Answer Key

Coordinate Geometry

Pack A — Answers

#	Question	Answer
1	Find the midpoint of $A(2, 4)$ and $B(8, 4)$.	$(5, 4)$
2	Find the midpoint of $C(6, 2)$ and $D(6, 10)$.	$(6, 6)$
3	Find the gradient of the line through $(1, 3)$ and $(4, 9)$.	$m = 2$
4	Find the gradient of the line through $(1, 8)$ and $(5, 2)$.	$m = -\dfrac{3}{2}$
5	Find the distance between $(2, 3)$ and $(9, 3)$.	7
6	State the $y$-intercept of the line $y = 3x + 5$.	$y$-intercept = 5 (point $(0, 5)$)
7	State the gradient of the line $y = 4x + 7$.	$m = 4$
8	A line has gradient $2$ and $y$-intercept $5$. Write the equation of the line.	$y = 2x + 5$
9	Does the point $(3, 11)$ lie on the line $y = 2x + 5$?	Yes
10	Find $y$ when $x = 4$ on the line $y = 2x + 3$.	$y = 11$
11	Find the distance between $(1, 2)$ and $(5, 5)$.	5
12	Find the exact distance between $(1, 1)$ and $(4, 5)$.	5
13	Find the midpoint of $(-3, 4)$ and $(5, -2)$.	$(1, 1)$
14	Find the gradient of the line through $(-2, 3)$ and $(4, -3)$.	$m = -1$
15	A line has gradient $3$ and passes through $(2, 5)$. Find its equation in the form $y = mx + c$.	$y = 3x - 1$
16	Find the equation of the line through $(1, 3)$ and $(4, 9)$.	$y = 2x + 1$
17	A line is parallel to $y = 4x + 7$. State its gradient.	Gradient = 4
18	A line is perpendicular to $y = 2x + 5$. State its gradient.	Gradient = $-\dfrac{1}{2}$
19	Rewrite the equation $2x + 1y = 6$ in the form $y = mx + c$.	$y = -2x + 6$
20	Find the $x$-intercept of the line $y = 2x + -6$.	$x = 3$ (point $(3, 0)$)
21	Find the exact distance between $(-3, 2)$ and $(5, -4)$.	10
22	What is the gradient of the line through $(3, 2)$ and $(3, 7)$?	Undefined (vertical line)
23	Find the equation of the line through $(4, 7)$ that is parallel to the $x$-axis.	$y = 7$
24	A line passes through $(4, 6)$ and is perpendicular to $y = 2x + 1$. Find its equation.	$y = -\dfrac{1}{2}x + 8$
25	The midpoint of $AB$ is $(5, 4)$ where $A = (2, 1)$. Find $B$.	$B = (8, 7)$
26	Find the gradient of the line through $(0.5, 1.5)$ and $(2.5, 5.5)$.	$m = 2$
27	Are the points $(1, 2)$, $(3, 8)$, and $(5, 14)$ collinear?	Yes (all on $y = 3x - 1$)
28	Find the equation of the line passing through $(4, 3)$ with gradient $-\dfrac{2}{3}$.	$y = -\dfrac{2}{3}x + \dfrac{17}{3}$
29	Find the point of intersection of $y = 2x + 1$ and $y = -1x + 7$.	$(2, 5)$
30	$ABCD$ is a rectangle with $A(1, 2)$ and $B(5, 5)$. State the gradient of $CD$.	$m_{CD} = \dfrac{3}{4}$ (parallel to $AB$)
31	Find the equation of the perpendicular bisector of $A(1, 2)$ and $B(7, 10)$.	$y = -\dfrac{3}{4}x + 9$
32	A triangle has vertices $A(0, 0)$, $B(6, 0)$, $C(0, 4)$. Show that $\triangle ABC$ is right-angled at $A$.	$AB \perp AC$ since $AB$ is horizontal and $AC$ is vertical
33	Show that the triangle with vertices $A(0, 0)$, $B(6, 0)$, $C(3, 4)$ is isosceles. Find its area.	Isosceles ($AC = BC = 5$); area = 12
34	The points $(1, 3)$ and $(5, k)$ lie on a line with gradient $2$. Find $k$.	$k = 11$
35	Three vertices of a parallelogram $ABCD$ are $A(1, 1)$, $B(5, 2)$, $C(7, 6)$. Find $D$.	$D = (3, 5)$
36	In triangle $ABC$, $A(0, 0)$, $B(8, 0)$, $C(2, 6)$. Find the equation of the altitude from $C$ to $AB$.	$x = 2$
37	For what value of $k$ are the points $(1, 2)$, $(3, k)$ and $(5, 14)$ collinear?	$k = 8$
38	A point $P$ lies on the $x$-axis and is equidistant from $A(1, 4)$ and $B(7, 2)$. Find $P$.	$P = (3, 0)$
39	A line passes through $(0, 6)$ and $(4, 0)$. (a) Find its equation. (b) Find the area of the triangle formed by the line and the axes.	(a) $y = -\dfrac{3}{2}x + 6$ (b) 12
40	A point $B$ is $5$ units from $A(2, 1)$ along the line $y = 0x + 1$. Find one possible position of $B$.	$B = (7, 1)$ or $(-3, 1)$

