y_1)\text{and}(x_2,y_2)$ is given by
$m=\frac{y_2-y_1}{x_2-x_1}$
$\therefore$ slope of the straight line passing through the points $(3,-2)\text{and}(-1,4)$ is
$m=\frac{4-(-2)}{-1-3}=\frac{4+2}{-1-3}=-\frac{3}{2}$

$\left(ii\right)$ Slope of the straight line passing through the points $(x_1,y_1)\text{and}(x_2,y_2)$ is given by
$m=\frac{y_2-y_1}{x_2-x_1}$
$\therefore$ slope of the straight line passing through the points $(3,-2)\text{and}(7,-2)$ is
$m=\frac{(-2)-(-2)}{7-3}=\frac{-2+2}{4}=\frac{0}{4}=0$

$\left(iii\right)$ Slope of the straight line passing through the points $(x_1,y_1)\text{and}(x_2,y_2)$ is given by
$m=\frac{y_2-y_1}{x_2-x_1}$
$\therefore$ slope of the straight line passing through the points $(3,-2)\text{and}(3,4)$ is
$m=\frac{4-(-2)}{3-3}=\frac{4+2}{0}=\frac{6}{0}$, which is not defined.

$\left(iv\right)$ Inclination of the line is ${60}^{\circ }$.
$\therefore \theta ={60}^{\circ }$
Slope of a line making an angle of $\theta$ with positive $x-$axis in and anti-clockwise direction $=\tan \theta$
$\therefore$ The slope of the line = $\tan \theta =\tan {60}^{\circ }=\sqrt{3}$