Where lines and graphs meet · Coordinate geometry

Where two lines cross

The point where two lines cross lies on both, so its coordinates fit both equations at once. To find it, solve the equations as simultaneous equations. On a graph, read it off where the lines cross; the algebra is a check.

Two lines crossingThe crossing point fits both equations

Two lines, one answerMove the lines

y = x + 1 and y = −2x + 4

(1, 2)where they crossx = 1, y = 2the solution

gradient of line 1 = 1line 1 cuts the y-axis at = 1gradient of line 2 = −2line 2 cuts the y-axis at = 4

Every point on a line satisfies its equation. Only the crossing point, (1, 2), is on both lines, so it's the only pair of values that makes both equations true. Check: 1 × 1 + 1 = 2.

Change the lines and watch the crossing point move. When the gradients are equal the lines are parallel and never meet.

Worked example · WAEC 2024

WAEC 2024 · Paper 2 · Q11 (a)

Find the equation of the line that passes through the origin and the point of intersection of the lines $x + 2y = 7$ and $x - y = 4$ . (Give $y$ in terms of $x$ .)

Find the crossing point
$(x + 2y) - (x - y) = 7 - 4$ , so $3y = 3$ and $y = 1$ . Then $x = 4 + 1 = 5$ : the lines cross at $(5, 1)$ .
Think first. Which letter disappears if you subtract the second equation from the first?
The gradient to the origin
$m = \frac{1 - 0}{5 - 0} = \frac15$ .
Think first. The line goes through (0, 0) and (5, 1).
The equation
$y = \frac15x$ , which can be written $x - 5y = 0$ .
Think first. A line through the origin has c = 0.

Where a line meets a curve

A line meets a curve where their $y$ -values are equal. Set the two expressions for $y$ equal and solve: with a quadratic curve this gives a quadratic equation, so there can be two meeting points, one, or none.

The same idea runs the other way. If the graph of a curve is already drawn, you can solve a new equation by drawing a straight line: rearrange the new equation so that one side is the curve’s expression. The other side is the line to draw.

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q6

The graph shows the relation of the form $y = mx^2 + nx + r$ , where $m$ , $n$ and $r$ are constants. Using the graph:

Scale: 2 cm to 2 units on the x-axis and 2 cm to 10 units on the y-axis.

State the scale used on both axes.

Find the values of $m$ , $n$ and $r$ .

Find the gradient of the line through $P$ and $Q$ .

State the range of values of $x$ for which $y > 0$ .

(a) The scales
$x$ -axis: 2 cm to 2 units; $y$ -axis: 2 cm to 10 units.
Think first. How many units does each 2 cm stand for on each axis?
(b) Use the roots
It crosses at $x = -2$ and $x = 4$ , and it has a highest point, so $y = -(x + 2)(x - 4) = -x^2 + 2x + 8$ . So $m = -1$ , $n = 2$ , $r = 8$ . (Check: the curve crosses the $y$ -axis at 8 ✓.)
Think first. Where does the curve cross the x-axis? What factors does that give?
(c) The gradient of PQ
$P(-5, -27)$ and $Q(3, 5)$ : gradient $= \frac{5 - (-27)}{3 - (-5)} = \frac{32}{8} = 4$ .
Think first. Read P and Q from the graph, then rise over run.
(d) Where y > 0
Between the roots: $-2 < x < 4$ .
Think first. Which part of the curve is above the x-axis?

Regions and loci as equations

A line splits the plane into two sides, so an inequality such as $y \le -3x + 3$ describes a region: find the boundary line, then test a point such as the origin to decide which side, as in inequalities on graphs.

y ≤ mx + cSolid boundary; the region below the line

A locus can also be written as an equation. The points $(x, y)$ equidistant from two points $A$ and $B$ lie on the perpendicular bisector of $AB$ : it passes through the midpoint of $AB$ with the perpendicular gradient, so you can find its equation with this lesson and the last. (See loci.)

Your turn

WAEC 2012 · Paper 2 · Q7 (a)

(a)
(i) Using a scale of 2 cm to 1 unit on both axes, draw on the same graph sheet the graphs of $y - \frac{3x}{4} = 3$ and $y + 2x = 6$ . (ii) From your graph, find the coordinates of the point of intersection of the two graphs. (iii) Show, on the graph sheet, the region satisfied by the inequality $y - \frac34x \ge 3$ .
Separate values with commas, e.g. 3, −2

Try it on a graph

The two lines; the shaded region is y − ¾x ≥ 3.

Worked solution (try it first)

(a)(i)

Rearrange each equation for $y$ : $y = \frac34x + 3$ and $y = 6 - 2x$ .
Plot points for each.
For $y = \frac34x + 3$ : $(-4, 0)$ , $(0, 3)$ , $(4, 6)$ .
For $y = 6 - 2x$ : $(0, 6)$ , $(1, 4)$ , $(3, 0)$ .
Join each set with a straight line.

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