# What is the equation of the line with x and y intercepts 4 and 7 respectively.

*print*Print*list*Cite

The x and y intercepts of the line are 4 and 7. Therefore the line passes through the points (4, 0) and (0, 7)

The equation of the line between the points (4, 0) and (0, 7) is given as (y – 0) = [(7- 0)/ (0 – 4)] (x – 4)

=> y = (7/-4) (x- 7)

=> y = (-7/ 4) (x – 7)

=> 4y = -7x + 28

=> 7x + 4y – 28 = 0

**The required line with x and y intercepts 4 and 7 is 7x + 4y – 28 = 0.**

The equation of the line with x and y in the standard double intercept form is x/a+y/b = a. Where intercept is a, and y intercept is b.

Therefore the equation of the line with x intercept 4 and y interceppt 7 is x/4 +y/7 = 1.

We check whether this is true:

x intercept is got by putting y= 0 in x/4+y/7 = 1:

So x/4 + 0/7 = 1,, So x/4 = 1, or x= 4.

We check whether y intercept is 7: We get y intercept by putting x = 0 in x/4+y/7 = 1. So, 0/4+y/7 = 1, or y/7 = 1. So y = 7.

Therefore x/4+y/7 = 1 is the required equation. We write the equation in ax+by+c = 0 form by mutiplying x/4+y/7 = 1 by 4*7 = 28 :

28(x/4+y/7) = 1*28. This gives us the equation,7x +4y = 28. Or 7x+4y-28 = 0.