Mathematics 9709/12 — October/November 2022

Walkthrough

We start with the coordinates of $A(5,2)$ and $B(10,-1)$. The perpendicular bisector of a segment is the line perpendicular to the segment and passing through its midpoint.

First calculate the midpoint: average the $x$-coordinates and $y$-coordinates:

Next, find the gradient of $AB$ using $\frac{\text{change in } y}{\text{change in } x}$:

A perpendicular line has gradient $m$ satisfying $m \cdot m_{AB} = -1$, so $m = \frac{5}{3}$.

Substitute the point $\left( \frac{15}{2}, \frac{1}{2} \right)$ into the point-slope form $y - y_1 = m(x - x_1)$:

Expand and simplify to get $y = \frac{5}{3}x - 12$.

Key Takeaways

The perpendicular bisector passes through the midpoint of the segment. Perpendicular gradients multiply to $-1$, so take the negative reciprocal. The point-slope form is an efficient way to write the equation of a line through a known point with a known gradient.

Common Mistakes

A common mistake is using one of $A$ or $B$ as the point on the perpendicular bisector instead of the midpoint. Another is forgetting to change the sign when taking the reciprocal of the gradient; the negative reciprocal of $-\frac{3}{5}$ is $\frac{5}{3}$, not $-\frac{5}{3}$ or $\frac{3}{5}$. Also, avoid putting the $x$ and $y$ changes in inconsistent orders when computing the gradient, as this can produce a sign error.

Things to Be Careful About

The midpoint averages must be computed carefully with signed coordinates: the $y$-coordinate is $\frac{2 + (-1)}{2} = \frac{1}{2}$, not $\frac{3}{2}$. The final equation may be written in several equivalent forms, such as $y = \frac{5}{3}x - 12$ or $5x - 3y = 36$; either is acceptable. Marking may award the first method mark for gradient use and for the perpendicular condition, so show both clearly.

Walkthrough

For a circle with centre $A$ passing through $B$, the radius is the distance from $A$ to $B$. We only need $r^2$ for the equation, so we compute the squared distance:

The centre is $A(5,2)$, so $h = 5$ and $k = 2$. Substitute into $(x-h)^2 + (y-k)^2 = r^2$:

This is the required equation. Expanding gives the equivalent form $x^2 + y^2 - 10x - 4y - 5 = 0$.

Key Takeaways

The equation of a circle with centre $(h,k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. The radius of a circle through a point is the distance from the centre to that point. Using the squared distance avoids introducing a square root before it is needed.

Common Mistakes

A common mistake is to substitute the centre coordinates with the wrong signs, for example writing $(x+5)^2 + (y+2)^2$. Another is forgetting to square the distance and writing $(x-5)^2 + (y-2)^2 = \sqrt{34}$ instead of $= 34$; unless the radius is written as $\sqrt{34}$ and then squared in the equation.

Things to Be Careful About

Be careful with the signed difference in the $y$-coordinates: $-1-2 = -3$, so the squared term is $(-3)^2 = 9$, not $-9$. The centre-radius form is often easiest, but an expanded form such as $x^2 + y^2 - 10x - 4y - 5 = 0$ is also acceptable. If a mark scheme awards separate marks for the radius, the centre form, and the final equation, write each step explicitly.