3.2 Parallel and perpendicular lines — Practise

1. The decision and the reason

Answer: Both lines have gradient $7$, and the difference is $0$. Parallel lines have equal gradients, so the lines are parallel. The intercepts, $2$ and $-5$, are different, so they are not the same line.

2. The perpendicular gradient

Answer: The perpendicular gradient is the negative reciprocal: invert $\frac{5}{2}$ to get $\frac{2}{5}$, then change the sign to get $-\frac{2}{5}$. Check: $\frac{5}{2} \times \left(-\frac{2}{5}\right) = -1$, so the lines are perpendicular.

3. The equation

Answer: The given gradient is $2$, so the perpendicular gradient is $-\frac{1}{2}$. Substitute $(-1, 4)$ into $y - 4 = -\frac{1}{2}(x + 1)$: $y = -\frac{1}{2}x - \frac{1}{2} + 4 = -\frac{1}{2}x + \frac{7}{2}$. Check the point: $-\frac{1}{2} \times (-1) + \frac{7}{2} = 4$.

4. The equation

Answer: Rearrange the given line: $y = -4x + 1$, so its gradient is $-4$. A parallel line has the same gradient. Substitute $(2, -3)$ into $y - (-3) = -4(x - 2)$: $y + 3 = -4x + 8$, so $y = -4x + 5$. Check the point: $-4 \times 2 + 5 = -3$.

5. The equation

Answer: Rearrange the given line: $2y = 3x - 8$, so $y = \frac{3}{2}x - 4$ and the gradient is $\frac{3}{2}$. The perpendicular gradient is $-\frac{2}{3}$. Substitute $(5, 1)$ into $y - 1 = -\frac{2}{3}(x - 5)$: $y = -\frac{2}{3}x + \frac{10}{3} + 1 = -\frac{2}{3}x + \frac{13}{3}$. Multiplying by $3$ gives $2x + 3y - 13 = 0$.

6. The midpoint

Answer: The midpoint averages the coordinates: $M = \left(\dfrac{-3 + 9}{2}, \dfrac{7 + (-1)}{2}\right) = (3, 3)$. It is the same distance from each endpoint.

7. The test

Answer: Multiply the two gradients: $\frac{2}{5} \times \left(-\frac{5}{2}\right) = -1$. Since the product of the gradients is $-1$, the lines meet at a right angle.

8. The distance

Answer: Write the line as $3x + 4y - 22 = 0$, so $a = 3$, $b = 4$ and $c = -22$. Then the distance is $\dfrac{|3 \times 2 + 4 \times 1 - 22|}{\sqrt{3^2 + 4^2}} = \dfrac{12}{5}$. The shortest path to the line is along the perpendicular, so this is the perpendicular distance.

9. The distance

Answer: Write the line as $x + y - 19 = 0$. Then the distance is $\dfrac{|6 + 8 - 19|}{\sqrt{1^2 + 1^2}} = \dfrac{5}{\sqrt{2}} = \dfrac{5\sqrt{2}}{2}$, about $3.54$. The point $(6, 8)$ will be the centre of the lake in Section 3.3, so this number decides whether the causeway crosses the water.

10. The foot of the perpendicular

Answer: The line has gradient $-1$, so the perpendicular has gradient $1$. Through $(6, 8)$: $y - 8 = 1(x - 6)$, which is $y = x + 2$. Solving $x + y = 19$ with $y = x + 2$ gives $2x + 2 = 19$, so $x = \frac{17}{2}$ and $y = \frac{21}{2}$. The foot is $\left(\frac{17}{2}, \frac{21}{2}\right)$. It is the midpoint of the chord the line makes with the circle, which Section 3.3 proves.

11. The equation

Answer: The segment has gradient $\dfrac{9 - 1}{6 - 2} = 2$, so the perpendicular bisector has gradient $-\frac{1}{2}$. The midpoint is $\left(\frac{2 + 6}{2}, \frac{1 + 9}{2}\right) = (4, 5)$. Substituting into $y - 5 = -\frac{1}{2}(x - 4)$ gives $y = -\frac{1}{2}x + 7$. Every point on this line is the same distance from $(2, 1)$ and $(6, 9)$.

12. The line and its gradient

Answer: Rearrange the given line: $5y = 2x - 10$, so $y = \frac{2}{5}x - 2$ and its gradient is $\frac{2}{5}$. The perpendicular gradient is the negative reciprocal, $-\frac{5}{2}$. That is line (a). Working through the rearrangement first is what makes the choice clear: answering from the printed coefficients would give the wrong sign.