# find the equation of the line. check the answer. through (12/5,1)forming the axes a triangle with area 5.hello, i got this problem in my book titled "analytic geometry" by love rainville chapter...

find the equation of the line. check the answer. through (12/5,1)forming the axes a triangle with area 5.

hello, i got this problem in my book titled "analytic geometry" by *love rainville* chapter 5:"the straight line",page 64 exercise number 15 and the topic was all about the "*intercept form*". hope you could help me.

### 2 Answers | Add Yours

The equation of the line which has x intercept = a and y intercept = b is of the form:

x/a+y/b = 1.

It is given that A(a,0), B(0,b) and the origin O(0,0) makes a triangle whose area is (1/2)ab which is given to be 5 by data.

=> (1/2)ab) = 5, Or ab = 10..

Therefore b = 10/a....(1)

Also the point (12/5, 1) is on the line x/a+y/b = 1. So (12/5)/a + 1/b = 1.....(2).

Substituting b = 10/a in (2), we get:

12/5a + a/10 = 1. Mutiplying by 10a , we get:

24+a^2 = 10a.

a^2 -10a +24 = 0.

(a-6)(a-4) = 0.

Therefore a-6 = 0, or a-4 = 0

a= 6, or a= 4.

If a= 6, then b = 10/6 = 5/3.

If a = 4, then b = 10/4 = 5/2.

Therefore the required equation is x/6+y/(5/3) = 1, Or 5x+18y = 30. Or x/4+y/(5/2) = 1, Or 5x+8y = 20

**Therefore the required equations of the lines satisfying the conditions are 5x+18y = 30, or 5x+8y = 20.**

The slope intercept form of the line is:

y = mx + b, where m is the slope of the line and b is y intercept.

We'll find the y intercept of the line:

x = 0 => y = b

We'll determine x intercept:

mx + b = 0

x = -b/m

We know that the area of the triangle formed by x and y axes and the line is 5 square units.

Since the triangle is right, the area is half from the product of thelegs of triangle.

A = x*y/2

Since we know x and y, we'll get:

A = -b^2/2m

5 = -b^2/2m

10m = -b^2 (1)

We also know that the point (12/5,1) is on the line, so it's coordinates verify the equation of the line:

1 = 12m/5 + b

5 = 12m + 5b

12m = 5 - 5b

m = (5 - 5b)/12 (2)

We'll substitute (2) in (1):

10(5 - 5b)/12 = -b^2

25(1 - b)/6 = -b^2

25(1 - b) = -6b^2

6b^2 - 25b + 25 = 0

We'll apply quadratic formula:

b1 = [25+sqrt(625 - 600)]/12

b1 = (25 + 5)/12

b1 = 30/12

b1 = 15/6

b1 = 2.5

b2 = (25 - 5)/12

b2 = 10/6

b2 = 1.6

m1 = (5 - 5b1)/12

m1 = -0.625

m2 = (5 - 5b2)/12

m2 = -1/4

**The equations of the line are:**

**y = -0.625x + 2.5**

**y2 = -x/4 + 10/6**