Straight Lines – OCR FSMQ Additional Maths

1 The Distance Between Two Points

CG1 asks you to calculate the distance between two points. The formula is simply Pythagoras' theorem applied to the horizontal and vertical gaps.

The distance formula

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

The order does not matter here. Because the differences are squared, $(x_2-x_1)^2 = (x_1-x_2)^2$. Distance is the one formula in this topic where you cannot get the sign wrong.

Worked Example 1 — A basic distance

Find the distance between $A(-2, 1)$ and $B(4, 9)$.

①$x$ difference: $4 - (-2) = 6$

②$y$ difference: $9 - 1 = 8$

③$d = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10$

Mind the double negative. $4 - (-2) = 6$, not $2$. Write the subtraction out rather than doing it in your head.

Worked Example 2 — An exact answer

Find the exact distance between $P(1, 3)$ and $Q(5, 6)$… and then between $P(1,3)$ and $R(4, 8)$.

①$PQ = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$ — a whole number.

②$PR = \sqrt{3^2 + 5^2} = \sqrt{34}$

③$34$ has no square factor, so $\sqrt{34}$ is already in simplest form. Leave it exact.

"Exact" means do not round. $\sqrt{34}$ is exact; $5.83$ is not. If the answer simplifies, though, do simplify it: $\sqrt{72} = 6\sqrt2$.

2 The Mid-Point of a Line Segment

CG2 — the mid-point

$M = \left(\dfrac{x_1+x_2}{2},\; \dfrac{y_1+y_2}{2}\right)$

In words: average the $x$-coordinates, average the $y$-coordinates. Nothing is squared and nothing is square-rooted, so this is the simplest formula in the topic.

Worked Example 3 — A mid-point with negatives

Find the mid-point of the segment joining $(-5, 8)$ and $(3, -2)$.

①$x$: $\dfrac{-5+3}{2} = \dfrac{-2}{2} = -1$

②$y$: $\dfrac{8+(-2)}{2} = \dfrac{6}{2} = 3$

③$M = (-1, 3)$

Sense-check by confirming the mid-point lies between the two given points in both coordinates: $-5 < -1 < 3$ ✓ and $-2 < 3 < 8$ ✓

Worked Example 4 — Working backwards to an endpoint

$M(4, 1)$ is the mid-point of $AB$, where $A$ is $(1, 7)$. Find $B$.

①Going from $A$ to $M$, the $x$-coordinate increases by $3$. The same step again reaches $B$.

②$x_B = 4 + 3 = 7$

③From $7$ to $1$ the $y$-coordinate falls by $6$, so $y_B = 1 - 6 = -5$.

④$B = (7, -5)$

⑤Or algebraically: $x_B = 2x_M - x_A = 8 - 1 = 7$ and $y_B = 2y_M - y_A = 2 - 7 = -5$ ✓

The reverse formula

$B = (2x_M - x_A,\; 2y_M - y_A)$

3 Gradient

Gradient

$m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{\text{change in } y}{\text{change in } x}$

Be consistent. Whichever point you call "point 2" on top, use the same one on the bottom. Mixing them reverses the sign.

Gradient	The line
Positive	Slopes upwards, left to right
Negative	Slopes downwards
Zero	Horizontal, $y = c$
Undefined	Vertical, $x = k$ — the denominator is zero

Parallel and perpendicular

Parallel: $m_1 = m_2$
Perpendicular: $m_1 \times m_2 = -1$, i.e. $m_2 = -\dfrac{1}{m_1}$

Worked Example 5 — Testing perpendicularity

Are the lines through $A(0,1)$, $B(4,3)$ and through $C(1,5)$, $D(3,1)$ perpendicular?

①$m_{AB} = \dfrac{3-1}{4-0} = \dfrac{2}{4} = \dfrac12$

②$m_{CD} = \dfrac{1-5}{3-1} = \dfrac{-4}{2} = -2$

③$\dfrac12 \times (-2) = -1$ ✓

④Yes, the lines are perpendicular.

