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