Coordinate Geometry Worked Examples (Medium)

Q: How do I know whether to set gradients equal or make their product -1?

Read the relationship in the question. Parallel means equal gradients; perpendicular means the product of the gradients is −1-1−1. Naming which one applies before you start is half the work.

What these examples cover

These medium Coordinate Geometry examples take the basics one clear step further. You already know how to read a gradient; now you use it to decide when two lines are perpendicular or parallel and to build a line's equation from that condition, and you turn a set of vertices straight into an area.

Each question uses small, exact numbers so the method stays visible, but the working has more moving parts than the easy set, a reciprocal gradient here, a run of signs to track there. Attempt each fully on paper before reading the solution, and when you check, look closely at the join between steps: that is where medium questions are usually won or lost.

Worked examples

Work through all three. Attempt each fully before reading the matching solution, and notice how a single reliable habit, write the condition first, then substitute one value at a time, carries you through perpendicular lines, parallel lines and an area alike.

Q1[3 marks]

A straight line $L_{1}$ has equation $y=3x-4$ . The line $L_{2}$ is perpendicular to $L_{1}$ and passes through the point $(6, 1)$ .

Find the equation of $L_{2}$ in the form $y=mx+c$ .

Show worked solution

Read the gradient of $L_{1}$ straight from its equation: in $y=3x-4$ the coefficient of $x$ is the gradient, so $m_{1}=3$ .

Perpendicular gradients multiply to $-1$ , so the gradient of $L_{2}$ is the negative reciprocal of $3$ :

m_{1}\,m_{2}=-1 \;\Rightarrow\; m_{2}=-\frac{1}{m_{1}}=-\frac{1}{3}

Now use the point–gradient form $y-y_{1}=m(x-x_{1})$ with $m_{2}=-\tfrac{1}{3}$ and the point $(6, 1)$ :

y-1=-\frac{1}{3}(x-6)

Expand the bracket, note that $-\tfrac{1}{3}\times(-6)=+2$ , and make $y$ the subject:

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

Answer

The line is $y=-\tfrac{1}{3}x+3$ . Two quick checks: the gradients satisfy $3\times(-\tfrac{1}{3})=-1$ , so the lines are perpendicular; and at $x=6$ , $y=-\tfrac{1}{3}(6)+3=-2+3=1$ , so $L_{2}$ passes through $(6, 1)$ .

Q2[3 marks]

The straight line joining $P(2, k)$ and $Q(6, 11)$ is parallel to the straight line joining $R(1, 0)$ and $S(3, 4)$ . Find the value of $k$ .

Show worked solution

Parallel lines have equal gradients, so the plan is to write both gradients and set them equal. Start with the line $RS$ , where both endpoints are fully known:

m_{RS}=\frac{4-0}{3-1}=\frac{4}{2}=2

Now write the gradient of $PQ$ in terms of $k$ , subtracting in the same order top and bottom:

m_{PQ}=\frac{11-k}{6-2}=\frac{11-k}{4}

Set the two gradients equal because the lines are parallel, then solve the resulting equation for $k$ :

\frac{11-k}{4}=2 \;\Rightarrow\; 11-k=8 \;\Rightarrow\; k=3

Answer

$k=3$ . Check by substituting back: with $k=3$ , $m_{PQ}=\dfrac{11-3}{4}=\dfrac{8}{4}=2$ , which equals $m_{RS}$ , so $PQ$ is parallel to $RS$ as required.

Q3[3 marks]

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

Show worked solution

Use the formula that reads the area straight from the vertices. List the coordinates in order around the triangle and apply the 'multiply across, then subtract' pattern:

\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|

Substitute $A(1, 1)$ , $B(5, 2)$ and $C(3, 6)$ , taking care with each difference inside the brackets:

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

Work out the brackets, then the products, then add inside the modulus:

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

Answer

The area is $9$ square units. Check by a second route using edge vectors from $A$ : $\vec{AB}=(4, 1)$ and $\vec{AC}=(2, 5)$ , so the area is $\tfrac{1}{2}\,|4\times5-1\times2|=\tfrac{1}{2}\,|20-2|=9$ , which agrees.

Q4[3 marks]

The distance between the points $A(1, 2)$ and $B(4, k)$ is $5$ units. Find the possible values of $k$ .

Show worked solution

Use the distance formula $AB=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ , substitute the coordinates of $A$ and $B$ , and square both sides to remove the root.

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

Simplify the left side, then isolate the squared term.

9+(k-2)^{2}=25 \;\Rightarrow\; (k-2)^{2}=16

Take the square root of both sides, keeping both the positive and negative root.

k-2=\pm4 \;\Rightarrow\; k=6 \text{ or } k=-2

Answer

$k=6$ or $k=-2$ . Check: with $k=6$ , $B(4,6)$ gives $AB^{2}=(4-1)^{2}+(6-2)^{2}=9+16=25$ , so $AB=5$ ; the same check works for $k=-2$ since $(-2-2)^{2}=16$ too.

Q5[4 marks]

The point $P$ divides the line segment joining $A(2, 3)$ and $B(8, 15)$ such that $AP:PB=1:2$ . Find the coordinates of $P$ .

Show worked solution

Use the formula for a point dividing a line segment in a given ratio. If $P$ divides $AB$ so that $AP:PB=m:n$ , then $P=\left(\dfrac{mx_{2}+nx_{1}}{m+n},\ \dfrac{my_{2}+ny_{1}}{m+n}\right)$ .

Here $m=1$ , $n=2$ , $A(2,3)$ and $B(8,15)$ .

P=\left(\frac{1(8)+2(2)}{1+2},\ \frac{1(15)+2(3)}{1+2}\right)

Work out each numerator, then divide by $3$ .