The shortcut: to get a perpendicular gradient, turn the fraction upside down and change its sign. $\tfrac23 \to -\tfrac32$; $-4 \to \tfrac14$.

4 The Equation of a Line

Two standard forms

$y = mx + c$ — gradient $m$, $y$-intercept $c$
$y - y_1 = m(x - x_1)$ — gradient $m$ through the point $(x_1, y_1)$

Use the second form whenever you are given a point. It avoids having to find $c$ separately, and there is one fewer step to get wrong.

Worked Example 6 — A line through two points

Find the equation of the line through $(2, -1)$ and $(6, 7)$, giving your answer in the form $ax + by + c = 0$.

①$m = \dfrac{7-(-1)}{6-2} = \dfrac{8}{4} = 2$

②$y - (-1) = 2(x - 2)$

③$y + 1 = 2x - 4$

④$y = 2x - 5$

⑤In the requested form: $2x - y - 5 = 0$

⑥Check with the other point: $2(6) - 7 - 5 = 0$ ✓

Worked Example 7 — A perpendicular bisector

Find the perpendicular bisector of the segment joining $A(-1, 2)$ and $B(5, 6)$.

①Mid-point: $\left(\tfrac{-1+5}{2}, \tfrac{2+6}{2}\right) = (2, 4)$

②$m_{AB} = \dfrac{6-2}{5-(-1)} = \dfrac{4}{6} = \dfrac23$

③Perpendicular gradient: $-\dfrac32$

④$y - 4 = -\tfrac32(x - 2)$

⑤$y = -\tfrac32 x + 3 + 4 = -\tfrac32 x + 7$, or $2y + 3x = 14$.

The perpendicular bisector is the set of points equidistant from $A$ and $B$. That makes it the key to finding a circle's centre from two points on its circumference.

Worked Example 8 — Where two lines meet

Find the point of intersection of $y = 3x - 4$ and $2x + y = 11$.

①Substitute the first into the second: $2x + (3x - 4) = 11$

②$5x = 15$, so $x = 3$.

③$y = 3(3) - 4 = 5$

④They meet at $(3, 5)$.

⑤Check: $2(3) + 5 = 11$ ✓

5 Quick Reference

Distance

$\sqrt{(\Delta x)^2 + (\Delta y)^2}$.

Mid-point

Average both coordinates.

Other endpoint

$(2x_M - x_A,\; 2y_M - y_A)$.

Gradient

$\dfrac{\Delta y}{\Delta x}$, same order top and bottom.

Vertical line

$x = k$; gradient undefined.

Parallel

Equal gradients.

Perpendicular

Flip and change sign.

Best line form

$y - y_1 = m(x-x_1)$.

Bisector

Through the mid-point, perpendicular gradient.

Intersection

Solve the two equations simultaneously.

6 Practice Questions

Question 1

Find the distance between $(1, 2)$ and $(7, 10)$.

▶ Show solution

$\sqrt{6^2 + 8^2} = \sqrt{100} = 10$

Question 2

Find the exact distance between $(-3, 1)$ and $(2, 5)$.

▶ Show solution

$\Delta x = 5$, $\Delta y = 4$

$d = \sqrt{25 + 16} = \sqrt{41}$

$41$ is prime, so this cannot be simplified.

Question 3

Find the mid-point of the segment joining $(8, -3)$ and $(-2, 7)$.

▶ Show solution

$\left(\dfrac{8+(-2)}{2}, \dfrac{-3+7}{2}\right) = (3, 2)$

Question 4

$M(1, -2)$ is the mid-point of $PQ$, where $P = (-3, 4)$. Find $Q$.

▶ Show solution

$x_Q = 2(1) - (-3) = 5$

$y_Q = 2(-2) - 4 = -8$

$Q = (5, -8)$

Question 5

Find the gradient of the line through $(2, 9)$ and $(6, 1)$.

