1.3.1 Equation of a Straight Line

In this topic we will learn how to:

• find the equation of a straight line given sufficient information
• interpret and use any of the forms $y = mx + c$, $y - y_{1} = m(x - x_{1})$, $ax + by + c = 0$ in solving problems
• Calculate distances, gradients, midpoints, points of intersection and use the relationship between the gradients of parallel and perpendicular lines.

A straight line is one that can be written in the form,
$$y = mx + c$$To be able to find the equation of a straight line, you need to find the gradient, $m$, and the y-intercept, $c$. To do that you either need two points that lie on the line or one point and the gradient. The formula for find the gradient is,
$$m = \frac{y_{1} - y_{2}}{x_{1} - x_{2}}$$The equation of a straight line can be written in many forms. The most common being,
$$y = mx + c$$Other common forms are,
$$ax + by + c = 0$$Where $a$, $b$ and $c$ are constants.

And,
$$y - y_{1} = m(x - x_{1})$$Where $m$ is the gradient and $x_{1}$ and $y_{1}$ represent the a point on the line $(x_{1}, y_{1})$.$$\textbf{\textcolor{gray}{Length of a Line}}$$To find the length of a line $AB$ use the formula,
$$AB = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$$$$\textbf{\textcolor{gray}{Midpoint}}$$To find the midpoint of a line use the formula,
$$M = \left(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}\right)$$$$\textbf{\textcolor{gray}{Parallel Lines}}$$Parallel lines are lines that never meet. They have the same gradient.$$\textbf{\textcolor{gray}{Perpendicular Lines}}$$Perpendicular lines are lines that intersect each other at $90^{\circ}$. Their gradients are negative reciprocals of each other i.e the product of their gradients is equal to $-1$,
$$m_{1} \times m_{2} = -1$$Let’s walk through some past paper questions to further understand.

1. The point $M$ is the mid-point of the line joining the points $(3, 7)$ and $(-1, 1)$. Find the equation of the line through $M$ which is parallel to the line $\frac{x}{3} + \frac{y}{2} = 1$. (9709/12/O/N/19 number 2)

Let’s call the line whose equation we’re looking for, $l$.

To find the equation of $l$ we can find one point and the gradient. We can get the gradient from the line parallel to $l$. $l$ passes through the mid-point, $M$, we can use this as our point.

Let’s find the coordinates of the midpoint,
$$M = \left(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}\right)$$$$M = \left(\frac{3 + (-1)}{2}, \frac{7 + 1}{2}\right)$$$$M = (1, 4)$$Therefore, $l$ passes through,
$$M(1, 4)$$To find the gradient of $l$, let’s use the equation of the line parallel to $l$,
$$\frac{x}{3} + \frac{y}{2} = 1$$$$\frac{1}{3}x + \frac{1}{2}y = 1$$Let’s write it in the form $y = mx + c$ by making $y$ the subject of the formula,
$$\frac{1}{3}x + \frac{1}{2}y = 1$$$$\frac{1}{2}y = -\frac{1}{3}x + 1$$$$y = -\frac{2}{3}x + 2$$Therefore, the gradient of $l$ is,
$$m = -\frac{2}{3}$$Now that we have a point and the gradient, let’s find the equation of $l$,
$$m = -\frac{2}{3} \textmd{ and the line passes through } M(1, 4)$$$$y - y_{1} = m(x - x_{1})$$$$y - 4 = -\frac{2}{3}(x - 1)$$You can leave it here, but it’s advised you simplify it to the form $ax + by + c = 0$ unless told otherwise,
$$y - 4 = -\frac{2}{3}\left(x - 1\right)$$$$-\frac{3}{2}(y - 4) = x - 1$$$$-\frac{3}{2}y + 6 = x - 1$$$$x + \frac{3}{2}y - 6 - 1 = 0$$$$x + \frac{3}{2}y - 7 = 0$$Therefore, the final answer is,
$$x + \frac{3}{2}y - 7 = 0$$2. Two points $A$ and $B$ have coordinates $(-1, 1)$ and $(3, 4)$ respectively. The line $BC$ is perpendicular to $AB$ and intersects the $x$-axis at $C$. Find the equation of $BC$ and the $x$-coordinate of $C$. (9709/13/O/N/18 number 4)

To understand the question, let’s sketch an oversimplified diagram of the problem,

To find the equation of $BC$ we will use the, one point and gradient, route. $AB$ is perpendicular to $BC$ so we can use the gradient of $AB$ to find the gradient of $BC$. Let’s find the gradient of $AB$,
$$\textmd{Gradient of }AB = \frac{4 - 1}{3 + 1}$$$$\textmd{Gradient of }AB = \frac{3}{4}$$For perpendicular lines,
$$m_{1} \times m_{2} = -1$$$$\frac{3}{4} \times m_{2} = -1$$$$m_{2} = -\frac{4}{3}$$Therefore, the gradient of line $BC$ is,
$$-\frac{4}{3}$$Note: Line $BC$ passes through $B(3, 4)$.

