Quick questions on Coordinate geometry of straight lines explained: O-Level A-Maths

A perpendicular bisector of a segment passes through the midpoint and is perpendicular to the segment. Find the midpoint, take the negative reciprocal of the segment's gradient, and write the line through the midpoint.

A common A-Maths extension is to find the foot of the perpendicular from a point to a line, the closest point on the line. The method chains the tools: take the negative reciprocal of the line's gradient to get the perpendicular's gradient, write the perpendicular through the given point, then solve it simultaneously with the original line to find their intersection. That intersection is the foot, and the distance from the point to it is the shortest distance to the line. Setting up the perpendicular and intersecting it with the line is the standard route, and it reuses the gradient, equation, and simultaneous-solving skills together.

Beyond the midpoint, A-Maths asks for a point dividing a segment in a given ratio. A point $P$ dividing $AB$ in the ratio $m : n$ has coordinates $\left(\tfrac{n x_1 + m x_2}{m + n}, \tfrac{n y_1 + m y_2}{m + n}\right)$, which reduces to the midpoint formula when $m = n$. So the point dividing $A(1, 2)$ and $B(7, 8)$ in the ratio $2 : 1$ is $\left(\tfrac{1(1) + 2(7)}{3}, \tfrac{1(2) + 2(8)}{3}\right) = (5, 6)$. Recognising the midpoint as the special case $1 : 1$ of this general section formula ties the two ideas together and handles the ratio problems that go beyond simple bisection.

A perpendicular gradient is the negative reciprocal; $\dfrac{1}{m}$ alone is wrong.

The distance formula squares both the $x$ and $y$ differences before the root.

The midpoint averages (adds and halves) the coordinates; subtraction gives the gap, not the midpoint.

Find the gradient of the line joining $(1, 4)$ and $(3, 10)$. [2 marks]

State the gradient of a line perpendicular to $y = 2x + 1$. [1 mark]

Find the equation of the line through $(0, 3)$ parallel to $y = 4x - 7$. [2 marks]