Coordinate Geometry Worked Examples (KBAT)

Q: Why do I square both sides in a locus question?

The distance formula contains a square root, and squaring PA=PBPA=PBPA=PB (or PA=2 PBPA=2\,PBPA=2PB) clears it so you can expand with ordinary algebra. Because distances are never negative, squaring here introduces no false solutions.

Q: How can I tell whether a locus will be a line or a circle before I finish?

An equidistant condition (ratio 1:11:11:1) always gives a straight line, and any other constant ratio of distances gives a circle. In the algebra, if the x2x^{2}x2 and y2y^{2}y2 terms cancel you have a line; if they survive with equal coefficients you have a circle.

Q: In the ratio AD:DB=1:2, is D closer to A or to B?

Closer to AAA, because ADADAD is the shorter part, one share out of three. So DDD sits one-third of the way along the segment from AAA towards BBB.

What these examples cover

These hard Coordinate Geometry examples combine two ideas at once, the way the longer paper questions do. The first two build the equation of a locus, the path of a moving point that obeys a fixed rule, one that turns out to be a straight line, and one that turns out to be a circle.

The third weaves a division point into an area calculation, so the answer to part (a) feeds straight into part (b). None of them needs a large calculation; they need a clear plan and steady algebra.

Attempt each fully on paper first, expand every bracket carefully, and keep a running check on your signs, that discipline is what makes hard questions manageable rather than daunting.

Worked examples

Work through all three. Attempt each fully before reading the matching solution, and notice how the plan stays the same even as the answers become richer: name the condition, translate it into algebra, expand and simplify without rushing, then confirm the result by an independent route.

Q1[4 marks]

A point $P(x, y)$ moves so that it is always equidistant from the points $A(1, 2)$ and $B(5, 4)$ . Find the equation of the locus of $P$ .

Show worked solution

'Equidistant' means $PA=PB$ . Squaring both sides removes the roots and keeps the algebra clean, so work from $PA^{2}=PB^{2}$ using the distance formula:

(x-1)^{2}+(y-2)^{2}=(x-5)^{2}+(y-4)^{2}

Expand every bracket carefully, keeping each squared term in place:

x^{2}-2x+1+y^{2}-4y+4=x^{2}-10x+25+y^{2}-8y+16

The $x^{2}$ and $y^{2}$ terms appear on both sides, so they cancel. Collect what remains:

-2x-4y+5=-10x-8y+41

Move every term to one side, simplify, then divide through by the common factor $4$ :

8x+4y-36=0 \;\Rightarrow\; 2x+y-9=0

Answer

The locus is the straight line $2x+y-9=0$ . It is the perpendicular bisector of $AB$ , which is a neat check: its gradient is $-2$ , the negative reciprocal of $m_{AB}=\tfrac{4-2}{5-1}=\tfrac{1}{2}$ , and it passes through the midpoint $(3, 3)$ since $2(3)+3-9=0$ .

Q2[5 marks]

A point $P(x, y)$ moves so that its distance from $A(1, 0)$ is always twice its distance from $B(4, 0)$ . Find the equation of the locus of $P$ , and state the type of curve it represents.

Show worked solution

The condition is $PA=2\,PB$ . Squaring both sides gives $PA^{2}=4\,PB^{2}$ , which clears the roots.

Write each squared distance with the distance formula:

(x-1)^{2}+y^{2}=4\bigl[(x-4)^{2}+y^{2}\bigr]

Expand both sides, remembering to multiply everything inside the brackets on the right by $4$ :

x^{2}-2x+1+y^{2}=4x^{2}-32x+64+4y^{2}

Bring every term to one side. Collecting like terms gives:

0=3x^{2}+3y^{2}-30x+63

Divide through by $3$ so the squared terms have coefficient $1$ :

x^{2}+y^{2}-10x+21=0

Answer

The locus is $x^{2}+y^{2}-10x+21=0$ . Because the $x^{2}$ and $y^{2}$ coefficients are equal and there is no $xy$ term, it is a circle; completing the square gives $(x-5)^{2}+y^{2}=4$ , a circle with centre $(5, 0)$ and radius $2$ .

