Coordinate Geometry: Equation of a Line

The coordinates of $P \left(2, 6\right)$ and $Q \left(-3, 5\right)$ are given. Find

1. the gradient of $PQ$;
2. the equation of $PQ$;
3. the coordinates of the point where $PQ$ intersects the $X$ axis.

Given: $\;\;$ $P \left(x_1, y_1\right) = \left(2, 6\right)$, $\;\;$ $Q \left(x_2, y_2\right) = \left(-3, 5\right)$

1. Gradient of $PQ$ $= m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{5 - 6}{-3 - 2} = \dfrac{-1}{-5} = \dfrac{1}{5}$



2. 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. $\;$ $x - 5y + 28 = 0$



3. 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 $\left(-28, 0\right)$.

Coordinate Geometry: Section Formula, Equation of a Line

Find the equation of a line whose slope is $\dfrac{3}{2}$ and passes through the point $P$ which divides the line segment joining $A \left(-2, 6\right)$ and $B \left(3, -4\right)$ in the ratio $2 : 3$.

Point $P \left(x,y\right)$ divides the line segment $A \left(-2, 6\right)$ and $B \left(3, -4\right)$ in the ratio $2 : 3$.

$\therefore \;$ By section formula

$x = \dfrac{2 \times 3 + 3 \times \left(-2\right)}{2 + 3} = 0$; $\;\;\;$ $y = \dfrac{2 \times \left(-4\right) + 3 \times 6}{2 + 3} = 2$

$\therefore \;$ $P \left(x, y\right) = P \left(0, 2\right)$

Equation of line passing through the point $P$ and slope $= \dfrac{3}{2}$ is

$y - 2 = \dfrac{3}{2} \left(x - 0\right)$

i.e. $\;$ $3x - 2y + 4 = 0$