Coordinate Geometry Practice

Q: In the ratio formula, which numbers go where?

For AP:PB=m:nAP:PB=m:nAP:PB=m:n, the point is (nx1+mx2m+n, ny1+my2m+n)\left(\frac{n x_{1}+m x_{2}}{m+n},\ \frac{n y_{1}+m y_{2}}{m+n}\right)(m+nnx1+mx2, m+nny1+my2). Notice the cross: mmm multiplies the second point and nnn the first. Getting this the wrong way round is the most common slip.

Q: How do I know two lines are perpendicular?

Their gradients multiply to −1-1−1, so m1m2=−1m_{1}m_{2}=-1m1m2=−1. If one gradient is 222, the perpendicular gradient is −12-\tfrac{1}{2}−21, flip it over and change the sign.

How to use these practice questions

The six questions below rise in difficulty across the whole Coordinate Geometry chapter, from a straightforward distance and midpoint up to the equation of a locus. Give yourself roughly five to eight minutes per question and work on paper first, writing every line the way you would in the real exam, a method line stating the formula you are using, a clear substitution, then the final answer.

Resist the urge to peek. Only once you have committed to a full answer should you open the solution and mark yourself line by line.

When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost. Because Add Math is marked analytically, a correct formula and a correct substitution still earn credit even when the final arithmetic slips, so always write the substitution in full before you simplify.

Treat this page as a rehearsal, not a test, the point is to find weak steps now, while there is still time to fix them.

Six practice questions

Q1[3 marks]

The points are $A(1,\,2)$ and $B(7,\,10)$ . Find (a) the distance $AB$ , and (b) the midpoint of $AB$ .

Show worked solution

(a) Use the distance formula $AB=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ . Substitute the two coordinates and simplify under the root:

AB=\sqrt{(7-1)^{2}+(10-2)^{2}}=\sqrt{36+64}=\sqrt{100}=10

(b) The midpoint is the average of the coordinates, $\left(\dfrac{x_{1}+x_{2}}{2},\ \dfrac{y_{1}+y_{2}}{2}\right)$ :

\left(\frac{1+7}{2},\ \frac{2+10}{2}\right)=(4,\ 6)

Answer

$AB=10$ units and the midpoint is $(4,\,6)$ . The midpoint should sit exactly halfway, so it is $3$ across and $4$ up from $A$ , and the same again to $B$ , which it is.

Q2[3 marks]

A straight line passes through $P(2,\,3)$ and $Q(5,\,9)$ . Find the equation of the line.

Show worked solution

First find the gradient using $m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$ :

m=\frac{9-3}{5-2}=\frac{6}{3}=2

Now use the point-gradient form $y-y_{1}=m(x-x_{1})$ with the point $P(2,\,3)$ , then simplify to $y=mx+c$ :

y-3=2(x-2)\;\Rightarrow\; y=2x-4+3\;\Rightarrow\; y=2x-1

Answer

The equation is $y=2x-1$ . Check with the other point $Q(5,\,9)$ : $2(5)-1=9$ , which matches, so the line really does pass through both points.

Q3[4 marks]

The point $P$ divides the line segment joining $A(1,\,2)$ and $B(11,\,17)$ in the ratio $AP:PB=2:3$ . Find the coordinates of $P$ .

Show worked solution

Use the ratio (section) formula. If $P$ divides $AB$ so that $AP:PB=m:n$ , then

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

Here $m=2$ , $n=3$ , with $A(1,\,2)$ and $B(11,\,17)$ . Substitute carefully:

P=\left(\frac{3(1)+2(11)}{2+3},\ \frac{3(2)+2(17)}{2+3}\right)=\left(\frac{25}{5},\ \frac{40}{5}\right)=(5,\ 8)

Answer

$P=(5,\,8)$ . Check by walking $\tfrac{2}{5}$ of the way from $A$ to $B$ : the step $A\to B$ is $(10,\,15)$ , and $\tfrac{2}{5}$ of it is $(4,\,6)$ , so $P=(1+4,\,2+6)=(5,\,8)$ , as found.

Q4[4 marks]

The straight line $L_{1}$ has equation $y=2x-1$ . Find the equation of the straight line $L_{2}$ that passes through $(4,\,5)$ and is perpendicular to $L_{1}$ .

