Coordinate geometry — practice questions

Step 1: Find the gradient of line $L_1$. Gradient of $L_1$, $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 5}{8 - 2} = \frac{-3}{6} = -\frac{1}{2}$. [M1 for gradient formula]

Step 2: Find the gradient of the perpendicular line $L_2$. Since $L_2$ is perpendicular to $L_1$, its gradient $m_2$ is the negative reciprocal of $m_1$. $m_2 = -\frac{1}{m_1} = -\frac{1}{-1/2} = 2$. [M1 for perpendicular gradient]

Step 3: Use the point-gradient form to find the equation of $L_2$. $L_2$ has gradient $m_2 = 2$ and passes through point $A(2, 5)$. Using $y - y_1 = m(x - x_1)$: $y - 5 = 2(x - 2)$ [M1 for using point and gradient] $y - 5 = 2x - 4$

Step 4: Rearrange the equation into the required form $ax+by+c=0$. $y = 2x + 1$ $2x - y + 1 = 0$ [A1 for correct final form]

Step 1: Find the midpoint of $PQ$. Midpoint $M = \left( \frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2} \right) = \left( \frac{-1 + 3}{2}, \frac{6 + 0}{2} \right) = \left( \frac{2}{2}, \frac{6}{2} \right) = (1, 3)$. [M1 for midpoint formula]

Step 2: Find the gradient of the line segment $PQ$. Gradient $m_{PQ} = \frac{0 - 6}{3 - (-1)} = \frac{-6}{4} = -\frac{3}{2}$. [M1 for gradient formula]

Step 3: Find the gradient of the perpendicular bisector. The perpendicular bisector has a gradient that is the negative reciprocal of $m_{PQ}$. Gradient of bisector, $m_{\perp} = -\frac{1}{-3/2} = \frac{2}{3}$. [M1 for perpendicular gradient]

Step 4: Find the equation of the perpendicular bisector. The line passes through the midpoint $M(1, 3)$ and has gradient $\frac{2}{3}$. Using $y - y_1 = m(x - x_1)$: $y - 3 = \frac{2}{3}(x - 1)$ [M1 for substitution] To clear the fraction, multiply by 3: $3(y - 3) = 2(x - 1)$ $3y - 9 = 2x - 2$ $2x - 3y + 7 = 0$. [A1 for correct equation in any form, then A1 for integer form if required]