Straight Line – Part 2

Point–Gradient Formula

\[ y - b = m(x - a) \]

where \( (a, b) \) is a point on the line.

Proof

\[ \text{Let } P(x, y) \text{ be any point on the line except } A(a, b). \] \[ \text{The gradient of a line is } m = \frac{\text{y step}}{\text{x step}} = \frac{y_2 - y_1}{x_2 - x_1}, \qquad x_2 \ne x_1. \] \[ \text{So the gradient of } AP \text{ is } m = \frac{y - b}{x - a}. \] \[ \Rightarrow\; m(x - a) = y - b. \] \[ \text{When } x = a, \] \[ 0 = y - b \] \[ \Rightarrow\; y = b \quad\text{Hence the point } (a, b) \text{ is also on the line.} \] \[ \Rightarrow\; y - b = m(x - a). \]

Example

Find the equation of the line of gradient 3 which passes through the point (1,5).

\[ y - b = m(x - a) \] \[ \Rightarrow\; y - 5 = 3(x - 1) \] \[ \Rightarrow\; y - 5 = 3x - 3 \] \[ \Rightarrow\; y = 3x + 2 \]

Gradient and Tangent

\[ \text{Gradient} = \tan \theta \]

where \( \theta \) is measured anticlockwise from the x‑axis.

\[ m = \frac{V}{H} = \frac{\text{y step}}{\text{x step}} = \frac{y_2 - y_1}{x_2 - x_1} \] \[ = \frac{\text{opposite}}{\text{adjacent}} = \tan\theta \]

Proof

\[ \text{In } \triangle ABC \] \[ \sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AB} \] \[ \cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AC}{AB} \] \[ \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sin\theta \times \text{Hypotenuse}} {\cos\theta \times \text{Hypotenuse}} \] \[ \Rightarrow\; \tan\theta = \frac{\sin\theta}{\cos\theta} \] \[ \Rightarrow\; \tan\theta = \frac{BC}{AC} \] \[ \text{But } m_{ab} = \frac{y\text{ step}}{x\text{ step}} = \frac{BC}{AC} \] \[ \Rightarrow\; m_{ab} = \tan\theta \]

Example

What is the gradient of the line shown?

\[ m = \tan\theta \] \[ = \tan 60^\circ \] \[ = 1.732 \; (3\text{ d.p.}) \]

Or using exact values:

\[ m = \frac{\sqrt{3}}{1} = \sqrt{3} \]

Example

The line AB makes an angle of π /6 radians with the y‑axis. as shown in the diagram.
a) Find the exact value of the gradient of AB.
b) Given that the point H (3,5) lies on AB, find the equation of the line AB

Remember, angle is measured anticlockwise!

\[ \text{a) The angle between the } x\text{-axis and line AB is } \frac{\pi}{2} - \frac{\pi}{6} \] \[ = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]\[ m_{ab} = \frac{\pi}{3} \]

\[ \text{b) } y - b = m(x - a) \quad\text{has point } (3,5) \text{ and gradient } \frac{\pi}{3} \] \[ \Rightarrow\; y - 5 = \frac{\pi}{3}(x - 3) \] \[ \Rightarrow\; y - 5 = \frac{\pi}{3}x - \pi \] \[ \Rightarrow\; y - 5 = \pi\left(\frac{1}{3}x - 1\right) \] \[ \Rightarrow\; y = \pi\left(\frac{1}{3}x - 1\right) + 5 \]

Differentiation and gradient

Distance Formula

This formula is used to find the length of the line between two points.

\[ \text{From Pythagoras' Theorem} \] \[ AB^2 = AC^2 + BC^2 \] \[ \Rightarrow\; AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \] \[ \Rightarrow\; AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] \[ \text{This is known as the distance formula.} \]

Example

Do the points A(4,6), B(5,-1)and C(10,4) form an isosceles triangle?

An isosceles triangle has two equal sides. Use the distance formula to find their lengths.

\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] \[ \Rightarrow\; AB = \sqrt{(5 - 4)^2 + \big((-1) - 6\big)^2} \] \[ \Rightarrow\; AB = \sqrt{1^2 + (-7)^2} \] \[ \Rightarrow\; AB = \sqrt{1 + 49} \] \[ \Rightarrow\; AB = \sqrt{50} \]

\[ AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] \[ \Rightarrow\; AC = \sqrt{(10 - 4)^2 + (4 - 6)^2} \] \[ \Rightarrow\; AC = \sqrt{6^2 + (-2)^2} \] \[ \Rightarrow\; AC = \sqrt{36 + 4} \] \[ \Rightarrow\; AC = \sqrt{40} \]

\[ BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] \[ \Rightarrow\; BC = \sqrt{(10 - 5)^2 + \big(4 - (-1)\big)^2} \] \[ \Rightarrow\; BC = \sqrt{5^2 + 5^2} \] \[ \Rightarrow\; BC = \sqrt{25 + 25} \] \[ \Rightarrow\; BC = \sqrt{50} \] \[ AB = BC \quad\text{so } \triangle ABC \text{ is isosceles.} \]

Midpoint Formula

\[ \text{The midpoint of } AB \text{ is} \] \[ M\!\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] \[ \text{This is known as the midpoint formula.} \]

Example

Find the coordinates of:

G, the midpoint of BC
F, the midpoint of CD
The midpoint of AC
The midpoint of BD

