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

The gradients of two perpendicular lines multiply to give \(-1\), so the gradient you want is \(-1\) divided by the gradient of \(L\).

\(L\) has gradient \(4\), so the perpendicular gradient is

\(-1 \div 4 = -\frac{1}{4}\)

Dividing \(-1\) by a fraction is the same as turning the fraction upside down and changing its sign.

\(\frac{2}{3}\) turned upside down is \(\frac{3}{2}\), and changing the sign gives

\(-\frac{3}{2}\)

Check it: \(\frac{2}{3} \times (-\frac{3}{2}) = -1\)

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

\(y + 3x = -6\)

\(y = -3x - 6\)

The gradient is \(-3\). Any line \(y = -3x + k\) is parallel to it, so for example \(y = -3x + 6\)

The line has gradient \(3\), so it is \(y = 3x + c\).

Three of the equations have gradient \(3\), so the point decides between them. Put \(x = -3\) and \(y = -2\) into \(y = 3x + c\):

\(-2 = 3 \times (-3) + c\)

\(-2 = -9 + c\)

Add \(9\) to both sides:

\(c = 7\)

So the equation is \(y = 3x + 7\)

Check the others with gradient \(3\). At \(x = -3\), the equation \(y = 3x\) gives \(y = -9\) and the equation \(y = 3x - 2\) gives \(y = -11\). Neither is \(-2\). The remaining equation has gradient \(-4\), not \(3\).

Compare the two gradients. Parallel lines have the same gradient; perpendicular gradients multiply to give \(-1\).

The first line has gradient \(2\) and the second has gradient \(2\).

Are they the same? They are equal, so the lines are parallel.

Do they multiply to \(-1\)? \(2 \times 2 = 4\). So the lines are not perpendicular.

Answer: Parallel

Two lines are perpendicular exactly when their gradients multiply to \(-1\). Reading the gradients out of the two equations gives \(-2\) and \(\frac{1}{2}\).

\(-2 \times \frac{1}{2} = -1\)

So the answer is Yes — the two lines are perpendicular.

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

The first line is already \(y = \frac{ 7 }{ 3 }x - 9\), so its gradient is \(\frac{7}{3}\).

The second line has \(9y\), so divide every term by \(9\):

\(y = \frac{ 21 }{ 9 }x - \frac{ 2 }{ 9 }\)

So the gradient of the second line is \(\frac{7}{3}\).

Compare: \(\frac{7}{3}\) and \(\frac{7}{3}\), the gradients are equal, so the lines are parallel.

Answer: Yes

First the gradient. Perpendicular gradients multiply to \(-1\), and \(L\) has gradient \(2\), so the new line has gradient

\(-1 \div 2 = -\frac{1}{2}\)

So the new line is \(y = -\frac{1}{2}x + c\) for some number \(c\). It passes through \((4, -4)\), so those coordinates fit the equation:

\(-4 = -\frac{1}{2} \times 4 + c\)

\(-4 = -2 + c\)

Add \(2\) to both sides:

\(c = -2\)

The equation is \(y = -\frac{1}{2}x - 2\)

The gradient can only be read off once the equation is in the form \(y = mx + c\), so rearrange it to make \(y\) the subject.

\(6x + 7y + 2 = 0\)

\(7y = -6x - 2\)

Now divide every term by \(7\):

\(y = -\frac{6}{7}x - \frac{2}{7}\)

The gradient is the number multiplying \(x\), so it is \(-\frac{6}{7}\).

Turn the gradient upside down and change its sign:

\(-1 \div (-\frac{6}{7}) = \frac{7}{6}\)

Check it: \(-\frac{6}{7} \times \frac{7}{6} = -1\)

Perpendicular lines have gradients that multiply to \(-1\).

\(2 \times a = -1\)

\(a = -\frac{1}{ 2 }\)

That is the negative reciprocal of 2, which is what "perpendicular" always means for gradients.