Find the point of intersection of the lines shown on the cover page.
Step-by-step solution
Idea: The point of intersection is the one point lying on both lines. Read it from the graph, then check it in the equation of the labelled line. Only one line has its equation printed, so the graph reading is checked with that equation.
- Line 1 is 15x − 10y − 115 = 0 (slope 32, meeting the x-axis at x = 233). Line 2 falls gently from left to right. Its equation is not printed in the picture; from the drawing it fits x + 6y = 21, but that equation is inferred from the picture, not given.½ mark
- From the graph, the two lines cross at x = 9, y = 2.1 mark
- Check in line 1: 15(9) − 10(2) − 115 = 135 − 20 − 115 = 0 ✓. So the point of intersection is (9, 2).½ mark
Check: (9, 2) also fits the second line as read from the graph: it passes through (0, 3.5), and 9 + 6(2) = 21 = 0 + 6(3.5) ✓. (The equation x + 6y = 21 for that line is inferred from the picture.)
Answer to write in the exam
Line 1: 15x − 10y − 115 = 0
From the graph, the lines meet at x = 9, y = 2
Check: 15(9) − 10(2) − 115 = 135 − 20 − 115 = 0 ✓
∴ Point of intersection = (9, 2)
Common mistakes that cost marks
- Reading the crossing point as (2, 9): the x-coordinate comes first.
- Giving a point read from the picture without checking it in the labelled equation.
- Treating an equation for the unlabelled line as if it were printed. Only 15x − 10y − 115 = 0 is given; anything about the other line is read from the drawing.
How this can come in the exam
Which point lies on the line 15x − 10y − 115 = 0?
- (9, 2)
- (2, 9)
- (7, 1)
- (5, 4)
Show answer
(A) (9, 2)
Substitute each point: (9, 2) gives 135 − 20 − 115 = 0 ✓. (2, 9) gives −175, (7, 1) gives −20 and (5, 4) gives −80.
Try one yourself
Find where 15x − 10y = 115 meets x + 6y = 21.
Show answer
x = 21 − 6y ⇒ 15(21 − 6y) − 10y = 115 ⇒ 315 − 100y = 115 ⇒ y = 2, x = 9: (9, 2).
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