Pack B — Answers

#	Question	Answer
1	Find the midpoint of $A(3, 5)$ and $B(9, 5)$.	$(6, 5)$
2	Find the midpoint of $C(4, 3)$ and $D(4, 11)$.	$(4, 7)$
3	Find the gradient of the line through $(2, 1)$ and $(5, 10)$.	$m = 3$
4	Find the gradient of the line through $(2, 10)$ and $(6, 2)$.	$m = -2$
5	Find the distance between $(4, 5)$ and $(12, 5)$.	8
6	State the $y$-intercept of the line $y = 2x + -4$.	$y$-intercept = $-4$ (point $(0, -4)$)
7	State the gradient of the line $y = -3x + 2$.	$m = -3$
8	A line has gradient $3$ and $y$-intercept $-1$. Write the equation of the line.	$y = 3x - 1$
9	Does the point $(4, 11)$ lie on the line $y = 3x + -1$?	Yes
10	Find $y$ when $x = 5$ on the line $y = -1x + 7$.	$y = 2$
11	Find the distance between $(2, 3)$ and $(8, 11)$.	10
12	Find the exact distance between $(0, 0)$ and $(2, 5)$.	$\sqrt{29}$
13	Find the midpoint of $(-6, 1)$ and $(4, -7)$.	$(-1, -3)$
14	Find the gradient of the line through $(-4, 2)$ and $(2, 14)$.	$m = 2$
15	A line has gradient $2$ and passes through $(4, 3)$. Find its equation in the form $y = mx + c$.	$y = 2x - 5$
16	Find the equation of the line through $(2, 1)$ and $(5, 10)$.	$y = 3x - 5$
17	A line is parallel to $y = -2x + 3$. State its gradient.	Gradient = $-2$
18	A line is perpendicular to $y = 3x + -1$. State its gradient.	Gradient = $-\dfrac{1}{3}$
19	Rewrite the equation $3x + 2y = 12$ in the form $y = mx + c$.	$y = -\dfrac{3}{2}x + 6$
20	Find the $x$-intercept of the line $y = 3x + -12$.	$x = 4$ (point $(4, 0)$)
21	Find the exact distance between $(-1, -2)$ and $(5, 6)$.	10
22	What is the gradient of the line through $(5, 2)$ and $(5, 7)$?	Undefined (vertical line)
23	Find the equation of the line through $(2, -3)$ that is parallel to the $x$-axis.	$y = -3$
24	A line passes through $(6, 5)$ and is perpendicular to $y = 3x + -2$. Find its equation.	$y = -\dfrac{1}{3}x + 7$
25	The midpoint of $AB$ is $(3, 6)$ where $A = (-1, 2)$. Find $B$.	$B = (7, 10)$
26	Find the gradient of the line through $(1.5, 2)$ and $(4.5, 11)$.	$m = 3$
27	Are the points $(1, 2)$, $(4, 5)$, and $(7, 8)$ collinear?	Yes (all on $y = x + 1$)
28	Find the equation of the line passing through $(6, 5)$ with gradient $-\dfrac{3}{4}$.	$y = -\dfrac{3}{4}x + \dfrac{19}{2}$
29	Find the point of intersection of $y = 3x + -2$ and $y = -1x + 6$.	$(2, 4)$
30	$ABCD$ is a rectangle with $A(1, 2)$ and $B(4, 8)$. State the gradient of $CD$.	$m_{CD} = 2$ (parallel to $AB$)
31	Find the equation of the perpendicular bisector of $A(1, 2)$ and $B(5, 14)$.	$y = -\dfrac{1}{3}x + 9$
32	A triangle has vertices $A(0, 0)$, $B(8, 0)$, $C(0, 5)$. Show that $\triangle ABC$ is right-angled at $A$.	$AB \perp AC$ since $AB$ is horizontal and $AC$ is vertical
33	Show that the triangle with vertices $A(0, 0)$, $B(6, 0)$, $C(3, 6)$ is isosceles. Find its area.	Isosceles ($AC = BC = 3\sqrt{5}$); area = 18
34	The points $(1, 3)$ and $(5, k)$ lie on a line with gradient $-3$. Find $k$.	$k = -9$
35	Three vertices of a parallelogram $ABCD$ are $A(1, 1)$, $B(5, 2)$, $C(6, 8)$. Find $D$.	$D = (2, 7)$
36	In triangle $ABC$, $A(0, 0)$, $B(8, 0)$, $C(2, 6)$. Find the equation of the altitude from $C$ to $AB$.	$x = 2$
37	For what value of $k$ are the points $(1, 2)$, $(3, k)$ and $(5, 14)$ collinear?	$k = 7$
38	A point $P$ lies on the $x$-axis and is equidistant from $A(1, 4)$ and $B(7, 2)$. Find $P$.	$P = (2, 0)$
39	A line passes through $(0, 8)$ and $(6, 0)$. (a) Find its equation. (b) Find the area of the triangle formed by the line and the axes.	(a) $y = -\dfrac{4}{3}x + 8$ (b) 24
40	A point $B$ is $5$ units from $A(2, 1)$ along the line $y = 0x + 1$. Find one possible position of $B$.	$B = (7, 1)$ or $(-3, 1)$