Check on the $x$ -axis: $x=3$ gives $PA=2$ , $PB=1$ ; $x=7$ gives $PA=6$ , $PB=3$ ; both satisfy $PA=2\,PB$ .

Q3[5 marks]

A triangle has vertices $A(-1, 2)$ , $B(5, 8)$ and $C(4, 0)$ . The point $D$ lies on $AB$ such that $AD:DB=1:2$ .

Find (a) the coordinates of $D$ , and (b) the area of triangle $ACD$ .

Show worked solution

(a) Use the division-point formula for a point dividing $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ in the ratio $m:n$ , with $m=1$ (the $AD$ part) and $n=2$ (the $DB$ part):

D=\left(\frac{n x_{1}+m x_{2}}{m+n},\; \frac{n y_{1}+m y_{2}}{m+n}\right)

Substitute $A(-1, 2)$ , $B(5, 8)$ , $m=1$ and $n=2$ , then simplify each coordinate:

D=\left(\frac{2(-1)+1(5)}{1+2},\; \frac{2(2)+1(8)}{1+2}\right)=\left(\frac{3}{3},\; \frac{12}{3}\right)=(1, 4)

(b) Now find the area of triangle $ACD$ with vertices $A(-1, 2)$ , $C(4, 0)$ and $D(1, 4)$ . Apply the vertex-area formula:

\text{Area}=\frac{1}{2}\bigl|\,x_{A}(y_{C}-y_{D})+x_{C}(y_{D}-y_{A})+x_{D}(y_{A}-y_{C})\,\bigr|

Substitute the three points and simplify inside the modulus:

\text{Area}=\frac{1}{2}\bigl|\,-1(0-4)+4(4-2)+1(2-0)\,\bigr|=\frac{1}{2}\bigl|\,4+8+2\,\bigr|=7

Answer

$D=(1, 4)$ and the area of triangle $ACD$ is $7$ square units. Check the area a second way with edge vectors from $A$ : $\vec{AC}=(5, -2)$ and $\vec{AD}=(2, 2)$ , so the area is $\tfrac{1}{2}\,|5\times2-(-2)\times2|=\tfrac{1}{2}\,|10+4|=7$ , which agrees.

Q4[5 marks]

$PQRS$ is a rhombus. Two opposite vertices are $P(1, 1)$ and $R(7, 5)$ , and a third vertex is $Q(2, 6)$ .

Find the coordinates of $S$ , and show that the diagonals $PR$ and $QS$ are perpendicular.

Show worked solution

In a rhombus, the diagonals bisect each other, so $PR$ and $QS$ share the same midpoint. Find the midpoint of $PR$ first:

M=\left(\frac{1+7}{2},\; \frac{1+5}{2}\right)=(4, 3)

Let $S=(x, y)$ . Since that same point must also be the midpoint of $QS$ , with $Q(2, 6)$ , set the midpoint of $QS$ equal to $(4, 3)$ and solve for each coordinate:

\frac{2+x}{2}=4 \;\text{and}\; \frac{6+y}{2}=3 \;\Rightarrow\; x=6,\ y=0

Now confirm the diagonals meet at right angles by comparing their gradients:

m_{PR}=\frac{5-1}{7-1}=\frac{2}{3}, \qquad m_{QS}=\frac{0-6}{6-2}=-\frac{3}{2}

Answer

$S=(6, 0)$ , and the diagonals are perpendicular because $m_{PR}\times m_{QS}=\tfrac{2}{3}\times\left(-\tfrac{3}{2}\right)=-1$ . As a further check, all four sides equal $\sqrt{26}$ : $PQ=\sqrt{(2-1)^{2}+(6-1)^{2}}=\sqrt{26}$ and $RS=\sqrt{(6-7)^{2}+(0-5)^{2}}=\sqrt{26}$ , confirming $PQRS$ is genuinely a rhombus.

