Coordinates and straight-line graphs

(a) A is 2 across and 3 up: $(2, 3)$.

(b) B is on the y-axis, 1 below the origin: $(0, -1)$.

(c) Go 2 left, then 1 up, and mark C there.

(a) Go across to $-2$, then down to $-1$: $(-2, -1)$

(b) Go 2 right and 3 up and mark B.

(c) Every point on $x = -4$ has $x$-coordinate $-4$, so draw the vertical line through $-4$ on the $x$-axis.

(a) A is 3 right and 1 up: $(3, 1)$.

(b) Go 3 right and 4 down and mark B there.

(c) C is $(-3, 3)$. Midpoint $= \left(\frac{3 + (-3)}{2}, \frac{1 + 3}{2}\right) = (0, 2)$

(a) $B$ is 5 across and 2 up: $(5, 2)$.

(b) Go 4 right and 2 down, and mark $C$.

(c) $A = (-3, 4)$. Midpoint $= \left(\frac{-3+5}{2}, \frac{4+2}{2}\right) = (1, 3)$

(d) $y = -4$ is the horizontal line through every point with $y$-coordinate $-4$.

(a) A is 1 left of O and 2 up, so A is $(-1, 2)$.

(b) Go 1 right, then 4 up, and mark B.

(c) $y = -3$ is the horizontal line where every point has y-coordinate −3.

(a) Go 3 right from $O$ then 2 up, and mark $A$.

(b) $B = (-3, 2)$ and $C = (1, -2)$.

Midpoint $= \left(\frac{-3 + 1}{2}, \frac{2 + (-2)}{2}\right) = (-1, 0)$

(a) A is 2 across and 1 up: $(2, 1)$.

(b) B is $(-2, -5)$. Midpoint $= \left(\frac{2 + (-2)}{2}, \frac{1 + (-5)}{2}\right) = (0, -2)$

(c) Go 4 left from $O$, then 2 up, and mark a cross labelled C.

Make a table using $y = 3x - 2$:

| $x$ | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| $y$ | −8 | −5 | −2 | 1 | 4 | 7 |

Plot the points and join them with a ruler from $x = -2$ to $x = 3$.

$P = (-3, -2)$ and $Q = (2, 4)$.

$x$: $\frac{-3 + 2}{2} = -0.5$

$y$: $\frac{-2 + 4}{2} = 1$

Midpoint $(-0.5, 1)$

Make a table of values using $y = 4 - x$:

$x$: $-2, -1, 0, 1, 2, 3, 4$

$y$: $6, 5, 4, 3, 2, 1, 0$

Plot the points and join them with a ruler from $(-2, 6)$ to $(4, 0)$.

(a) Put each $x$ into $4x - 6$:

$x=-1$: $-4-6=-10$; $x=0$: $-6$; $x=2$: $2$; $x=3$: $6$

(b) Plot the six points from $(-1,-10)$ to $(4,10)$ and join them with a single ruled straight line.

(a) C is on the $y$-axis at $-1$: $(0, -1)$

(b) From B to A is 3 right, 1 up. Do the same from C: D is at $(3, 0)$

(c) Midpoint of B$(-1, 2)$ and C$(0, -1)$: $\left(\frac{-1+0}{2}, \frac{2+(-1)}{2}\right) = (-0.5, 0.5)$

Make a table of values using $y = 4x - 1$:

| $x$ | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| $y$ | −9 | −5 | −1 | 3 | 7 |

Plot the points and join them with one straight ruled line from $x = -2$ to $x = 2$.

Make a table using $y = 2x - 2$:

| $x$ | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| $y$ | −6 | −4 | −2 | 0 | 2 | 4 |

Plot the points and join them with a ruler from $x = -2$ to $x = 3$.

Rewrite as $y = -4x + 3$, which is $y = mx + c$.

(i) Gradient $m = -4$

(ii) The intercept $c = 3$, so it crosses at $(0, 3)$

In $y = mx + c$ the gradient is $m$.

For $y = 2x + 3$, the gradient is 2.

Make a table of values using $y = 2x - 3$:

| $x$ | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| $y$ | −7 | −5 | −3 | −1 | 1 | 3 | 5 |

Plot the points and join them with a single ruled line from $(-2,-7)$ to $(4,5)$.