What do you notice about your answers for the last two questions?

\[ \text{The midpoint of } BC \text{ is} \] \[ M\!\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] \[ \Rightarrow\; G\!\left( \frac{0 + 6}{2}, \frac{4 + 4}{2} \right) = G\!\left( \frac{6}{2}, \frac{8}{2} \right) = G(3,4) \]

\[ \text{The midpoint of } CD \text{ is} \] \[ M\!\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] \[ \text{so} \] \[ F\!\left( \frac{6 + 3}{2}, \frac{4 + 0}{2} \right) = F\!\left( \frac{9}{2}, \frac{4}{2} \right) \] \[ = F\!\left(4\tfrac{1}{2},\, 2\right) \]

\[ \text{The midpoint of } AC \text{ is} \] \[ M\!\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] \[ = \left( \frac{(-3) + 6}{2}, \frac{0 + 4}{2} \right) \] \[ = \left( \frac{3}{2}, \frac{4}{2} \right) \] \[ = \left( 1\tfrac{1}{2},\; 2 \right) \]

\[ \text{The midpoint of } BD \text{ is} \] \[ M\!\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] \[ = \left( \frac{0 + 3}{2}, \frac{4 + 0}{2} \right) \] \[ = \left( \frac{3}{2}, \frac{4}{2} \right) \] \[ = \left( 1\tfrac{1}{2},\; 2 \right) \]

The diagonals intersect at their midpoint.

Perpendicular Lines

OA is perpendicular to OB.

\[ m_{OA} = \frac{b}{a} \qquad\qquad m_{OB} = \frac{a}{-b} \] \[ m_{OA} \times m_{OB} = \frac{b}{a} \times \frac{a}{-b} = \frac{b}{-b} = -1 \]

\[ \text{If } m_1 \times m_2 = -1 \text{ for two gradients } m_1, m_2, \] \[ \text{then the two lines are perpendicular.} \] \[ \text{If } m_1 = m_2,\; \text{the lines are parallel.} \]

Example

Are the lines \(y = 3x + 7\) and \(3y + x = 54\) perpendicular?

\[ \text{Call } y = 3x + 7 \text{ equation (1)} \] \[ \text{so } m_1 = 3 \] \[ \text{Call } 3y + x = 54 \text{ equation (2)} \] \[ \Rightarrow\; y = \frac{54 - x}{3} \] \[ \Rightarrow\; y = -\frac{x}{3} + 18 \] \[ \text{so } m_2 = -\frac{1}{3} \] \[ m_1 \times m_2 = 3 \times \left(-\frac{1}{3}\right) = -1 \] \[ \text{So the two lines are perpendicular.} \]

Example

The points A(-2,-4), B(1,2) , C(-5,0) D(1,t) are plotted on a chart.

\[ \text{If } AB \perp CD,\; \text{find the value of } t. \]

Start by finding the gradient of AB. Then use this to find the gradient of CD.

\[ m_{AB} = \frac{2 - (-4)}{\,1 - (-2)\,} = \frac{6}{3} = 2 \] \[ AB \perp CD \] \[ m_{AB} \times m_{CD} = -1 \] \[ \text{so } m_{CD} = -\frac{1}{2} \]

\[ \text{Find the equation of line } CD \] \[ y = mx + c \] \[ y = -\frac{1}{2}x + c \]

\[ \text{Use known point } C(-5,0) \text{ to find } c \] \[ 0 = -\frac{1}{2} \times (-5) + c \] \[ \Rightarrow\; c = \frac{5}{2} \]

\[ y = -\frac{1}{2}x + \frac{5}{2} \] \[ y = \frac{1}{2}(-x + 5) \]

\[ \text{To find } t,\; \text{substitute } x = 1 \text{ since } D(1,t) \] \[ y = \frac{1}{2}(-1 + 5) \] \[ \Rightarrow\; y = \frac{1}{2} \times 4 \] \[ \Rightarrow\; y = 2 \] \[ \text{so } t = 2 \]

Collinearity

Points that lie on the same straight line and share a common point are collinear.

Example

Are the points A(-5,3), B(-1,0)and C(7,-8)collinear?

\[ m_{AB} = \frac{0 - 3}{\,(-1) - (-5)\,} = \frac{-3}{4} \] \[ m_{BC} = \frac{-8 - 0}{\,7 - (-1)\,} = \frac{-8}{8} = -1 \] \[ m_{AB} \ne m_{BC} \quad\text{so the points are not collinear.} \]

Example

Are the points A(-2,1), B(-1,0) and C(7,-8) collinear?

\[ m_{AB} = \frac{0 - 1}{\,(-1) - (-2)\,} = \frac{-1}{1} = -1 \] \[ m_{BC} = \frac{-8 - 0}{\,7 - (-1)\,} = \frac{-8}{8} = -1 \]

\[ m_{AB} = m_{BC} \quad\text{so yes, the points are collinear,} \] \[ \text{since they have the same gradient and share the common point } B(-1,0). \]

Showing equal gradients only proves the lines are parallel. To be collinear, they must share a common point.

General Equation of a Straight Line

\[ Ax \;+\; By \;+\; C \;=\; 0 \]

Example

\[ y = 2x + 5 \text{ written in the form } Ax + By + C = 0 \text{ is} \] \[ -2x + y - 5 = 0 \] \[ \text{with coefficients } A = -2,\; B = 1,\; C = -5 \]