Question

Find the equation of the straight lines passing through (8,3)and having intercepts whose sum is 1​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Equation of întercepts is x/a + y/(1 - a) = 1

As it pâss through the line (8, 3). So, x and y are 8 and 3 respectively.

8/a + 3/(1 - a) = 1

[8(1 - a) + 3a]/(1 - a) = 1

8 - 8a + 3a = 1a(1 - a)

8 - 5a = a - a²

a² - 5a - a + 8 = 0

a² - 6a + 8 = 0

Split the middle term in such a way that it's sum is 6 and product is 8.

a² - 2a - 4a + 8 = 0

a(a - 2) - 4(a - 2) = 0

(a - 4) (a - 2) = 0

a = 4, 2

When a = 2 then equation of the line is:

x/2 + y/(1 - 2) = 1

x/2 + y/(- 1) = 1

x/2 - y = 1

x - 2y = 2

When a = 4 then equation of the line is:

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

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

x/4 - y/3 = 1

(3x - 4y)/12 = 1

3x - 4y = 12