Show worked solution

Read the gradient of $L_{1}$ straight from $y=2x-1$ , so $m_{1}=2$ . For perpendicular lines the gradients satisfy $m_{1}m_{2}=-1$ :

m_{1}m_{2}=-1\;\Rightarrow\; 2\,m_{2}=-1\;\Rightarrow\; m_{2}=-\frac{1}{2}

Now use point-gradient form through $(4,\,5)$ with $m_{2}=-\dfrac{1}{2}$ :

y-5=-\frac{1}{2}(x-4)\;\Rightarrow\; y=-\frac{1}{2}x+2+5\;\Rightarrow\; y=-\frac{1}{2}x+7

Clearing the fraction gives a tidy form for the answer:

x+2y=14

Answer

$L_{2}:\ x+2y=14$ (equivalently $y=-\tfrac{1}{2}x+7$ ). Check the point $(4,\,5)$ : $4+2(5)=14$ .

Check perpendicularity: $2\times(-\tfrac{1}{2})=-1$ .

Q5[5 marks]

A triangle has vertices $A(1,\,1)$ , $B(5,\,2)$ and $C(3,\,6)$ . Find the area of triangle $ABC$ .

Show worked solution

Use the coordinate area formula. Taking the vertices in order, the area is

\text{Area}=\frac{1}{2}\,\bigl|\,x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\,\bigr|

Substitute $A(1,\,1)$ , $B(5,\,2)$ , $C(3,\,6)$ , keeping every bracket:

=\frac{1}{2}\,\bigl|\,1(2-6)+5(6-1)+3(1-2)\,\bigr|

Work the inside step by step before taking the absolute value:

=\frac{1}{2}\,\bigl|\,-4+25-3\,\bigr|=\frac{1}{2}\,(18)=9

Answer

The area is $9$ square units. The absolute value keeps the area positive whichever order you list the vertices; if you had gone clockwise the inside would read $-18$ , and $\tfrac{1}{2}|{-18}|=9$ all the same.

Q6[5 marks]

A point $P(x,\,y)$ moves such that its distance from $A(1,\,0)$ is always twice its distance from $B(4,\,0)$ ; that is, $PA=2\,PB$ . Find the equation of the locus of $P$ .

Show worked solution

Square both sides at the start to remove the roots: $PA=2\,PB$ becomes $PA^{2}=4\,PB^{2}$ . Write each squared distance in coordinates:

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

Expand both sides fully:

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

Bring everything to one side and collect like terms:

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

Divide through by $3$ to reach the simplest form:

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

Answer

The locus is $x^{2}+y^{2}-10x+21=0$ , a circle. Completing the square gives $(x-5)^{2}+y^{2}=4$ , centre $(5,\,0)$ , radius $2$ .

Check the point $(3,\,0)$ : $PA=2$ , $PB=1$ , so $PA=2PB$ ; and $(7,\,0)$ : $PA=6$ , $PB=3$ , again $PA=2PB$ .

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for the correct formula quoted, the right values put into it, and a clean final statement with units where needed.

Method mark: did you write the correct formula first, distance, midpoint, gradient, the ratio formula, the area formula, or $m_{1}m_{2}=-1$ for perpendicular lines?
Substitution mark: are the correct coordinates in the correct slots, with brackets kept around every term, especially negatives?
Answer mark: is the final value stated clearly, simplified, and does it survive a check by substituting back?
For the locus question, you only earn full marks by squaring first, then expanding and simplifying, leaving a stray root loses the method.
If your final number is wrong but the formula and substitution lines are right, give yourself those marks, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, coordinates swapped in the distance formula, the ratio numbers placed the wrong way round, or a dropped negative when expanding a locus, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each step. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

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Frequently asked questions

How long should each of these questions take me?

Aim for roughly five to eight minutes each, rising with the mark value. If a question takes far longer, note it and bring it to a lesson, the time it steals in the exam is often the real problem, not the topic itself.

In the ratio formula, which numbers go where?

For $AP:PB=m:n$ , the point is $\left(\frac{n x_{1}+m x_{2}}{m+n},\ \frac{n y_{1}+m y_{2}}{m+n}\right)$ . Notice the cross: $m$ multiplies the second point and $n$ the first.

Getting this the wrong way round is the most common slip.

How do I know two lines are perpendicular?

Their gradients multiply to $-1$ , so $m_{1}m_{2}=-1$ . If one gradient is $2$ , the perpendicular gradient is $-\tfrac{1}{2}$ , flip it over and change the sign.

Do I lose all the marks if my final answer is wrong?

No. Because marking is analytic, a correct formula line and a correct substitution still earn marks even if the arithmetic slips at the end.

That is why you should always show full working.

Source:SRC-DSKP-EN

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

Coordinate Geometry, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

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