Quick questions on Distance, midpoint and gradient explained: O-Level E-Maths

The distance comes from Pythagoras theorem applied to the horizontal and vertical gaps:

The midpoint of the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is the average of the coordinates:

The gradient of the segment is the change in $y$ over the change in $x$:

Used together, these formulas verify geometric facts: equal distances show equal sides, equal midpoints show diagonals that bisect each other, and gradient relationships show parallel or perpendicular sides. This is how coordinate geometry proves shapes are squares, parallelograms or right-angled.

Combining the three formulas lets you classify a triangle given its vertices. Compute the three side lengths with the distance formula: all three equal means equilateral, exactly two equal means isosceles, and all different means scalene. To test for a right angle, check whether the gradients of two sides multiply to $-1$, or equivalently whether the side lengths satisfy Pythagoras. So a triangle with sides $5$, $5$ and $\sqrt{50}$ is isosceles, and because $5^2 + 5^2 = 50$, it is also right-angled.

A common twist gives you one endpoint and the midpoint, and asks for the other endpoint. Because the midpoint is the average of the endpoints, you rearrange: if $M = (m_x, m_y)$ is the midpoint of $A(x_1, y_1)$ and $B$, then $B = (2m_x - x_1,\ 2m_y - y_1)$. So if $A = (1, 2)$ and the midpoint is $(4, 5)$, then $B = (2(4) - 1, 2(5) - 2) = (7, 8)$. Doubling the midpoint and subtracting the known endpoint reverses the averaging, a neat application of the midpoint formula that appears often in E-Maths problems.

Subtracting a negative adds; handle negative coordinates carefully in all three formulas.

Find the distance between $(0, 0)$ and $(6, 8)$. [2 marks]

Find the midpoint of the segment joining $(2, 5)$ and $(8, 1)$. [1 mark]

Find the gradient of the segment joining $(1, 7)$ and $(4, 1)$. [2 marks]