▶ Show solution

$m = \dfrac{1-9}{6-2} = \dfrac{-8}{4} = -2$

Question 6

Find the equation of the line through $(-1, 5)$ parallel to $y = 4x - 7$.

▶ Show solution

Parallel means the same gradient, $m = 4$.

$y - 5 = 4(x + 1)$

$y = 4x + 9$

Question 7

Find the equation of the line through $(6, 2)$ perpendicular to $y = \tfrac12 x + 3$.

▶ Show solution

Perpendicular gradient: $-\dfrac{1}{1/2} = -2$

$y - 2 = -2(x - 6)$

$y = -2x + 14$

Question 8

Find the point of intersection of $y = 2x + 1$ and $3x + 2y = 16$.

▶ Show solution

$3x + 2(2x+1) = 16$

$7x + 2 = 16$, so $x = 2$.

$y = 2(2) + 1 = 5$. They meet at $(2, 5)$.

Question 9

Show that $A(1,1)$, $B(4,2)$, $C(3,5)$ and $D(0,4)$ form a parallelogram, and determine whether it is a rectangle.

▶ Show solution

$m_{AB} = \dfrac{2-1}{4-1} = \dfrac13$ and $m_{DC} = \dfrac{5-4}{3-0} = \dfrac13$ — equal, so $AB \parallel DC$.

$m_{AD} = \dfrac{4-1}{0-1} = -3$ and $m_{BC} = \dfrac{5-2}{3-4} = -3$ — equal, so $AD \parallel BC$.

Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram.

Now test the corner: $\tfrac13 \times (-3) = -1$, so adjacent sides are perpendicular.

Therefore it is also a rectangle. (In fact $AB = \sqrt{10}$ and $AD = \sqrt{10}$, so it is a square.)

Question 10

$A(-2, 1)$, $B(6, 5)$ and $C(4, 9)$ are three points.

(a) Find the lengths of all three sides, in exact form. (b) Show that the triangle is right-angled, and state at which vertex. (c) Find the area of the triangle. (d) Find the equation of the perpendicular bisector of $AB$, and verify that $C$ does not lie on it.

▶ Show solution

(a) $AB = \sqrt{8^2 + 4^2} = \sqrt{80} = 4\sqrt5$

$BC = \sqrt{(4-6)^2 + (9-5)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt5$

$AC = \sqrt{(4+2)^2 + (9-1)^2} = \sqrt{36+64} = \sqrt{100} = 10$

(b) $AB^2 + BC^2 = 80 + 20 = 100 = AC^2$

By the converse of Pythagoras' theorem the triangle is right-angled, at $B$ (the vertex opposite the longest side $AC$).

Confirming with gradients: $m_{AB} = \tfrac{4}{8} = \tfrac12$, $m_{BC} = \tfrac{4}{-2} = -2$, and $\tfrac12 \times (-2) = -1$ ✓

(c) The right angle is at $B$, so $AB$ and $BC$ are the base and height.

Area $= \tfrac12 \times 4\sqrt5 \times 2\sqrt5 = \tfrac12 \times 8 \times 5 = 20$ square units.

(d) Mid-point of $AB$: $\left(\tfrac{-2+6}{2}, \tfrac{1+5}{2}\right) = (2, 3)$

$m_{AB} = \tfrac12$, so the bisector has gradient $-2$:

$y - 3 = -2(x - 2)$, giving $y = -2x + 7$.

Testing $C(4, 9)$: $-2(4) + 7 = -1 \neq 9$, so $C$ does not lie on the bisector.

This tells us $C$ is not equidistant from $A$ and $B$ — consistent with part (a), where $AC = 10$ but $BC = 2\sqrt5 \approx 4.47$.

Straight Lines (CG1–CG2) · OCR FSMQ Additional Maths · Created with MathJax