Make a table with $y = 1 - 4x$:

$x$: −3, −2, −1, 0, 1, 2, 3

$y$: 13, 9, 5, 1, −3, −7, −11

Plot the points and join them with a ruler.

$y = 2$ is a horizontal line through 2 on the $y$-axis: graph D.

$y = x$ goes through the origin sloping upwards: graph F.

$x + y = 2$ gives $y = 2 - x$, crossing both axes at 2 and sloping down: graph A.

The line crosses the $y$-axis at $(0, 1)$, so $c = 1$.

It also passes through $(2, 5)$. Gradient $= \frac{5 - 1}{2 - 0} = 2$

So $y = 2x + 1$

The line crosses the $y$-axis at 3, so $c = 3$.

It passes through $(0, 3)$ and $(1, 1)$: $y$ goes down 2 for each 1 across, so $m = -2$.

$y = -2x + 3$

The line crosses the $y$-axis at 3, so $c = 3$.

From (0, 3) to (1, 1): down 2 for every 1 across, so gradient $= -2$.

Equation: $y = -2x + 3$

Rearrange $L_2$: $5y = 10x - 4$, so $y = 2x - \frac{4}{5}$

$L_1$: $y = 2x + 3$ has gradient 2. $L_2$ has gradient 2.

Equal gradients, so the lines are parallel.

Divide $6y = 3x + 7$ by 6: $y = \frac{1}{2}x + \frac{7}{6}$

The first line is $y = \frac{1}{2}x - 6$.

Both gradients are $\frac{1}{2}$, so the lines are parallel. Oscar is correct.

Width $= 38 - 6 = 32$, covering four square widths, so each side is $32 \div 4 = 8$.

C is two sides right of A: $x = 6 + 2 \times 8 = 22$

C is two sides below B: $y = 36 - 2 \times 8 = 20$

So $C = (22, 20)$

Total width $= 38 - 6 = 32$. The four squares sit side by side across, so each side is $32 \div 4 = 8$.

C is two squares to the right of A: $x = 6 + 2 \times 8 = 22$

C is two squares below the top of B: $y = 36 - 2 \times 8 = 20$

So C is $(22, 20)$.

Gradient $= \frac{\text{change in } y}{\text{change in } x}$

Change in $y = 15 - 9 = 6$, so change in $x = 6 \div 3 = 2$

$d = 5 + 2 = 7$

The line crosses the $y$-axis at $-6$, so $c = -6$.

From $(0, -6)$ to $(2, 0)$: up 6, across 2, so gradient $= 6 \div 2 = 3$.

$y = 3x - 6$

The line crosses the y-axis at 1, so $c = 1$.

It passes through (0, 1) and (2, 5): up 4 for 2 across, so the gradient is $4 \div 2 = 2$.

$y = 2x + 1$

Rearrange each into $y = mx + c$:

A: $y = 2x + 4$ (gradient 2)

B: $y = \frac{1}{2}x + 2$ (gradient $\frac{1}{2}$)

C: $y = -x + 2$ (gradient −1)

D: $y = 2x - 4$ (gradient 2)

A and D have the same gradient, so they are parallel.

Rewrite each line as $y = mx + c$:

A: $y = 2x + 4$ (gradient 2)

B: $y = \frac{1}{2}x + 2$ (gradient $\frac{1}{2}$)

C: $y = -x + 2$ (gradient $-1$)

D: $y = 2x - 4$ (gradient 2)

Parallel lines have equal gradients, so A and D.

(a) The line crosses the $y$-axis at 3.

From $(0, 3)$ to $(2, 6)$: gradient $= \frac{6-3}{2-0} = \frac{3}{2}$

$y = \frac{3}{2}x + 3$

(b) Parallel lines have the same gradient, e.g. $y = 5x + 1$.

Rearrange: $y = \frac{3}{2}x - \frac{7}{2}$, so the gradient of $L$ is $\frac{3}{2}$

Perpendicular gradient: $-\frac{2}{3}$

$-5 = -\frac{2}{3} \times 6 + c = -4 + c$, so $c = -1$

$y = -\frac{2}{3}x - 1$

Gradient of $L_1$ is 3, so the perpendicular gradient is $-\frac{1}{3}$.

