Hint
Find the gradient, then determine the constant using either point. Keep coordinate pairs together.
Worked solution
(a)
The gradient is (−1−5)/(4−(−2))=−1. Substitution into y=−x+c gives c=3, so x+y=3.
Answer: \(x+y=3\).
(b)
Setting y=0 gives (3,0); setting x=0 gives (0,3). The right triangle with O has area (1/2)(3)(3)=9/2 square coordinate units.
Answer: \((3,0),(0,3)\); area \(9/2\).
(c)
Substitute into x+y=3: x+2x+6=3, so x=−1 and y=4. Both line equations are satisfied.
Answer: \((-1,4)\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Obtain gradient −1.
- 1 mark: Use either point to obtain x+y=3.
Alternative methods: A determinant or two-point line equation is accepted.
Common error: Using a gradient from unmatched coordinate differences.
(b) · 3 marks
- 1 mark: Find both intercept coordinates.
- 1 mark: Use one half times the two axis lengths.
- 1 mark: Obtain area 9/2 square coordinate units.
Alternative methods: A coordinate-area formula is accepted.
Common error: Using the sloping segment as both base and height.
(c) · 2 marks
- 1 mark: Solve a correct simultaneous system.
- 1 mark: Give the matched point (−1,4).
Alternative methods: Elimination instead of substitution is accepted.
Common error: Stopping after finding x or pairing it with the wrong y-value.
Total: 7 marks.