SL 2.1—Equations of straight lines, parallel and perpendicular

The equation of $L_1$ is $y=4x-3$, which is in gradient-intercept form $y=mx+c$. The gradient is $m=4$.

Parallel lines have the same gradient, so $L_2$ also has gradient $m=4$.

Substituting into $y-y_1=m(x-x_1)$: $y-5=4(x-1)$ $y-5=4x-4$ $y=4x+1$

The equation $y=4x+1$ shows a $y$-intercept of $1$. We can verify: $5eq4(1)-3=1$, confirming $Q$ is not on $L_1$, so a distinct parallel line exists.

We need the $y$-intercept of a line parallel to $y=4x-3$ through $(1,5)$. The parallel line must have gradient $4$ and satisfy $5=4(1)+c$.

The option $-3$ is the $y$-intercept of $L_1$ itself. A line with $y$-intercept $-3$ and gradient $4$ would be identical to $L_1$, not a distinct parallel line.

The option $5$ is just the $y$-coordinate of point $Q$. This would give $y=4x+5$. Checking: $4(1)+5=9eq5$, so point $Q$ does not lie on this line.

The option $4$ is the gradient of $L_1$, not a $y$-intercept value that works. Checking: $4(1)+4=8eq5$, so $Q$ does not lie on $y=4x+4$.

Solving $5=4(1)+c$ gives $c=1$. Checking: $4(1)+1=5$ ✓. The correct answer is $1$.

From $y=-3x+2$, the gradient is $m_1=-3$.

The perpendicular gradient is $m_2=-\frac{1}{m_1}=-\frac{1}{-3}=\frac{1}{3}$. Check: $(-3)\times \frac{1}{3}=-1$ ✓

$y-0=\frac{1}{3}(x-6)$ $y=\frac{1}{3}x-2$

The equation of $L_2$ is $y=\frac{1}{3}x-2$. Verify: when $x=6$, $y=\frac{1}{3}(6)-2=0$ ✓

The correct line must have gradient $\frac{1}{3}$ (perpendicular to gradient $-3$) and pass through $(6,0)$.

This has gradient $-3$, the same as $L_1$. A line through $(6,0)$ with $m=-3$ would be parallel to $L_1$, not perpendicular.

This has gradient $3$, not $\frac{1}{3}$. The product $(-3)(3)=-9\ne -1$, so this line is not perpendicular to $L_1$.

This has the correct gradient $\frac{1}{3}$, but checking the point $(6,0)$: $\frac{1}{3}(6)+6=8\ne 0$. This line does not pass through $(6,0)$.

This has gradient $\frac{1}{3}$ and passes through $(6,0)$: $\frac{1}{3}(6)-2=0$ ✓. This is the correct answer.