Hint
Rearrange each equation for y before comparing gradients. Use P to determine each new constant.
Worked solution
(a)
Rearranging gives \(\displaystyle y= (\frac{3}{2})x- \frac{7}{2}\) and \(\displaystyle y= - (\frac{2}{3})x+ \frac{5}{6}\). The gradient product is −1, so the lines are perpendicular.
Answer: \(\displaystyle m_{l}= \frac{3}{2}\), \(\displaystyle m_{k}= - \frac{2}{3}\); product −1.
(b)
A parallel line has the form \(\displaystyle 3x-2y=c\). At \(\displaystyle P, c=3(2)-2(-1)=8\).
Answer: \(3x-2y=8\).
(c)
The perpendicular gradient is \(\displaystyle - \frac{2}{3}\). Hence \(\displaystyle 2x+3y=c\) and P gives \(\displaystyle c=4-3=1\).
Answer: \(2x+3y=1\).
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) · 3 marks
- 1 mark: Obtain gradient \(\displaystyle \frac{3}{2}\) for l.
- 1 mark: Obtain gradient \(\displaystyle - \frac{2}{3}\) for k.
- 1 mark: Use their product −1 to establish perpendicularity.
Alternative methods: A correct normal/direction-vector calculation is accepted.
Common error: Comparing the coefficients of x before dividing by the coefficient of y.
(b) · 2 marks
- 1 mark: Use the same line direction as l.
- 1 mark: Use P to obtain \(\displaystyle 3x-2y=8\).
Alternative methods: The equivalent form \(\displaystyle y= (\frac{3}{2})x- 4\) is accepted.
Common error: Keeping the original constant, which does not place the line through P.
(c) · 2 marks
- 1 mark: Use gradient \(\displaystyle - \frac{2}{3}\) or an equivalent direction.
- 1 mark: Obtain \(\displaystyle 2x+3y=1\).
Alternative methods: A point-gradient equation with correct substitution is accepted.
Common error: Using \(\displaystyle - \frac{3}{2}\) instead of the negative reciprocal.
Total: 7 marks.