The gradient of a line including the connection to the angle it makes with the x-axis, the equation of a line through a point with a given gradient, parallel and perpendicular lines, and the medians, altitudes and perpendicular bisectors of a triangle.

What this dot point is asking

The SQA wants you to find a gradient and connect it to the angle a line makes with the positive direction of the x-axis, write the equation of a line through a point with a given gradient, recognise parallel and perpendicular lines, and use these tools to find the medians, altitudes and perpendicular bisectors of a triangle.

Gradient and angle

The gradient between two points is $m = \dfrac{y_2 - y_1}{x_2 - x_1}$. It is linked to the angle $\theta$ the line makes with the positive direction of the x-axis by $m = \tan\theta$. A positive gradient gives an acute angle; a negative gradient gives an obtuse angle. To go from a gradient back to an angle, use $\theta = \tan^{-1} m$, but add $180^\circ$ when the gradient is negative so the angle is measured correctly between $0^\circ$ and $180^\circ$.

The equation of a line

The point-gradient form $y - b = m(x - a)$ is the workhorse: every line problem in the Higher reduces to finding a gradient and a point, then substituting. You can leave the answer in this form or rearrange to $y = mx + c$ or $ax + by + c = 0$ as asked.

Lines inside a triangle

The strategy is identical for all three: identify the point the line passes through and the gradient it needs, then use point-gradient form. A median needs the midpoint of the opposite side (use the midpoint formula) and the gradient from the vertex to that midpoint. An altitude and a perpendicular bisector both need the perpendicular gradient (negative reciprocal of the side's gradient); they differ only in the point they pass through.

Examples in context

Coordinate geometry locates a point equidistant from several others, useful in planning. If two transmitter masts stand at $A(0, 0)$ and $B(8, 4)$, any receiver equally distant from both lies on the perpendicular bisector of $AB$. Its midpoint is $(4, 2)$ and the gradient of $AB$ is $\dfrac{1}{2}$, so the bisector has gradient $-2$ and equation $y - 2 = -2(x - 4)$. Every point on this line is a valid equidistant siting position.

Try this

Q1. Find the gradient of the line making an angle of $60^\circ$ with the positive x-axis. [2 marks]

• Cue. $m = \tan 60^\circ = \sqrt{3}$.

Q2. Find the equation of the line through $(2, -1)$ perpendicular to a line of gradient $4$. [3 marks]

• Cue. Perpendicular gradient $-\frac{1}{4}$, giving $y + 1 = -\frac{1}{4}(x - 2)$.

Q3. Find the equation of the median from $A(0, 0)$ to the midpoint of $BC$, where $B(2, 6)$ and $C(8, 2)$. [3 marks]

• Cue. Midpoint of $BC$ is $(5, 4)$; gradient $\dfrac{4}{5}$; equation $y = \dfrac{4}{5}x$.