**Gradients and intercepts** are the only piece of information we need in order to plot the graph of a straight line and know its properties. Let us explore what they are in this article.

## Gradient and intercepts of a line meaning

### Gradient of a line meaning

Suppose a man is walking up a hill, and for keeping things simple, the hill is straight in its path with no bumps and curves.

If the man travels 100 meters in the horizontal direction and finds himself 200 meters above the ground, how can we measure the steepness of the hill that is, how inclined the hill is?

To be more rigorous, the sloppiness, in mathematical terms, is known as the **Gradient **of the hill.

The g**radient **of a straight line is the measure of how steep it is, or how sloppy it is.

### Intercepts of a line meaning

A linear function of the form $ax+by+c=0$ is represented by a straight line that extends infinitely in either direction. Hence, at one point it will intersect either the y-axis or the x-axis or more commonly, both the axes of the coordinate plane.

The coordinates of the point where a straight line intersects either of the axes is known as the i**ntercept **of that straight line.

#### Types of intercepts of a line

And as there are two axes in the coordinate plane, a straight line may have intersections with the x-axis or with the y-axis, or with both of them.

The x-intercepts are the points where a straight line intersects the x-axis.

The y-intercepts are the points where a straight line intersects the y-axis.

## Gradient and intercepts calculation and formula

### Gradient calculation and formula

The **gradient **of a line is the ratio of the change in the y-direction to the change in the x-direction.

To measure the steepness of the hill, an intuitive way is to know how the change in the vertical direction is with respect to the change in a horizontal direction, which is essentially what we want to know. And this is exactly how the gradient is defined mathematically.

Translating this definition into an equation, with *m *denoting the gradient, we get

$m=\frac{\u2206y}{\u2206x}=\frac{{y}_{f}-{y}_{i}}{{x}_{f}-{x}_{i}}$

where ${y}_{f}$, ${y}_{i}$ denote the final and initial y-coordinates, and ${x}_{f},{x}_{i}$denote the final and initial x-coordinates.

#### Geometrical interpretation of the gradient

On close inspection, it can be observed that the formula of the gradient which is equal to the ratio between the change in y over the change in x, is nothing but the tangent of the angle formed between the line and the positive x-axis.

$m=\mathrm{tan}\theta $

Now if we go back to our man on the top of the hill, we can calculate the gradient of the hill with our formula,

$m=\frac{200-0}{100-0}=\frac{200}{100}=2$

Hence, the hill has a gradient of 2.

But how can we find the gradient of a line if we are given nothing but only the equation of the line? According to the definition of the gradient, we recall

$m=\frac{\u2206y}{\u2206x}=\frac{{y}_{f}-{y}_{i}}{{x}_{f}-{x}_{i}}$.

Hence we need the coordinates of any two points on the line to calculate the gradient.

But there are infinite of them to choose from, and how do we know which ones to choose? Recall that we do know the coordinates of two specific points on the line, **the x and y-intercepts!**

So we can choose the two points as $\left({x}_{i},{y}_{i}\right)=\left(\frac{-c}{a},0\right)$ and $\left({x}_{f},{y}_{f}\right)=\left(0,\frac{-c}{b}\right)$. After substitution we get,

$m=\frac{\frac{-c}{b}-0}{0-\left(\frac{-c}{a}\right)}=\frac{\frac{-c}{b}}{\frac{c}{a}}=\frac{-c}{b}\times \frac{a}{c}=\frac{-a}{b}$

Therefore, the gradient (or slope) of a line given by $ax+by+c=0$ is $m=-\frac{a}{b}$.

Remember that **slope **and **gradient **share the same meaning in this context of a straight line.

### Intercepts calculation

We consider the graph of a straight line intercepting the axes of the coordinate plane.

We notice that as per the definition, the intercepts are formed on the axes themselves.

The x-intercept will lie on the x-axis and so the y coordinate will be 0 at that point.

- To calculate the x-intercept, set y=0 and solve for x.

Consider a straight line, modeled by the linear equation, where *a, b *and *c *are real-valued constants, with a and b not being simultaneously 0.

To find the x-intercepts of this line, we need to substitute *y *by 0 since the y-coordinate vanishes at that point. Hence, we get

$ax+b\left(0\right)+c=0\phantom{\rule{0ex}{0ex}}ax+c=0\phantom{\rule{0ex}{0ex}}ax=-c\phantom{\rule{0ex}{0ex}}x=-\frac{c}{a}$

Thus, the x-intercept of the line $ax+by+c=0$, is $\left(-\frac{c}{a},0\right)$.

- To calculate the y-intercept, set x=0 and solve for y.

To find the y-intercept of the line, we need to substitute x* *by 0, which yields

$a\left(0\right)+by+c=0\phantom{\rule{0ex}{0ex}}by+c=0\phantom{\rule{0ex}{0ex}}by=-c\phantom{\rule{0ex}{0ex}}y=\frac{-c}{b}$

Thus, the y-intercept of the line $ax+by+c=0$ is $\left(0,\frac{-c}{b}\right)$.

## Gradient and Intercepts examples

Find the x-intercept and the y-intercept of the line of equation$2x+4y-1=0$.

**Solution**

Comparing the given equation of the line with the general form: $ax+by+c=0$,

We get $a=2,b=4,c=-1$.