Problems — Worked Solutions

**Triangle in the plane.** Points $A(1, 2)$, $B(7, 2)$, and $C(4, 6)$ form a triangle. (a) Find the lengths $AB$, $BC$, $AC$. (b) Is the triangle isosceles, equilateral, or scalene? (c) Find the area of the triangle.

Answer

(a) $AB = 6$, $BC = 5$, $AC = 5$ (b) Isosceles (c) 12

(a) $AB = |7 - 1| = 6$ (horizontal). $BC = \sqrt{(4-7)^2 + (6-2)^2} = \sqrt{9 + 16} = 5$. $AC = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = 5$. (b) $AC = BC = 5$, $AB = 6$. Two sides equal → **isosceles**. (c) Base $AB = 6$ on line $y = 2$. Height = $|6 - 2| = 4$ (vertical distance from $C$). Area $= \frac{1}{2}(6)(4) = 12$.

**Parallelogram.** Three vertices of a parallelogram are $A(1, 1)$, $B(5, 3)$, $C(7, 7)$. (a) Find the gradient of $AB$ and $BC$. (b) The fourth vertex $D$ is such that $ABCD$ is a parallelogram (in order). Find $D$. (c) Find the perimeter of the parallelogram in exact form.

Answer

(a) $m_{AB} = 1/2$; $m_{BC} = 2$ (b) $D = (3, 5)$ (c) $8\sqrt{5}$

(a) $m_{AB} = \frac{3-1}{5-1} = \frac{1}{2}$; $m_{BC} = \frac{7-3}{7-5} = 2$. (b) In parallelogram $ABCD$ (vertices in order), $\overrightarrow{AD} = \overrightarrow{BC}$. So $D = A + (C - B) = (1, 1) + (2, 4) = (3, 5)$. (c) $|AB| = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$. $|BC| = \sqrt{4 + 16} = 2\sqrt{5}$. (Both pairs equal in parallelogram.) Perimeter $= 2(|AB| + |BC|) = 2(2\sqrt{5} + 2\sqrt{5}) = \mathbf{8\sqrt{5}}$.

**Perpendicular bisector.** Find the equation of the perpendicular bisector of the line segment joining $A(-2, 3)$ and $B(6, 7)$.

Answer

$y = -2x + 9$

Midpoint: $M = \left(\frac{-2+6}{2}, \frac{3+7}{2}\right) = (2, 5)$. Gradient of $AB$: $m_{AB} = \frac{7-3}{6-(-2)} = \frac{4}{8} = \frac{1}{2}$. Perpendicular gradient: $-2$. Line through $(2, 5)$ with gradient $-2$: $y - 5 = -2(x - 2) = -2x + 4$, so $\mathbf{y = -2x + 9}$.

**Right-angled triangle.** Show that the triangle with vertices $P(-1, -1)$, $Q(5, 1)$, $R(4, 4)$ is right-angled. At which vertex is the right angle?

