Question

the equation of the line whose intercepts are a,b such that a+b=4 and a-b=2 what is the equation of this

Answer 1

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$\large\mathbb{a\:+\:b\:=\:4....... 1.}$

$\large\mathbb{a\:-\:b\:=\:2....... 2.}$

----------------------------- Adding

2a = 6

a = 3

putting this value in eq. 1.

$\mathbb{3\:+\:b\:=\:4}$

b = 1

Now you have two points

A = (3,0)

B = (0,1)

Using two point form :--

(y - y0) = $\large\frac{y2-y1}{x2-x1} × (x - x0)$

(y - 3) = $\large\frac{1-0}{0-3} × (x - 0)$

(y - 3) = $\large\frac{1}{-3} × x$

(y - 3) = $\large\frac{-1}{3} x$

3(y-3) = -x

3y - 9 = -x

x + 3y - 9 = 0

This is the required equation.

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Answer 2

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a+b=4.......1.

\large\mathbb{a\:-\:b\:=\:2....... 2.}a−b=2.......2.

----------------------------- Adding

2a = 6

a = 3

putting this value in eq. 1.

\mathbb{3\:+\:b\:=\:4}3+b=4

b = 1

Now you have two points

A = (3,0)

B = (0,1)

Using two point form :--

(y - y0) = \large\frac{y2-y1}{x2-x1} × (x - x0)

x2−x1

y2−y1

×(x−x0)

(y - 3) = \large\frac{1-0}{0-3} × (x - 0)

0−3

1−0

×(x−0)

(y - 3) = \large\frac{1}{-3} × x

−3

1

×x

(y - 3) = \large\frac{-1}{3} x

3

−1

x

3(y-3) = -x

3y - 9 = -x

x + 3y - 9 = 0