Coordinate Geometry Worked Examples (Easy)

Q: Does it matter which point I subtract first in the distance formula?

No. Both differences are squared, so (x2−x1)2(x_{2}-x_{1})^{2}(x2−x1)2 equals (x1−x2)2(x_{1}-x_{2})^{2}(x1−x2)2. Either order gives the same positive distance, just stay consistent within one calculation.

Q: How is the midpoint different from a dividing point?

The midpoint is the special case where the ratio is 1:11:11:1, so it sits exactly halfway. A general dividing point can sit anywhere on the segment, closer to whichever end carries the larger part of the ratio.

Q: Which point should I use in y-y_{1}=m(x-x_{1})?

Either point on the line works. The final equation y=mx+cy=mx+cy=mx+c is the same whichever you choose, because both points satisfy it, picking the tidier coordinates just keeps the arithmetic light.

Q: In the ratio AP:PB=3:1, which number multiplies which point?

The formula crosses the weights over: the 3 (the APAPAP part) multiplies BBB, and the 1 (the PBPBPB part) multiplies AAA. That places PPP closer to BBB, which matches APAPAP being the longer piece.

What these examples cover

These easy Coordinate Geometry examples build the everyday moves that open almost every question in the chapter: measuring the distance between two points, locating the midpoint of a segment, turning a gradient into the equation of a straight line, and finding the point that divides a segment in a given ratio. Each one uses small, clean numbers, so you can follow every line without a calculator getting in the way.

Use the set the honest way, cover the solution, attempt the question in full on paper, and only then check line by line against our working. Where your answer differs, find the exact step where the two solutions part company; that single line is usually where the real learning is.

A quick sketch of the points, drawn to no particular scale, keeps the signs under control.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and notice how the same discipline, substitute carefully, keep every bracket, simplify one step at a time, runs through the distance, the midpoint, the line equation and the dividing point alike.

Q1[2 marks]

Find the distance between the points $A(1, 2)$ and $B(4, 6)$ .

Show worked solution

The distance uses the horizontal gap and the vertical gap between the two points, combined by Pythagoras' theorem. Substitute the coordinates, keeping each subtraction inside a bracket:

AB=\sqrt{(4-1)^{2}+(6-2)^{2}}

Work out each bracket, square it, then add the two squares before taking the root:

AB=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5

Answer

The distance $AB=5$ units. Because each gap is squared, it does not matter which point you subtract first, $(1-4)^{2}$ gives the same $9$ as $(4-1)^{2}$ .

Q2[2 marks]

The points $P(-3, 5)$ and $Q(7, 1)$ are the ends of a line segment. Find the coordinates of the midpoint $M$ of $PQ$ .

Show worked solution

The midpoint is the average of the two $x$ -coordinates and the average of the two $y$ -coordinates. Substitute carefully, keeping the negative sign on $-3$ :

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

Simplify each coordinate on its own:

M=\left(\frac{4}{2},\; \frac{6}{2}\right)=(2, 3)

Answer

The midpoint is $M(2, 3)$ . A quick check: $2$ sits halfway between $-3$ and $7$ , and $3$ sits halfway between $5$ and $1$ , exactly as a midpoint should.

Q3[4 marks]

A straight line passes through the points $A(-1, 1)$ and $B(3, 9)$ . Find (a) the gradient of the line, and (b) the equation of the line in the form $y=mx+c$ .

Show worked solution

(a) The gradient is the change in $y$ divided by the change in $x$ . Subtract the two points in the same order on top and bottom:

m=\frac{9-1}{3-(-1)}=\frac{8}{4}=2

(b) Use the point–gradient form $y-y_{1}=m(x-x_{1})$ with $m=2$ and the point $A(-1, 1)$ :

y-1=2\bigl(x-(-1)\bigr)=2(x+1)

Expand the bracket and make $y$ the subject:

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

Answer

The gradient is $2$ and the line is $y=2x+3$ . Check with $B(3, 9)$ : $2(3)+3=9$ , which matches, so the line does pass through both points.

Q4[3 marks]

The point $P$ divides the line segment joining $A(1, 2)$ and $B(9, 10)$ in the ratio $AP:PB=3:1$ . Find the coordinates of $P$ .

Show worked solution

For a point dividing $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ in the ratio $m:n$ , use the division-point formula. The weights cross over, so $m$ multiplies the far point $B$ .

Here $m=3$ and $n=1$ :

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

Substitute $A(1, 2)$ , $B(9, 10)$ , $m=3$ and $n=1$ , then simplify each coordinate:

P=\left(\frac{1(1)+3(9)}{3+1},\; \frac{1(2)+3(10)}{3+1}\right)=\left(\frac{28}{4},\; \frac{32}{4}\right)=(7, 8)

Answer

The point is $P(7, 8)$ . Check by stepping along the segment: from $A$ to $B$ the move is $(8, 8)$ , and $AP:PB=3:1$ places $P$ three-quarters of the way, at $A+\tfrac{3}{4}(8, 8)=(1+6,\; 2+6)=(7, 8)$ .

Q5[2 marks]

A straight line has equation $3x+4y-12=0$ . Find (a) the gradient of the line, and (b) its $y$ -intercept.

