# Bring the equation of the line 3x-4y-12=0 to the intercept form and find the intercept on the axis .

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### 3 Answers

The standard for for the intercept is:

x/a + y/b = 1 where a is x-intercept and b is y-intercept;

We have the equation:

3x - 4y - 12 = 0

Move 12 to the right side:

==> 3x - 4y = 12

Now divide by 12:

==> 3x/12 - 4y/12 = 1

==> x/4 - y/3 = 1

==>** x/4 + y/-3 = 1**

**Then x-intercept = 4**

**and y-intercept = -3**

3x-4y-12 = 0. To bring to the intercept form.

Solution:

There are two types of intercept forms:

(i) slope and intercept form: y = mx+c , where m is slope and c is y intercept.

The given equation is 3x-4y -12 = 0. Add 4y.

3x-12 = 4y Divide by 4:

3x/4 -12/4 = y. Or

y = (3/4)x-3.

(ii) Double ntercept form: x/a +y/b =1, where a and b are the intercepts on x and y axis.

Given equation : 3x-4y -12 = 0. Add 12.

3x-4y = 12. Divide by 12:

3x/12 -4y/12 = 1. Re write as:

x / (12/3) + y/(12/-4) = 1. Or

x/4 + y/(-3) = 1.

Therefore 3x-4y -12 has the double intercept form x/4 +y/(-3) = 1.

So 4 is x intercept and -3 is y intercept.

x

The given form of the equation is the general form.

This general form of the equation of the line has to be reduced to the intercept form, that is:

x/a + y/b = 1

We'll write the given equation:

3x-4y-12=0

We'll add 12 both sides:

3x-4y = 12

We'll divide by 12 both sides, in order to obtain 1 to the right side:

3x/12 - 4y/12 = 12/12

x/4 - y/3 = 1

We'll re-write the equation:

**x/4 + y/-3 = 1**

**So, the x intercept is 4 and the y intercept is -3.**