Q5[4 marks]

The points $A(1, 2)$ , $B(3, k)$ and $C(5, 4)$ are the vertices of a triangle with area $8$ square units. Find the two possible values of $k$ .

Show worked solution

Substitute the three vertices into the area formula, keeping $k$ as an unknown, and write out the bracket before simplifying:

\text{Area}=\frac{1}{2}\bigl|\,x_{A}(y_{B}-y_{C})+x_{B}(y_{C}-y_{A})+x_{C}(y_{A}-y_{B})\,\bigr|=\frac{1}{2}\bigl|\,1(k-4)+3(4-2)+5(2-k)\,\bigr|

Expand and collect the $k$ terms inside the modulus:

\frac{1}{2}\bigl|\,k-4+6+10-5k\,\bigr|=\frac{1}{2}\bigl|\,-4k+12\,\bigr|=\bigl|\,6-2k\,\bigr|

Set this equal to the given area of $8$ . A modulus equation splits into two cases:

6-2k=8 \;\Rightarrow\; k=-1, \qquad\text{or}\qquad 6-2k=-8 \;\Rightarrow\; k=7

Answer

$k=-1$ or $k=7$ . Neither value is $3$ , the value that would make $A$ , $B$ , $C$ collinear ( $6-2k=0$ ), so both give a genuine triangle.

The two answers sit symmetrically $4$ either side of $k=3$ : with $A$ and $C$ fixed, $B$ can sit the same perpendicular distance from line $AC$ on either side and still enclose the same area.

Q6[5 marks]

The perpendicular bisector of the line segment joining $A(0, 0)$ and $B(4, 2)$ meets the line $y=3x$ at the point $Q$ . Find the coordinates of $Q$ , and verify that $Q$ is equidistant from $A$ and $B$ .

Show worked solution

First find the midpoint of $AB$ and the gradient of $AB$ , which the perpendicular bisector needs:

M=\left(\frac{0+4}{2},\; \frac{0+2}{2}\right)=(2, 1), \qquad m_{AB}=\frac{2-0}{4-0}=\frac{1}{2}

The perpendicular bisector passes through $M$ with gradient $-2$ , the negative reciprocal of $m_{AB}$ :

y-1=-2(x-2) \;\Rightarrow\; y=-2x+5

Solve this simultaneously with $y=3x$ to find $Q$ :

3x=-2x+5 \;\Rightarrow\; 5x=5 \;\Rightarrow\; x=1,\ y=3

Answer

$Q=(1, 3)$ . Checking distances confirms it: $QA=\sqrt{1^{2}+3^{2}}=\sqrt{10}$ and $QB=\sqrt{(1-4)^{2}+(3-2)^{2}}=\sqrt{10}$ ; since $QA=QB$ , $Q$ genuinely lies on the perpendicular bisector of $AB$ .

Q7[4 marks]

Find the coordinates of the points where the straight line $x+y=7$ intersects the circle $x^{2}+y^{2}=25$ .

Show worked solution

Make $y$ the subject of the line and substitute it into the circle's equation:

y=7-x \;\Rightarrow\; x^{2}+(7-x)^{2}=25

Expand and simplify to a quadratic in $x$ :

x^{2}+49-14x+x^{2}=25 \;\Rightarrow\; 2x^{2}-14x+24=0 \;\Rightarrow\; x^{2}-7x+12=0

Factorise and solve for $x$ , then find the matching $y$ from $y=7-x$ :

(x-3)(x-4)=0 \;\Rightarrow\; x=3 \text{ or } x=4

Answer

The line meets the circle at $(3, 4)$ and $(4, 3)$ . Both check out: $3^{2}+4^{2}=25$ and $3+4=7$ .

It makes sense there are two points, since the centre $(0,0)$ is a distance $\tfrac{7}{\sqrt{2}}\approx4.9$ from the line, which is less than the radius $5$ , so the line cuts right through the circle.