P=\left(\frac{8+4}{3},\ \frac{15+6}{3}\right)=\left(\frac{12}{3},\ \frac{21}{3}\right)=(4, 7)

Answer

$P=(4, 7)$ . Check: $AP=\sqrt{(4-2)^{2}+(7-3)^{2}}=\sqrt{20}$ and $PB=\sqrt{(8-4)^{2}+(15-7)^{2}}=\sqrt{80}=2\sqrt{20}$ , so $AP:PB=1:2$ as required.

Q6[3 marks]

Find the coordinates of the point of intersection of the straight lines $y=2x-1$ and $x+y=8$ .

Show worked solution

The point of intersection lies on both lines, so substitute $y=2x-1$ into $x+y=8$ and solve for $x$ .

x+(2x-1)=8 \;\Rightarrow\; 3x-1=8 \;\Rightarrow\; x=3

Substitute $x=3$ into $y=2x-1$ to find the matching $y$ -value.

y=2(3)-1=5

Answer

The lines meet at $(3, 5)$ . Check: substituting into the second equation gives $x+y=3+5=8$ , which is correct.

Q7[4 marks]

A point $P(x, y)$ moves such that its distance from the fixed point $A(3, -2)$ is always $5$ units. Find the equation of the locus of $P$ , giving your answer in the form $x^{2}+y^{2}+px+qy+r=0$ .

Show worked solution

Write the distance $PA$ using the distance formula and set it equal to $5$ , then square both sides to clear the root.

(x-3)^{2}+(y-(-2))^{2}=5^{2} \;\Rightarrow\; (x-3)^{2}+(y+2)^{2}=25

Expand both squared brackets.

x^{2}-6x+9+y^{2}+4y+4=25

Collect like terms and move the constant across so the equation equals zero.

x^{2}+y^{2}-6x+4y+13-25=0 \;\Rightarrow\; x^{2}+y^{2}-6x+4y-12=0

Answer

The locus is $x^{2}+y^{2}-6x+4y-12=0$ . Check: completing the square returns $(x-3)^{2}+(y+2)^{2}=25$ , a circle of radius $5$ centred at $A(3, -2)$ , which matches $PA=5$ .

Q8[3 marks]

The points $A(1, 2)$ , $B(3, 6)$ and $C(k, 14)$ are collinear. Find the value of $k$ .

Show worked solution

Collinear points lie on the same straight line, so the gradient of $AB$ must equal the gradient of $BC$ . Find $m_{AB}$ first.

m_{AB}=\frac{6-2}{3-1}=\frac{4}{2}=2

Write $m_{BC}$ in terms of $k$ and set it equal to $m_{AB}$ .

m_{BC}=\frac{14-6}{k-3}=\frac{8}{k-3}=2

Solve the resulting equation for $k$ .

8=2(k-3) \;\Rightarrow\; 8=2k-6 \;\Rightarrow\; 2k=14 \;\Rightarrow\; k=7

Answer

$k=7$ . Check: with $C(7, 14)$ , $m_{AC}=\frac{14-2}{7-1}=\frac{12}{6}=2$ , equal to $m_{AB}$ , confirming the three points are collinear.

Across these three, the pattern is the same: name the relationship first, perpendicular, parallel, or an area from vertices, write the matching formula, and only then substitute the numbers. Deciding the method before touching the arithmetic is exactly what keeps medium questions from turning into guesswork, and it is a habit worth rehearsing until it becomes automatic.

Key method points

These three examples rehearse the gradient and area tools that sit at the centre of Coordinate Geometry. Keep the following points in mind as you practise more.

Read a gradient straight from $y=mx+c$ : it is the coefficient of $x$ .
Parallel lines have equal gradients ( $m_{1}=m_{2}$ ); perpendicular lines have gradients whose product is $-1$ , so $m_{2}=-\tfrac{1}{m_{1}}$ , the negative reciprocal.
To find an unknown coordinate, write the gradient in terms of the unknown, set it equal to the known gradient, and solve.
The area of a triangle from its vertices is $\tfrac{1}{2}\left|x_{A}(y_{B}-y_{C})+x_{B}(y_{C}-y_{A})+x_{C}(y_{A}-y_{B})\right|$ ; the modulus keeps the area positive whatever order you list the points.
Always finish an equation-of-a-line answer by checking that the given point satisfies it.
Because marking is analytic, a correct condition or area formula with the right substitution earns method marks even if a later step slips.

How a teacher helps

At medium level the marks usually turn on a single decision, is the relationship parallel or perpendicular, and which gradient goes where. In a one-to-one lesson our teacher makes that decision explicit, so you name the condition before you reach for a formula, and slips like forgetting the negative reciprocal simply stop happening.

Because our teachers are experienced, you work with someone who can show two routes to the same area and let you pick the one you trust. 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

How do I find a perpendicular gradient quickly?

Take the negative reciprocal: flip the fraction and change the sign. So $3$ becomes $-\tfrac{1}{3}$ , and $-\tfrac{2}{5}$ becomes $+\tfrac{5}{2}$ .

A fast check is that the two gradients must multiply to $-1$ .

Does the order of the vertices matter in the area formula?

The modulus makes the final area positive whichever way you go round, so the answer is the same. What does matter is being consistent, use the same going-around order for all three terms, or a difference can end up with the wrong sign.

How do I know whether to set gradients equal or make their product

-1

Read the relationship in the question. Parallel means equal gradients; perpendicular means the product of the gradients is $-1$ .

Naming which one applies before you start is half the work.

What if the area is negative before I take the modulus?

That is normal. The sign only records the direction you listed the points (clockwise or anticlockwise); it is not part of the area.

Take the absolute value and the area is the positive result.

Source:SRC-DSKP-EN

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

Coordinate Geometry, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

Ready to get started?