Count the squares across and up along each line.

Line A: 2 across, 4 up. Line B: 2 across, 4 down.

Line C: 1 across, 2 up. Line D: 2 across, 3 up.

Lines A and C both go 1 across for every 2 up, so they have the same gradient and are parallel.

Line B has the same steps as one of them but runs the other way, and line D has different steps (2 across for 3 up), so neither is parallel.

Each point needs \(x + y = 5\), so \(y = 5 - x\), and it must stay on the grid, where \(x\) runs from \(-2\) to \(3\) and \(y\) from \(0\) to \(5\).

\(x = 0\) gives \(y = 5\), \(x = 1\) gives \(y = 4\), \(x = 2\) gives \(y = 3\) and \(x = 3\) gives \(y = 2\).

The next whole-number \(x\) either side of these would put a point off the grid, so the points are \((0, 5)\), \((1, 4)\), \((2, 3)\) and \((3, 2)\).

Parallel lines have the same gradient, so the new line keeps the gradient \(-\frac{1}{3}\) of \(y = -\frac{1}{3}x + 1\).

A different line needs a different intercept, so the equation is \(y = -\frac{1}{3}x + k\) for any number \(k\) other than \(1\).

For example, \(y = -\frac{1}{3}x - 1\)

Parallel lines have the same gradient, and the gradient can only be read off \(y = mx + c\), so make \(y\) the subject.

\(y - 4x = -5\)

\(y = 4x - 5\)

The gradient is \(4\). Any line \(y = 4x + k\) is parallel to it, so for example \(y = 4x + 5\)

The line has gradient \(-4\), so it is \(y = -4x + c\).

Three of the equations have gradient \(-4\), so the point decides between them. Put \(x = -3\) and \(y = 6\) into \(y = -4x + c\):

\(6 = -4 \times (-3) + c\)

\(6 = 12 + c\)

Subtract \(12\) from both sides:

\(c = -6\)

So the equation is \(y = -4x - 6\)

Check the others with gradient \(-4\). At \(x = -3\), the equation \(y = -4x\) gives \(y = 12\) and the equation \(y = -4x + 6\) gives \(y = 18\). Neither is \(6\). The remaining equation has gradient \(-3\), not \(-4\).

Parallel lines have the same gradient, so find both gradients.

The first line is already \(y = \frac{ 3 }{ 5 }x - 3\), so its gradient is \(\frac{3}{5}\).

The second line has \(15y\), so divide every term by \(15\):

\(y = \frac{ -9 }{ 15 }x + \frac{ 5 }{ 15 }\)

So the gradient of the second line is \(-\frac{3}{5}\).

Compare: \(\frac{3}{5}\) and \(-\frac{3}{5}\), the gradients are different, so the lines are not parallel.

Answer: No

Parallel lines have the same gradient, so write both equations as \(y = mx + c\).

The first line is already in this form: \(y = 2x + 3\), so its gradient is \(2\).

For the second line, add \(10x\) to both sides:

\(5y = 10x + 40\)

Now divide every term by \(5\):

\(y = 2x + 8\)

The gradient of the second line is \(2\).

Both lines have gradient \(2\), so they are parallel.

Put the \(x\)-coordinate of the point into the equation of the line to find where the line is at that \(x\).

\(y = 2 \times (-1) + 3 = 1\)

The point has \(y = -2\), and \(-2\) is less than \(1\), so the point is below the line.

Answer: Below

Count the squares across and up along each line.

Line A: 2 across, 1 up. Line B: 4 across, 2 up.

Line C: 4 across, 2 down. Line D: 3 across, 2 up.

Lines A and B both go 2 across for every 1 up, so they have the same gradient and are parallel.

Line C has the same steps as one of them but runs the other way, and line D has different steps (3 across for 2 up), so neither is parallel.

Each point needs \(x + y = 8\), so \(y = 8 - x\), and it must stay on the grid, where \(x\) runs from \(-1\) to \(6\) and \(y\) from \(-2\) to \(5\).

\(x = 3\) gives \(y = 5\), \(x = 4\) gives \(y = 4\), \(x = 5\) gives \(y = 3\) and \(x = 6\) gives \(y = 2\).

The next whole-number \(x\) either side of these would put a point off the grid, so the points are \((3, 5)\), \((4, 4)\), \((5, 3)\) and \((6, 2)\).

Parallel lines have the same gradient, so the new line keeps the gradient \(-\frac{5}{3}\) of \(y = -\frac{5}{3}x + 6\).

A different line needs a different intercept, so the equation is \(y = -\frac{5}{3}x + k\) for any number \(k\) other than \(6\).

For example, \(y = -\frac{5}{3}x - 6\)

Parallel lines have the same gradient, and the gradient can only be read off \(y = mx + c\), so make \(y\) the subject.

\(y + 4x = -9\)

\(y = -4x - 9\)

The gradient is \(-4\). Any line \(y = -4x + k\) is parallel to it, so for example \(y = -4x + 9\)

The line has gradient \(-6\), so it is \(y = -6x + c\).

Three of the equations have gradient \(-6\), so the point decides between them. Put \(x = -2\) and \(y = 4\) into \(y = -6x + c\):

\(4 = -6 \times (-2) + c\)

\(4 = 12 + c\)

Subtract \(12\) from both sides:

\(c = -8\)

So the equation is \(y = -6x - 8\)

Check the others with gradient \(-6\). At \(x = -2\), the equation \(y = -6x\) gives \(y = 12\) and the equation \(y = -6x + 4\) gives \(y = 16\). Neither is \(4\). The remaining equation has gradient \(5\), not \(-6\).

Parallel lines have the same gradient, so find both gradients.

The first line is already \(y = \frac{ 4 }{ 3 }x + 2\), so its gradient is \(\frac{4}{3}\).

The second line has \(12y\), so divide every term by \(12\):

\(y = \frac{ 9 }{ 12 }x + \frac{ 2 }{ 12 }\)

So the gradient of the second line is \(\frac{3}{4}\).

Compare: \(\frac{4}{3}\) and \(\frac{3}{4}\), the gradients are different, so the lines are not parallel.

Answer: No

Parallel lines have the same gradient, so write both equations as \(y = mx + c\).

The first line is already in this form: \(y = 3x + 7\), so its gradient is \(3\).

For the second line, add \(15x\) to both sides:

\(5y = 15x + 25\)

Now divide every term by \(5\):

\(y = 3x + 5\)

The gradient of the second line is \(3\).

Both lines have gradient \(3\), so they are parallel.

Put the \(x\)-coordinate of the point into the equation of the line to find where the line is at that \(x\).

\(y = 3 \times (-3) + 7 = -2\)

The point has \(y = 2\), and \(2\) is greater than \(-2\), so the point is above the line.

Answer: Above