Q8[6 marks]

Find (a) the coordinates of the foot of the perpendicular, $N$ , from the point $A(6, 1)$ to the line $2x-y=1$ , and (b) the shortest distance from $A$ to the line.

Show worked solution

(a) The line $2x-y=1$ , or $y=2x-1$ , has gradient $2$ , so the perpendicular from $A$ has gradient $-\tfrac{1}{2}$ . Write its equation through $A(6, 1)$ :

y-1=-\frac{1}{2}(x-6) \;\Rightarrow\; y=-\frac{1}{2}x+4

Solve this simultaneously with $y=2x-1$ to find $N$ :

2x-1=-\frac{1}{2}x+4 \;\Rightarrow\; \frac{5}{2}x=5 \;\Rightarrow\; x=2,\ y=3

(b) The shortest distance from $A$ to the line is the length $AN$ :

AN=\sqrt{(6-2)^{2}+(1-3)^{2}}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}

Answer

$N=(2, 3)$ and the shortest distance is $2\sqrt{5}$ units (about $4.47$ ). Check with the perpendicular-distance route: for $2x-y-1=0$ , the distance from $A(6,1)$ is $\dfrac{|2(6)-1-1|}{\sqrt{2^{2}+(-1)^{2}}}=\dfrac{10}{\sqrt{5}}=2\sqrt{5}$ , which agrees.

Look back at all three and the shared plan stands out: put the condition into symbols first, square away any roots, expand without rushing, and finish with a check by a second route. Hard questions in this chapter reward that patience, each one is really a short sequence of easy steps held together by careful algebra.

Key method points

These three examples pull together the locus, division and area tools into the kind of combined question the longer paper favours. Keep the following points in mind as you practise more.

For a locus, name the distance condition first ( $PA=PB$ , or $PA=2\,PB$ ), then square both sides to remove the roots before expanding.
An equidistant locus is always a straight line, the perpendicular bisector of the two fixed points; a constant ratio of distances other than $1:1$ gives a circle.
After expanding, the $x^{2}$ and $y^{2}$ terms cancel for a line but survive for a circle; equal $x^{2}$ and $y^{2}$ coefficients with no $xy$ term signal a circle.
Complete the square to read a circle's centre and radius, and to confirm the equation is a genuine circle.
In a two-part question, the point you find in part (a) is usually the input to part (b), carry it forward exactly, without rounding.
Verify an area independently with edge vectors from one vertex: $\tfrac{1}{2}\,|x_{1}y_{2}-x_{2}y_{1}|$ .

How a teacher helps

Hard questions rarely fail on a hard step; they fail on a rushed expansion or a lost sign three lines in. In a one-to-one lesson our teacher slows down exactly those risky lines, expanding $(x-5)^{2}$ , or multiplying a whole bracket by $4$ , and shows you how to spot a wrong answer with a quick check before you move on.

Because our teachers are experienced, you work with someone who knows which combined questions the paper favours and how to plan them. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you either way.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

Why do I square both sides in a locus question?

The distance formula contains a square root, and squaring $PA=PB$ (or $PA=2\,PB$ ) clears it so you can expand with ordinary algebra. Because distances are never negative, squaring here introduces no false solutions.

How can I tell whether a locus will be a line or a circle before I finish?

An equidistant condition (ratio $1:1$ ) always gives a straight line, and any other constant ratio of distances gives a circle. In the algebra, if the $x^{2}$ and $y^{2}$ terms cancel you have a line; if they survive with equal coefficients you have a circle.

In the ratio

AD:DB=1:2

, is

D

closer to

A

or to

B

Closer to $A$ , because $AD$ is the shorter part, one share out of three. So $D$ sits one-third of the way along the segment from $A$ towards $B$ .

Do I really need to check a high-mark answer twice?

It is not required, but a quick independent check, a point that must lie on the locus, or an area recomputed with vectors, is the cheapest insurance against a sign slip on a question worth several marks.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Coordinate Geometry, Worked Examples (KBAT)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

Ready to get started?