Straight-Line Graphs: Question 1

Worked solution

Part (a): Find the gradient

The gradient $m$ of the line through two points $(x_1, y_1)$ and $(x_2, y_2)$ is the change in $y$ divided by the change in $x$:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Take $A(-2, \, 1)$ as $(x_1, y_1)$ and $B(4, \, 13)$ as $(x_2, y_2)$:

$m = \frac{13 - 1}{4 - (-2)} = \frac{12}{4 + 2} = \frac{12}{6} = 2$

Notice that $4 - (-2)$ becomes $4 + 2 = 6$; subtracting a negative is the same as adding.

$\boxed{m = 2}$

Tip: It does not matter which point you call “1” and which you call “2”, as long as you are consistent in the numerator and denominator. Reversing both gives $\dfrac{1 - 13}{-2 - 4} = \dfrac{-12}{-6} = 2$, the same answer.

Part (b): Equation in the form $y = mx + c$

We already have $m = 2$, so the line is:

$y = 2x + c$

To find $c$, substitute the coordinates of either point into this equation (the line passes through both). Using $A(-2, \, 1)$:

$1 = 2(-2) + c$ $1 = -4 + c$ $c = 1 + 4 = 5$

So the equation is:

$\boxed{y = 2x + 5}$

Check with the other point $B(4, \, 13)$:

$2(4) + 5 = 8 + 5 = 13 \checkmark$

This matches the $y$-coordinate of $B$, confirming the equation is correct.

Alternative method (point–gradient form)

You can substitute straight into $y - y_1 = m(x - x_1)$ using $A(-2, 1)$:

$y - 1 = 2\big(x - (-2)\big)$ $y - 1 = 2(x + 2)$ $y - 1 = 2x + 4$ $y = 2x + 5$

Both routes give the same equation.

Part (c): Where the line crosses the $y$-axis

A line crosses the $y$-axis where $x = 0$. In the equation $y = mx + c$, the constant $c$ is the $y$-intercept, so here $c = 5$ tells us the line crosses the $y$-axis at $y = 5$.

The crossing point must be written as a coordinate pair:

$\boxed{(0, \, 5)}$