And we know that the y-intercept is given by

$-\frac{c}{b}=-\left(\frac{-1}{4}\right)$$=\frac{1}{4}$.Hence the y-intercept lies at $\left(0,\frac{1}{4}\right)$.

Similarly, the x-intercept is given by

$-\frac{c}{a}=-\left(\frac{-1}{2}\right)=\frac{1}{2}$.

Hence the x-intercept lies at$\left(\frac{1}{2},0\right)$.

If we were to plot the line, then we locate the x and y-intercepts and then connect them to get the desired line.

Find the gradient of the line of equation $3x-y+1=0$.

**Solution**

Comparing the given equation of line to the general form $ax+by+c=0$, we get $a=3,b=-1,c=1$.

The slope or the gradient of the line can be calculated via,

$m=\frac{-a}{b}=\frac{-3}{-1}=\frac{3}{1}=3$

Thus, the slope of the given line is 3.

The graph of this straight line would be,

where A and B lie at the x and y-intercepts of the line.

Recall that the coordinates of the x-intercept are $\left(-\frac{c}{a},0\right)$ and for the y-intercept $\left(0,\frac{-c}{b}\right)$.

Using this, the x-intercept of our line $3x-y+1=0$ is $\left(-\frac{c}{a},0\right)=\left(-\frac{1}{3},0\right)$ and the y-intercept is $\left(0,-\frac{c}{b}\right)=\left(0,1\right)$.

## Gradient and Intercepts applications: Slope-Intercept form of a line

Just as a straight line can be generally expressed by the form $ax+by+c=0$, we can also derive a general form determined by the slope and the intercept of the line.

If we rearrange the given equation to get *y *on one side of the equation, we have

$by=-ax-c$

$y=\frac{-a}{b}x-\frac{c}{b}$

where we observe that $-\frac{a}{b}$ is the slope of the line as we found out in the previous section. And let us name $-\frac{c}{b}=d$ where *d *is just another constant renamed in terms of *c and b*. Recall that $-\frac{c}{b}$ is the y-intercept itself, which here will be *d. *Hence, our equation is reduced to

$y=\underset{slope}{\underset{\u23df}{m}}x+\underset{y-intercept}{\underset{\u23df}{d}}$

Find the slope and the y-intercept of the line $3x-2y+1=0$.

**Solution**

To compare the given equation of the line with the slope- intercept form, we need to solve it explicitly for *y, *we have

$2y=3x+1$

Dividing by 2, we get,

$y=\frac{3}{2}x+\frac{1}{2}$

Comparing this with the standard form $y=mx+d$, where *m *is the slope and *d *is the y -intercept, we get $m=\frac{3}{2},d=\frac{1}{2}$.

Hence the slope is $\frac{3}{2}$ and the y-intercept is $\left(0,\frac{1}{2}\right)$.

To find the x-intercept, we set y=0, and we solve for x, and in this case we get,

$3x+1=0\phantom{\rule{0ex}{0ex}}3x=-1\phantom{\rule{0ex}{0ex}}x=-\frac{1}{3}$

and thus the x-intercept is $\left(\frac{-1}{3},0\right)$.

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

**Solution**

Comparing the given equation to the general form $ax+by+c=0$, we get $a=1,b=2,c=0$.

The slope-intercept form is given by $y=\frac{-a}{b}x-\frac{c}{b}$, which gives us $y=-\frac{x}{2}$.

Thus the slope is $\frac{-1}{2}$ and the y-intercept is (0,0).

In order to find the x-intercept, we set y=0 and solve for x. Thus, we get

$x=0$

and hence the x-intercept is (0,0).

## Gradient and Intercepts - Key takeaways

- The non-zero coordinates of the point where a straight line intersects the two axes are known as the intercepts of that line.
- For a line given by $ax+by+c=0$, the x-intercept is given by $-\frac{c}{a}$ and the y-intercept as $-\frac{c}{b}$.
- The gradient of a line is a measure of how steep it is (or sloppy it is). An alternative term for gradient is slope.
- The gradient of a straight line given by $m=\mathrm{tan}\theta =-\frac{a}{b}$ where $\theta $ is the angle the line makes with the positive x-axis.
- A straight line can be alternatively expressed in a slope-intercept form where we can deduce the slope and the y-intercept directly.

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##### Frequently Asked Questions about Gradient and Intercept

What is Gradient and Intercept?

Gradient is a measure of how steep a straight line is. The non-zero coordinate of the point where a straight line intersects either of the axes, is known as the **Intercept **of that straight line.

What is the relationship between Gradient and Intercept?

The gradient of any straight line is equal to the negative of the ratio between the y and x-intercepts.

What is the formula for solving gradient and intercept?

For a straight line given by ax+by+c=0, the x and y intercepts are given by -c/a and -c/b. The gradient of that line is given by -b/a.

How do you graph gradient and intercept?

Using the x and y-intercepts, one can plot the two points and then connect them in order to get the required line. And the gradient is simply the slope of that line.

What is an example of gradient and intercept?

Suppose a straight line is given by x-y+1=0, the y-intercept of this line will be at (0,1) and the x-intercept will be at (-1,0) and the gradient of this line will be 1.

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