Now that we have both the point and the gradient, let’s find the equation of line $BC$,
$$m = -\frac{4}{3} \textmd{ and the line passes through } B(3, 4)$$$$y = mx + c$$$$4 = -\frac{4}{3}(3) + c$$$$4 = -4 + c$$$$c = 4 + 4$$$$c = 8$$$$y = -\frac{4}{3}x + 8$$Now that we have the equation of line $BC$, let’s find the $x$-coordinate of $C$. $C$ is the $x$-intercept, so $y = 0$ at $C$,
$$\textmd{at } y = 0$$$$y = -\frac{4}{3}x + 8$$$$0 = -\frac{4}{3}x + 8$$$$\frac{4}{3}x = 8$$$$x = 6$$Therefore, the final answer is,
$$\textmd{The equation of line }BC\textmd{ is } y = -\frac{4}{3}x + 8$$$$\textmd{The }x\textmd{-coordinate of }C\textmd{ is }6$$3. The straight line with equation $y = x + 3$ intersects the curve $y = 4x^{\frac{1}{2}}$ at points $A$ and $B$. Find the length of $AB$. (9709/12/F/M/19 number 10)

To be able to find the length of $AB$, we have to find the coordinates of $A$ and $B$. To do that we have to start by solving simultaneously, since the two graphs intersect,
$$y = 4x^{\frac{1}{2}}\ \ \ \ \ \ y = x + 3$$$$4x^{\frac{1}{2}} = x + 3$$$$x - 4x^{\frac{1}{2}} + 3 = 0$$Solve the hidden quadratic,
$$x - 4x^{\frac{1}{2}} + 3 = 0$$$$(x^{\frac{1}{2}})^{2} - 4x^{\frac{1}{2}} + 3 = 0$$Let $v = x^{\frac{1}{2}}$,
$$v^{2} - 4v + 3 = 0$$$$(v - 1)(v - 3) = 0$$$$v = 1\ \ \ \ \ \ v = 3$$$$v = x^{\frac{1}{2}}$$$$x^{\frac{1}{2}} = 1\ \ \ \ \ \ x^{\frac{1}{2}} = 3$$$$\left(x^{\frac{1}{2}}\right)^{2} = 1^{2}\ \ \ \ \ \ \left(x^{\frac{1}{2}}\right)^{2} = 3^{2}$$$$x = 1\ \ \ \ \ \ x = 9$$Now let’s find the $y$ values. At $x = 1$,
$$y = x + 3$$$$y = 1 + 3$$$$y = 4$$At $x = 9$,
$$y = \textcolor{#2192ff}{x} + 3$$$$y = \textcolor{#2192ff}{9} + 3$$$$y = 12$$Therefore, the coordinates of $A$ and $B$ are,
$$A(1, 4) \textmd{ and } B(9, 12)$$Now let’s find the length of $AB$,
$$AB = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$$$$AB = \sqrt{(1 - 9)^{2} + (4 - 12)^{2}}$$$$AB = 8\sqrt{2}$$Therefore, the final answer is,
$$AB = 8\sqrt{2}$$4. The coordinates of two points $A$ and $B$ are $(-7, 3)$ and $(5, 11)$ respectively. Show that the equation of the perpendicular bisector of $AB$ is $3x + 2y = 11$. (9709/13/M/J/20 number 10)

A perpendicular bisector, is one that intersects a line at $90^{\circ}$ and it splits the line into two equal parts. This means it passes through the midpoint of the line. To find the equation of the perpendicular bisector, we need to find a point it passes through (midpoint) and the gradient. Let’s start by finding the gradient of line $AB$,
$$\textmd{Gradient of } AB = \frac{11 - 3}{5 + 7}$$$$\textmd{Gradient of } AB = \frac{2}{3}$$Let’s use $m_{1} \times m_{2} = -1$, to find the gradient of the perpendicular bisector,
$$m_{1} \times m_{2} = -1$$$$\frac{2}{3} \times m_{2} = -1$$$$m_{2} = -\frac{3}{2}$$Let’s find the midpoint of line $AB$,
$$M = \left(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}\right)$$$$M = \left(\frac{-7 + 5}{2}, \frac{3 + 11}{2}\right)$$$$M = (-1, 7)$$Now that we have a point and the gradient, we can find the equation of the perpendicular bisector,
$$\textmd{perpendicular bisector passes through } M(-1, 7)\\ \textmd{gradient is } -\frac{3}{2}$$$$y - y_{1} = m(x - x_{1})$$$$y - 7 = -\frac{3}{2}(x + 1)$$$$-2(y - 7) = 3(x + 1)$$$$-2y + 14 = 3x + 3$$$$3x + 2y + 3 - 14 = 0$$$$3x + 2y - 11 = 0$$$$3x + 2y = 11$$Therefore, the equation of the perpendicular bisector is,
$$3x + 2y = 11$$