Show worked solution

Rearrange the equation into the gradient form $y=mx+c$ by making $y$ the subject. Move the $x$ -term and the constant to the other side first:

4y=-3x+12

Divide every term by $4$ , the coefficient of $y$ , so $y$ stands alone:

y=-\frac{3}{4}x+3

Answer

The gradient is $-\frac{3}{4}$ and the $y$ -intercept is $3$ . Check by setting $x=0$ in the original equation: $4y-12=0$ gives $y=3$ , which matches.

Q6[3 marks]

Line $L_{1}$ has equation $y=2x-1$ . Line $L_{2}$ is parallel to $L_{1}$ and passes through the point $(3, 4)$ .

Find the equation of $L_{2}$ .

Show worked solution

Parallel lines always have the same gradient, so $L_{2}$ also has gradient $2$ . Use the point–gradient form $y-y_{1}=m(x-x_{1})$ with $m=2$ and the point $(3, 4)$ :

y-4=2(x-3)

Expand the bracket and make $y$ the subject:

y-4=2x-6 \;\Rightarrow\; y=2x-2

Answer

The line $L_{2}$ is $y=2x-2$ . Check: at $x=3$ , $y=2(3)-2=4$ , which matches the given point, and the gradient $2$ is the same as $L_{1}$ , so the lines are indeed parallel.

Q7[3 marks]

Line $L_{1}$ has equation $y=\frac{1}{2}x+3$ . Line $L_{2}$ is perpendicular to $L_{1}$ and passes through the point $(4, 1)$ .

Find the equation of $L_{2}$ .

Show worked solution

For perpendicular lines, the product of the gradients is $-1$ . The gradient of $L_{1}$ is $\frac{1}{2}$ , so solve for the gradient of $L_{2}$ :

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

Use the point–gradient form with $m_{2}=-2$ and the point $(4, 1)$ , then make $y$ the subject:

y-1=-2(x-4) \;\Rightarrow\; y=-2x+9

Answer

The line $L_{2}$ is $y=-2x+9$ . Check: at $x=4$ , $y=-2(4)+9=1$ , which matches the given point, and $\frac{1}{2}\times(-2)=-1$ confirms the two lines are perpendicular.

Q8[3 marks]

Find the area of the triangle with vertices $A(1, 1)$ , $B(5, 1)$ and $C(1, 4)$ .

Show worked solution

The area of a triangle with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is given by:

\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 $A(1, 1)$ , $B(5, 1)$ and $C(1, 4)$ , keeping each bracket separate:

\text{Area}=\frac{1}{2}\bigl|1(1-4)+5(4-1)+1(1-1)\bigr|=\frac{1}{2}\bigl|-3+15+0\bigr|=\frac{1}{2}(12)=6

Answer

The area is $6$ square units. Sense check: $AB$ is a horizontal side of length $4$ and $AC$ is a vertical side of length $3$ , meeting at a right angle at $A$ , so the area is also $\frac{1}{2}\times4\times3=6$ .

Notice how different these four questions look on the surface, yet how similar the discipline is underneath: read exactly what is asked, substitute one value at a time with every bracket in place, and simplify before you round or judge the answer. That steadiness is what turns Coordinate Geometry into a reliable block of marks rather than a place for small slips.

Key method points

These four examples rehearse the tools that open almost every Coordinate Geometry question in Add Math. Keep the following points in mind as you practise more.

The distance between two points comes from Pythagoras: $AB=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ ; the squares make the order of subtraction irrelevant.
The midpoint is the average of the coordinates: $M=\left(\frac{x_{1}+x_{2}}{2},\; \frac{y_{1}+y_{2}}{2}\right)$ .
The gradient is the change in $y$ over the change in $x$ ; subtract the two points in the same order top and bottom.
Turn a gradient and a point into a line with $y-y_{1}=m(x-x_{1})$ , then make $y$ the subject.
A dividing point in the ratio $m:n$ uses $\left(\frac{n x_{1}+m x_{2}}{m+n},\; \frac{n y_{1}+m y_{2}}{m+n}\right)$ , the weights cross over, so $m$ multiplies the second point.
A quick sketch, drawn to no particular scale, catches most sign slips before they cost a mark; and because marking is analytic, a clear substitution line still earns method marks even if the final arithmetic slips.

How a teacher helps

When a student drops a mark on questions like these, it is nearly always a small, fixable habit, a missing bracket around a negative coordinate, or the two points subtracted in a different order on top and bottom. In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it on the spot, before it settles into a routine.

Because our teachers are experienced, you work with someone who explains why each formula is built the way it is, not just how to use it. 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.

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Frequently asked questions

Does it matter which point I subtract first in the distance formula?

No. Both differences are squared, so $(x_{2}-x_{1})^{2}$ equals $(x_{1}-x_{2})^{2}$ .

Either order gives the same positive distance, just stay consistent within one calculation.

How is the midpoint different from a dividing point?

The midpoint is the special case where the ratio is $1:1$ , so it sits exactly halfway. A general dividing point can sit anywhere on the segment, closer to whichever end carries the larger part of the ratio.

Which point should I use in

y-y_{1}=m(x-x_{1})

Either point on the line works. The final equation $y=mx+c$ is the same whichever you choose, because both points satisfy it, picking the tidier coordinates just keeps the arithmetic light.

In the ratio

AP:PB=3:1

, which number multiplies which point?

The formula crosses the weights over: the 3 (the $AP$ part) multiplies $B$ , and the 1 (the $PB$ part) multiplies $A$ . That places $P$ closer to $B$ , which matches $AP$ being the longer piece.

Source:SRC-DSKP-EN

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

Coordinate Geometry, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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