Hint
Use the stated units and precision. Use the negative reciprocal of the nonzero original gradient.
Worked solution
(a)
Use the negative reciprocal of the nonzero original gradient.
\(\displaystyle M_{h}= - \frac{1}{M_{l}}= \frac{1}{4}\)
Answer: \(\displaystyle \frac{1}{4}\)
(b)
Substitute P into the perpendicular-line rule.
\(\displaystyle c_h=\frac{29}{4}\)
Answer: \(\displaystyle \frac{29}{4}\)
(c)
Set \(\displaystyle y=0\) in the new line equation.
\(\displaystyle 0=\frac{x}{4} + \frac{29}{4}\) gives \(\displaystyle x=\)\(-29\).
Answer: \(-29\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternative methods and consequential reasoning. An unexplained result does not earn a method mark.
(a) · 2 marks
- 1 mark: Use the negative reciprocal of the nonzero original 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: Keep intermediate values exact until the final step.
(b) · 2 marks
- 1 mark: Substitute P into the perpendicular-line rule.
- 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: Check the answer against the original conditions.
(c) · 2 marks
- 1 mark: Set \(\displaystyle y=0\) in the new line equation.
- 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: Keep intermediate values exact until the final step.
Total: 6 marks.