Is Slope The X Intercept

In geometry, the equation of a line tin can be written in different forms and each of these representations is useful in different ways. The equation of a straight line is written in either of the following methods:

1. Betoken slope course
2. Ii betoken form
3. Slope intercept course
4. Intercept form

In this article, you will larn about one of the most common forms of the equation of lines chosen gradient-intercept form along with derivation, graph and examples.

Larn what is the intercept of a line here.

Let's have a wait at the slope-intercept form definition.

What is the Gradient Intercept Form of a Line?

The graph of the linear equation y = mx + c is a line with m equally slope, thousand and c as the y-intercept. This course of the linear equation is called the slope-intercept form , and the values of k and c are real numbers.

The slope, m , represents the steepness of a line. The slope of the line is besides termed equally gradient, sometimes. The y-intercept, b, of a line, represents the y-coordinate of the point where the graph of the line intersects the y-axis.

Slope Intercept Class Equation

In this section, you will acquire the derivation of the equation of a line in the slope-intercept form.

Consider a line Fifty with slope m cuts the y-axis at a distance of c units from the origin.

Slope intercept form

Here, the distance c is called the y-intercept of the given line L.

So, the coordinate of a point where the line L meets the y-axis will be (0, c).

That means, line L passes through a fixed point (0, c) with slope 1000.

We know that, the equation of a line in signal slope grade, where (ten₁, y₁) is the betoken and slope m is:

(y – y₁) = m(x – x₁)

Hither, (x₁, y₁) = (0, c)

Substituting these values, we become;

y – c = m(x – 0)

y – c = mx

y = mx + c

Therefore, the bespeak (x, y) on the line with slope thou and y-intercept c lies on the line if and only if y = mx + c

Note: The value of c tin can be positive or negative based on the intercept is made on the positive or negative side of the y-axis, respectively.

Slope Intercept Form Formula

As derived above, the equation of the line in slope-intercept form is given by:

y = mx + c

Here,

(ten, y) = Every point on the line

m = Slope of the line

c = y-intercept of the line

Usually, 10 and y have to be kept equally the variables while using the above formula.

Slope Intercept Course ten Intercept

We tin can write the formula for the slope-intercept form of the equation of line L whose slope is m and x-intercept d as:

y = one thousand(10 – d)

Here,

m = Slope of the line

d = x-intercept of the line

Sometimes, the slope of a line may be expressed in terms of tangent angle such as:

m = tan θ

As well, endeavour: Slope Intercept Class Figurer

Derivation of Slope-Intercept Form from Standard Form Equation

We can derive the slope-intercept form of the line equation from the equation of a straight line in the standard form as given below:

Equally nosotros know, the standard class of the equation of a straight line is:

Ax + By + C = 0

Rearranging the terms as:

Past = -Ax – C

⇒y = (-A/B)x + (-C/B)

This is of the form y = mx + c

Here, (-A/B) represents the gradient of the line and (-C/B) is the y-intercept.

Slope Intercept Form Graph

When we plot the graph for gradient-intercept course equation we get a straight line. Slope-intercept is the best form. Since it is in the form "y=", hence it is piece of cake to graph it or solve give-and-take problems based on information technology. We just have to put the x-values and the equation is solved for y.

The all-time office of the gradient-intercept class is that nosotros can get the value of slope and the intercept directly from the equation.

Solved Examples

Example ane:

Notice the equation of the straight line that has gradient k = 3 and passes through the betoken (–two, –5).

Solution:

By the slope-intercept form we know;

y = mx+c

Given,

yard = 3

As per the given point, we take;

y = -v and 10 = -2

Hence, putting the values in the above equation, nosotros go;

-five = 3(-two) + c

-five = -6+c

c = -5 + 6 = one

Hence, the required equation volition be;

y = 3x+i

Example 2:

Find the equation of the straight line that has slope 1000 = -ane and passes through the point (two, -3).

Solution:

By the slope-intercept grade we know;

y = mx+c

Given,

m = -one

As per the given point, we have;

y = -3 and x = 2

Hence, putting the values in the higher up equation, we get;

-iii = -1(ii) + c

-iii = -2 + c

c = -3+2 = -1

Hence, the required equation will exist;

y = -x-1

Example three:

Discover the equation of the lines for which tan θ = 1/two, where θ is the inclination of the line such that:

(i) y-intercept is -5

(ii) x-intercept is 7/iii

Solution:

Given, tan θ = 1/2

So, slope = 1000 = tan θ = 1/two

(i) y-intercept = c = -5

Equation of the line using slope intercept form is:

y = mx + c

y = (ane/two)ten + (-v)

2y = x – 10

10 – 2y – 10 = 0

(two) x-intercept = d = vii/3

Equation of slope intercept form with ten-intercept is:

y = k(x – d)

y = (one/ii)[x – (7/3)]

2y = (3x – vii)/iii

6y = 3x – seven

3x – 6y – vii = 0

Practise Problems

Discover the gradient of the line y = 5x + 2.
Find the slope of the line which crosses the line at bespeak (-two,half dozen) and have an intercept of 3.
What is the equation of the line whose angle of inclination is 45 degrees and 10-intercept is -⅗?
Write the equation of the line passing through the betoken (0, 0) with gradient -four.