Answer

Right-angled at $Q$

Compute gradients: - $m_{PQ} = \frac{1 - (-1)}{5 - (-1)} = \frac{2}{6} = \frac{1}{3}$. - $m_{QR} = \frac{4 - 1}{4 - 5} = -3$. - $m_{PR} = \frac{4 - (-1)}{4 - (-1)} = 1$. Check products: - $m_{PQ} \times m_{QR} = \frac{1}{3} \times (-3) = -1$ ✓ → perpendicular at $Q$. - $m_{PQ} \times m_{PR} = \frac{1}{3} \neq -1$. - $m_{QR} \times m_{PR} = -3 \neq -1$. So the right angle is at **$Q$**.

**Find missing coordinates.** A line passes through $(2, -1)$ and has gradient 3. (a) Find its equation. (b) Where does this line cross the $x$-axis and $y$-axis? (c) Find the area of the triangle formed by this line and the two axes.

Answer

(a) $y = 3x - 7$ (b) $x$-int $(7/3, 0)$; $y$-int $(0, -7)$ (c) $49/6 \approx 8.17$

(a) Point-slope: $y - (-1) = 3(x - 2)$, so $y + 1 = 3x - 6$, giving $y = 3x - 7$. (b) $y$-intercept: $x = 0 \Rightarrow y = -7$ → $(0, -7)$. $x$-intercept: $y = 0 \Rightarrow 3x = 7 \Rightarrow x = \frac{7}{3}$ → $(\frac{7}{3}, 0)$. (c) Triangle with legs $\frac{7}{3}$ (on $x$-axis) and 7 (on $y$-axis, taking magnitudes): Area $= \frac{1}{2} \cdot \frac{7}{3} \cdot 7 = \mathbf{\frac{49}{6} \approx 8.17}$.

**Square's diagonals.** $ABCD$ is a square with $A(0, 0)$ and $C(6, 8)$. (a) Find the midpoint of $AC$ (the centre of the square). (b) The diagonals of a square are perpendicular and equal. Find the equation of the other diagonal $BD$. (c) Hence find $B$ and $D$ (given they are symmetric about the centre, each at half the diagonal length from the centre).

Answer

(a) $(3, 4)$ (b) $y = -\frac{3}{4}x + \frac{25}{4}$ (c) $B = (7, 1)$ and $D = (-1, 7)$

(a) Midpoint of $AC$: $(\frac{0+6}{2}, \frac{0+8}{2}) = (3, 4)$. (b) Gradient of $AC$: $\frac{8}{6} = \frac{4}{3}$. Perpendicular gradient: $-\frac{3}{4}$. Through $(3, 4)$: $y - 4 = -\frac{3}{4}(x - 3)$, so $y = -\frac{3}{4}x + \frac{9}{4} + 4 = -\frac{3}{4}x + \frac{25}{4}$. (c) Diagonal $AC$ length: $\sqrt{36 + 64} = 10$. So half-diagonal = 5 from centre. $B$ and $D$ lie on line $BD$ at distance 5 from $(3, 4)$. Direction along $BD$: $(\cos\theta, \sin\theta)$ where gradient $= -\frac{3}{4}$, so direction $(\frac{4}{5}, -\frac{3}{5})$ (unit vector). Points: $B = (3 + 4, 4 - 3) = (7, 1)$; $D = (3 - 4, 4 + 3) = (-1, 7)$. Verify $|AB| = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$ — wait, side of square should be $\frac{10}{\sqrt{2}} = 5\sqrt{2}$ ✓.

**Two lines and angle.** Line $\ell_1$ has equation $y = 2x + 1$. Line $\ell_2$ passes through $(0, 5)$ and is perpendicular to $\ell_1$. (a) Find the equation of $\ell_2$. (b) Find the point of intersection of $\ell_1$ and $\ell_2$. (c) Find the distance from $(0, 5)$ to the intersection point.

Answer

(a) $y = -\frac{1}{2}x + 5$ (b) $(1.6, 4.2)$ (c) $\sqrt{3.2} \approx 1.79$

(a) Perpendicular to gradient 2 → gradient $-\frac{1}{2}$. Through $(0, 5)$: $y = -\frac{1}{2}x + 5$. (b) Set $2x + 1 = -\frac{1}{2}x + 5$, so $\frac{5}{2}x = 4$, $x = \frac{8}{5} = 1.6$. Then $y = 2(1.6) + 1 = 4.2$. Intersection: $(1.6, 4.2)$. (c) Distance from $(0, 5)$ to $(1.6, 4.2)$: $\sqrt{1.6^2 + 0.8^2} = \sqrt{2.56 + 0.64} = \sqrt{3.2} \approx \mathbf{1.79}$.

