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The straight line through two points will have an equation in the form \(y = mx + c\).

We can find the value of \(m\), the gradient of the line, by forming a right-angled triangle using the coordinates of the two points.

Then, we can find the value of \(c\), the \(y\)-intercept, by substituting the coordinates of one point into the equation.

Sketch the two points and join them with a straight line. Draw a right-angled triangle to show the difference in the \(x\)-coordinates and the difference in the \(y\)-coordinates.

First, find the gradient of the line, \(m\). In the \(x\)-direction, the difference between 3 and −1 is equal to 3 – (−1) = 4. In the \(y\)-direction, the difference between 11 and 3 is equal to 11 – 3 = 8.

The gradient of the line through the points (−1, 3) and (3, 11) is given by \(\frac{\text{change in y}}{\text{change in x}} = \frac{8}{4} = 2\).

This is the value of \(m\) in the equation of the line, so the equation will be \(y = 2x + c\).

Next, find the value of \(c\). Using the \(x\) and \(y\) values from the point (3, 11), substitute into the equation.

\(y=2x+c\)

\(11 = 2 \times 3+c\)

\(11=6+c\)

\(c=5\)

So the equation of the line through the points (-1, 3) and (3, 11) is \(y = 2x + 5\).

Finally, check by substituting the \(x\)-coordinate of the other point into this equation:

\(y=2x+5\)

\(x=-1\)

\(y=2 \times(-1)+5= -2+5=3\)

This gives the correct \(y\)-coordinate for the point (−1, 3), so the equation is correct.