Hint
The line here has a finite nonzero gradient. Substitute A into y=mx+c.
Worked solution
(a)
\(y - 25 = x - 13\); expand if required.
Answer: \(y = x + 12\)
(b)
The perpendicular gradient is \(\displaystyle - \frac{1}{m}\); use the same point A.
Answer: \(y - 25 = 13 - x\)
(c)
Take the negative reciprocal of the nonzero gradient.
\(-1\)
Answer: \(-1\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 2 marks
- 1 mark: Use the stated point and gradient in the point-gradient equation.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Perpendicular gradients have product −1, not 1.
(b) · 2 marks
- 1 mark: Use the negative reciprocal gradient through the same point.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: The point coordinate \(\displaystyle y_{1}\) is not generally the y-intercept.
(c) · 2 marks
- 1 mark: Take the negative reciprocal of the nonzero gradient.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Perpendicular gradients have product −1, not 1.
Total: 6 marks.