The line meets the \(y\)-axis where \(x = 0\), which gives \(y = 9\).

The point is \((0, 9)\).

The line meets the \(x\)-axis where \(y = 0\), so put \(y = 0\) into the equation:

\(0 = -3x + 9\)

Subtract \(9\) from both sides:

\(-9 = -3x\)

Divide both sides by the number in front of x:

\(x = 3\)

The point is \((3, 0)\).

Put the \(x\)-coordinate of the point into the equation and see what \(y\) the line gives there. When \(x = -2\),

\(y = 6 - 4 = 2\)

So at \(x = -2\) the line is at \(y = 2\), and the point is at \(y = 2\). Those are the same, so the point does lie on the line.

Answer: Yes

\(L\) goes through the origin, so it crosses the \(y\)-axis at \(0\) and \(c = 0\). There is no number to add on the end.

One square to the right takes the line \(1\) up, so \(m = 1\).

The equation is \(y = x\)

Substitute \(x = -2\) into \(y = x\):

\(s = -2\)

You can check it on the graph: at \(x = -2\) the line is level with \(y = -2\).

Start with \(c\), the height at which the line crosses the \(y\)-axis. Here that is \(-1\), so \(c = -1\).

Now the gradient. Go one square to the right and see how far the line goes up: it moves \(1\), and the line slopes upwards, so \(m = 1\).

The equation is \(y = x - 1\)

\(L\) crosses the \(y\)-axis at \(3\), so \(c = 3\).

One square to the right takes the line \(1\) up, so \(m = 1\).

\(y = x + 3\)

Parallel lines have the same gradient, so the new line also has \(m = 1\).

It crosses the \(y\)-axis at \(-3\), so \(c = -3\).

\(y = x - 3\)

Gradient first — the change in \(y\) divided by the change in \(x\):

\(m = \frac{ 8 - (-7) }{ 1 - (-4) } = \frac{ 15 }{ 5 } = 3\)

So the line is \(y = 3x + c\) for some number \(c\).

Substitute one of the points, say \((-4, -7)\):

\(-7 = -12 + c\)

\(c = 5\)

The equation is \(y = 3x + 5\)

The gradient is given, so the line is \(y = -3x + c\) and only \(c\) is unknown.

The point \((-5, 21)\) is on the line, so its coordinates fit the equation. Put \(x = -5\) and \(y = 21\) into it:

\(21 = 15 + c\)

\(c = 6\)

The equation is \(y = -3x + 6\)

The equation is not in the form \(y = mx + c\) yet, because there is a \(4\) in front of the \(y\).

Divide every term by \(4\):

\(16 \div 4 = 4\) and \(-24 \div 4 = -6\)

\(y = 4x - 6\)

Once the equation is in the form \(y = mx + c\), the gradient is the number in front of \(x\):

\(m = 4\)

Reading \(16\) straight off the original equation is the trap: \(16\) is the gradient of \(y = 16x - 24\), which is a different line.

Take \(x\) away from both sides, so that \(y\) is on its own:

\(y = -5 - x\)

Written with the \(x\) term first, that is

\(y = -x - 5\)

The number in front of \(x\) is \(-1\), so the gradient is \(-1\).

The line slopes downwards, which is the check: a negative gradient always does.

The gradient is the change in \(y\) divided by the change in \(x\):

\(\frac{ -27 - (-19) }{ 5 - 3 } = \frac{ -8 }{ 2 } = -4\)

Find the equation of \(AB\). The gradient is \(-4\), so \(y = -4x + c\).

Substitute \(A\) to find \(c\):

\(-19 = -4 \times 3 + c\)

\(-19 = -12 + c\)

\(c = -7\)

So the line is \(y = -4x - 7\).

Put in the \(x\)-coordinate of \(C\):

\(y = -4 \times 255 - 7 = -1020 - 7 = -1027\)

The line is at \(y = -1027\) when \(x = 255\), and \(C\) is at \(y = -1027\). These are the same, so \(C\) lies on the line.

Alternative: the gradient of \(AC\) is \(\frac{ -1008 }{ 252 }\), which is equal to \(-4\), so \(C\) lies on the line.

Answer: Yes