$y = -\frac{1}{3}x + c$ through $(9, 5)$: $5 = -3 + c$, so $c = 8$.

$L_2$: $y = -\frac{1}{3}x + 8$

$3y = 4x + 7$ gives $y = \frac{4}{3}x + \frac{7}{3}$, gradient $\frac{4}{3}$.

Perpendicular gradient $= -\frac{3}{4}$.

Through $(3, -5)$: $-5 = -\frac{3}{4}(3) + c$, so $c = -5 + \frac{9}{4} = -\frac{11}{4}$.

$y = -\frac{3}{4}x - \frac{11}{4}$

$L_1$ has gradient 2, so a perpendicular line has gradient $-\frac{1}{2}$

Rearrange $L_2$: $6y = 12 - kx$, so $y = 2 - \frac{k}{6}x$, with gradient $-\frac{k}{6}$

$-\frac{k}{6} = -\frac{1}{2}$ gives $k = 3$

(a) From P to Q you go 6 right and 3 up. QR is $\frac{3}{2}$ of PQ, so from Q go $9$ right and $4.5$ up.

$R = (0 + 9, 3 + 4.5) = (9, 7.5)$.

(b) Gradient of L $= \frac{3}{6} = \frac{1}{2}$. Perpendicular gradient $= -2$.

The line passes through $(0, 3)$, so $y = -2x + 3$.

Gradient of L1 $= \frac{2 - 6}{12 - 4} = -\frac{1}{2}$

$6 = -\frac{1}{2}(4) + c$, so $c = 8$: L1 is $y = -\frac{1}{2}x + 8$

L2 is $y = -3x$.

$-\frac{1}{2}x + 8 = -3x \Rightarrow 2.5x = -8 \Rightarrow x = -3.2$, $y = 9.6$

$P = (-3.2, 9.6)$

Gradient of OB $= \frac{-4}{2} = -2$

DE is parallel to OB through $(12, -6.5)$: $y = -2x + c$, $-6.5 = -24 + c$, so $y = -2x + 17.5$

OD is perpendicular to OB through O: gradient $\frac{1}{2}$, so $y = \frac{1}{2}x$

$\frac{1}{2}x = -2x + 17.5 \Rightarrow 2.5x = 17.5 \Rightarrow x = 7$, $y = 3.5$

$D = (7, 3.5)$

L: $2y = 17 - 3x$, so $y = 8.5 - 1.5x$. Gradient $-\frac{3}{2}$ and $B = (0, 8.5)$.

M has gradient $\frac{2}{3}$ through $(0, 2)$: $y = \frac{2}{3}x + 2$.

Solve: $\frac{2}{3}x + 2 = 8.5 - 1.5x \Rightarrow \frac{13}{6}x = 6.5 \Rightarrow x = 3$, so $C = (3, 4)$.

Base $AB = 8.5 - 2 = 6.5$ on the y-axis, height = 3.

Area $= \frac{1}{2} \times 6.5 \times 3 = 9.75$

$3x + 2y = 7 \Rightarrow y = -\frac{3}{2}x + \frac{7}{2}$, gradient $-\frac{3}{2}$

PQ is perpendicular, so its gradient is $\frac{2}{3}$

$\frac{b - 4}{a - 3} = \frac{2}{3} \Rightarrow b - 4 = \frac{2}{3}(a - 3) = \frac{2}{3}a - 2$

$b = \frac{2}{3}a + 2$

$A$ is $(-2, 0)$ and $B$ is $(0, 4)$, so the gradient of $AB = \frac{4}{2} = 2$.

Perpendicular gradient $= -\frac{1}{2}$.

Through $C(5, -1)$: $-1 = -\frac{1}{2}(5) + c$, so $c = \frac{3}{2}$.

$y = -\frac{1}{2}x + \frac{3}{2}$

Gradient $PR = \frac{k - 2}{k + 4}$, gradient $QR = \frac{k + 9}{k - 6}$

Perpendicular: $\frac{k-2}{k+4} \times \frac{k+9}{k-6} = -1$

$(k-2)(k+9) = -(k+4)(k-6)$

$k^2 + 7k - 18 = -k^2 + 2k + 24 \Rightarrow 2k^2 + 5k - 42 = 0$

$(k + 6)(2k - 7) = 0$, so $k = -6$ or $k = 3.5$