The coordinates of $P \left(2, 6\right)$ and $Q \left(-3, 5\right)$ are given. Find
- the gradient of $PQ$;
- the equation of $PQ$;
- the coordinates of the point where $PQ$ intersects the $X$ axis.
Given: $\;\;$ $P \left(x_1, y_1\right) = \left(2, 6\right)$, $\;\;$ $Q \left(x_2, y_2\right) = \left(-3, 5\right)$
- Gradient of $PQ$ $= m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{5 - 6}{-3 - 2} = \dfrac{-1}{-5} = \dfrac{1}{5}$
- Equation of $PQ$ is
$y - y_1 = m \left(x - x_1\right)$
i.e. $\;$ $y - 6 = \dfrac{1}{5} \left(x - 2\right)$
i.e. $\;$ $5y - 30 = x - 2$
i.e. $\;$ $x - 5y + 28 = 0$ - When $PQ$ intersects the $X$ axis, the $Y$ coordinate of the point of intersection is $0$.
Let $PQ$ intersect the $X$ axis at the point $A\left(a, 0\right)$.
Substituting the coordinates of point $A$ in the equation of $PQ$ gives,
$a - 5\times 0 + 28 = 0$ $\implies$ $a = -28$
$\therefore \;$ $PQ$ intersects the $X$ axis at the point $\left(-28, 0\right)$.