**Reflection.** A line $\ell$ has equation $y = x$. Find the image of the point $P(3, 5)$ after reflection in $\ell$.

Answer

$P' = (5, 3)$

Reflection in $y = x$ swaps $x$ and $y$ coordinates. So $P(3, 5) \to P'(5, 3)$. **Check:** midpoint of $PP'$ is $(4, 4)$, which lies on $y = x$ ✓. Also, gradient of $PP'$ is $\frac{3-5}{5-3} = -1$, perpendicular to $y = x$ (gradient 1) since $1 \times -1 = -1$ ✓.

**Distance & perimeter problem.** A rectangular field has corners at $A(0, 0)$, $B(60, 0)$, $C(60, 40)$, $D(0, 40)$ (measurements in metres). A diagonal path runs from $A$ to $C$, and another from $B$ to $D$. (a) Find the length of each diagonal. (b) Find the coordinates where the two diagonals cross. (c) A jogger runs around the perimeter once. How far does she run?

Answer

(a) Both $\sqrt{5200} \approx 72.1$ m (b) $(30, 20)$ (c) 200 m

(a) Diagonal $AC = \sqrt{60^2 + 40^2} = \sqrt{3600 + 1600} = \sqrt{5200} \approx 72.1$ m. By symmetry, $BD$ has the same length. (b) In a rectangle, diagonals bisect each other, meeting at the centre. Centre = $(\frac{0+60}{2}, \frac{0+40}{2}) = (30, 20)$. (c) Perimeter = $2(60 + 40) = 200$ m.

**System of lines.** Two lines pass through the point $(2, 5)$. One has gradient $3$ and the other has gradient $-\frac{1}{3}$. (a) Find the equation of each line. (b) Without sketching, state the angle between them and justify. (c) The first line crosses the $x$-axis at $P$ and the second crosses the $x$-axis at $Q$. Find $|PQ|$.

Answer

(a) $y = 3x - 1$ and $y = -\frac{1}{3}x + \frac{17}{3}$ (b) 90° (perpendicular) (c) $|PQ| = \frac{50}{3} \approx 16.67$

(a) Line 1: $y - 5 = 3(x - 2) \Rightarrow y = 3x - 1$. Line 2: $y - 5 = -\frac{1}{3}(x - 2) \Rightarrow y = -\frac{1}{3}x + \frac{2}{3} + 5 = -\frac{1}{3}x + \frac{17}{3}$. (b) Product of gradients: $3 \times (-\frac{1}{3}) = -1$ → **perpendicular** (90°). (c) Line 1 $x$-intercept: $0 = 3x - 1 \Rightarrow x = \frac{1}{3}$. So $P = (\frac{1}{3}, 0)$. Line 2 $x$-intercept: $0 = -\frac{1}{3}x + \frac{17}{3} \Rightarrow x = 17$. So $Q = (17, 0)$. $|PQ| = 17 - \frac{1}{3} = \frac{51-1}{3} = \frac{50}{3} \approx 16.67$.

**Coordinate proof.** Show that the quadrilateral with vertices $A(0, 0)$, $B(4, 0)$, $C(5, 3)$, $D(1, 3)$ is a parallelogram. Is it a rhombus?

Answer

Parallelogram (opposite sides parallel & equal); NOT a rhombus (sides not all equal)

**Parallelogram check.** - $\overrightarrow{AB} = (4, 0)$; $\overrightarrow{DC} = (5-1, 3-3) = (4, 0)$. So $AB$ and $DC$ are parallel and equal length. - $\overrightarrow{AD} = (1, 3)$; $\overrightarrow{BC} = (1, 3)$. Parallel and equal. Both pairs of opposite sides parallel and equal → **parallelogram** ✓. **Rhombus check.** All four sides must be equal. $|AB| = 4$; $|AD| = \sqrt{1 + 9} = \sqrt{10} \approx 3.16$. Not equal → **not a rhombus**.

**Circle through three points (informal).** Show that the three points $A(0, 5)$, $B(3, 4)$, $C(4, -3)$ are all equidistant from the point $P(0, 0)$. What is the significance of this result?

Answer

All three are at distance 5 from $P$; they lie on a circle centred at $P$ with radius 5.

Distances from $P(0, 0)$: - $|PA| = \sqrt{0 + 25} = 5$. - $|PB| = \sqrt{9 + 16} = \sqrt{25} = 5$. - $|PC| = \sqrt{16 + 9} = \sqrt{25} = 5$. All three points are exactly 5 units from $P(0, 0)$. **Significance:** the three points lie on a circle centred at the origin with radius 5, i.e. the circle $x^2 + y^2 = 25$.