Coordinate Geometry: Question 2

Worked solution

Part (a): Equation of the line $PQ$

Gradient. For $P(1, 2)$ and $Q(9, 8)$:

$m = \frac{8 - 2}{9 - 1} = \frac{6}{8} = \frac{3}{4}.$

Checking again: the change in $y$ is $8 - 2 = 6$ and the change in $x$ is $9 - 1 = 8$, so $m = 6 \div 8 = \dfrac{3}{4}$, the same result.

Equation. Using point $P(1, 2)$ in $y - y_1 = m(x - x_1)$:

$y - 2 = \frac{3}{4}(x - 1) \implies y = \frac{3}{4}x - \frac{3}{4} + 2 = \frac{3}{4}x + \frac{5}{4}.$

Check with $Q(9, 8)$:

$y = \frac{3}{4}(9) + \frac{5}{4} = \frac{27}{4} + \frac{5}{4} = \frac{32}{4} = 8. \checkmark$

So the equation of $PQ$ is $\boxed{y = \dfrac{3}{4}x + \dfrac{5}{4}}$.

Part (b): Perpendicular bisector of $PQ$

Midpoint of $PQ$:

$M = \left(\frac{1 + 9}{2}, \frac{2 + 8}{2}\right) = (5, 5).$

Gradient of the perpendicular bisector. Since $PQ$ has gradient $\dfrac{3}{4}$, a perpendicular line has gradient satisfying $m_1 m_2 = -1$:

$m_2 = -\frac{1}{3/4} = -\frac{4}{3}.$

Equation. Using the midpoint $M(5, 5)$:

$y - 5 = -\frac{4}{3}(x - 5) \implies y = -\frac{4}{3}x + \frac{20}{3} + 5 = -\frac{4}{3}x + \frac{35}{3}.$

Multiplying through by $3$ to clear the fraction:

$3y = -4x + 35 \implies 4x + 3y = 35.$

Check with the midpoint $(5, 5)$: $4(5) + 3(5) = 20 + 15 = 35. \checkmark$

So the perpendicular bisector is $\boxed{4x + 3y = 35}$.

Part (c): Where the perpendicular bisector meets the $x$-axis

On the $x$-axis, $y = 0$. Substituting into $4x + 3y = 35$:

$4x + 3(0) = 35 \implies 4x = 35 \implies x = \frac{35}{4}.$

Checking: $4\left(\dfrac{35}{4}\right) + 3(0) = 35 + 0 = 35. \checkmark$

So the perpendicular bisector crosses the $x$-axis at $\boxed{\left(\dfrac{35}{4}, 0\right)}$, i.e. $(8.75, 0)$.

Final answers

• (a) $PQ$: $y = \dfrac{3}{4}x + \dfrac{5}{4}$
• (b) Perpendicular bisector: $4x + 3y = 35$
• (c) Crosses the $x$-axis at $\left(\dfrac{35}{4}, 0\right)$, i.e. $(8.75, 0)$