Slope-intercept and intercept forms

A line can be described by where it crosses the axes. Two such descrip­tions are espe­cially use­ful: the slope-inter­cept form, which shows the slope and the point where the line cuts the $y$-axis, and the inter­cept form, which shows where it cuts both axes. Each form makes cer­tain ques­tions very quick, so it pays to be able to move between them.

Inter­cepts#

If a line cuts the $x$-axis at $(a, 0)$, the num­ber $a$ is called its $x$-inter­cept. If it cuts the $y$-axis at $(0, b)$, the num­ber $b$ is its $y$-inter­cept. Inter­cepts can be pos­i­tive, neg­a­tive or zero; they are num­bers, not dis­tances.

To find the inter­cepts from an equa­tion: put $y = 0$ to get the $x$-inter­cept, and $x = 0$ to get the $y$-inter­cept.

Slope-inter­cept form#

Result. The line with slope $m$ and $y$-inter­cept $c$ is

$$y = mx + c$$

Why. The line passes through $(0, c)$ with slope $m$. By the point-slope form, $y - c = m(x - 0)$, that is, $y = mx + c$.

In this form you can read off the slope and the $y$-inter­cept at once: they are the coef­fi­cient of $x$ and the con­stant term, once the equa­tion has been solved for $y$.

A line through the ori­gin has $c = 0$, so its equa­tion is $y = mx$.

Exam­ple 1#

Find the line with slope $3$ and $y$-inter­cept $-2$.

$y = 3x - 2$, or $3x - y - 2 = 0$.

Exam­ple 2#

Find the slope and the $y$-inter­cept of $4x + 2y - 6 = 0$.

Solve for $y$: $2y = -4x + 6$, so $y = -2x + 3$. The slope is $-2$ and the $y$-inter­cept is $3$.

Exam­ple 3#

Find the line with $y$-inter­cept $-3$ that is par­al­lel to $2x - y + 7 = 0$.

The given line is $y = 2x + 7$, with slope $2$. A par­al­lel line has the same slope, so the required line is $y = 2x - 3$.

Inter­cept form#

Result. The line with $x$-inter­cept $a$ and $y$-inter­cept $b$ (both non-zero) is

$$\displaystyle \frac{x}{a} + \frac{y}{b} = 1$$

Why. The line passes through $(a, 0)$ and $(0, b)$, so its slope is $\displaystyle \frac{b - 0}{0 - a} = -\frac{b}{a}$, and by the slope-inter­cept form $\displaystyle y = -\frac{b}{a}x + b$. Divide by $b$ and rearrange: $\displaystyle \frac{y}{b} = -\frac{x}{a} + 1$, that is, $\displaystyle \frac{x}{a} + \frac{y}{b} = 1$.

Check: putting $y = 0$ gives $x = a$, and putting $x = 0$ gives $y = b$.

Area cut off. The line and the two axes enclose a right tri­an­gle with legs of lengths $|a|$ and $|b|$, so its area is $\displaystyle \frac{1}{2}|ab|$.

Exam­ple 4#

Write $3x + 4y = 12$ in inter­cept form, and find the area of the tri­an­gle it makes with the axes.

Divide by $12$: $\displaystyle \frac{3x}{12} + \frac{4y}{12} = 1$, that is,

$$\displaystyle \frac{x}{4} + \frac{y}{3} = 1$$

The inter­cepts are $4$ and $3$, and the area is $\displaystyle \frac{1}{2} \times 4 \times 3 = 6$ square units.

Exam­ple 5#

Find the line through $(2, 3)$ whose inter­cepts on the axes are equal.

Let both inter­cepts be $a$: $\displaystyle \frac{x}{a} + \frac{y}{a} = 1$, that is, $x + y = a$. Since $(2, 3)$ is on the line, $a = 2 + 3 = 5$. The line is $x + y = 5$.

Exam­ple 6#

The point $(3, 2)$ is the mid­point of the part of a line cut off between the axes. Find the line.

Let the line meet the axes at $(a, 0)$ and $(0, b)$. Their mid­point is $\displaystyle \left(\frac{a}{2}, \frac{b}{2}\right) = (3, 2)$, so $a = 6$ and $b = 4$. The line is

$$\displaystyle \frac{x}{6} + \frac{y}{4} = 1, \quad \text{that is,} \quad 2x + 3y = 12$$

Prac­tice#

1. Find the line with slope $2$ and $y$-inter­cept $5$.
2. Find the line with slope $\displaystyle -\frac{1}{2}$ and $y$-inter­cept $3$.
3. Find the slope and $y$-inter­cept of $4x - 2y + 6 = 0$.
4. Find the line whose $x$-inter­cept is $3$ and $y$-inter­cept is $-4$.
5. Find the inter­cepts of $5x + 2y = 10$ and the area of the tri­an­gle it makes with the axes.
6. Find the line through $(3, 4)$ with equal inter­cepts.
7. Find the line through $(2, -3)$ whose inter­cepts are equal in size but oppo­site in sign.
8. The point $(2, 3)$ is the mid­point of the part of a line between the axes. Find the line.
9. Find the line with $y$-inter­cept $4$ that is per­pen­dic­u­lar to $x + 3y - 1 = 0$.
10. Find the inter­cepts of $3x - 5y + 15 = 0$.

Answers#