Question

Find the equation of the line passing through the point (2,2) and cutting off intercept on the axes whose sum is 9 . ​

Answer 1

Solution : The equation of a line in the intercept is

x/a + y/b = 1

this line passing through (2,2) therefore

2/a + 2/b = 1 _______ (1)

it is given that ,

a+ b = 9 , i. e ,

b = 9- a _________(2)

from (1 ) and (2) , se get

2/a + 2/9-a = 1

or, a² - 9a + 18 = 0

or, (a -6) (a - 3) = 0

i. e , a = 6 or a = 3

if a = 6 and b = 9- 6 = 3 , then the equation of the line is

x/6 + y/3 = 1 or x + 2y - 6= 0

of a = 3 and b = 9- 3 = 6 then equation of the line

x /3 + y/6 = 1 or 2x +y - 6 = 0

Answer 2

AnswEr:

Let the equation of the line be

$\\ \qquad \sf \frac{x}{a} + \frac{y}{b} = 1$

It passes through (2,2) and the sum of the intercepts on the axis is 9. Therefore,

$\\ \qquad \sf \: \frac{2}{a} + \frac{2}{b} = 1 \: \: and \: \: a + b = 9 \\ \\ \\ \implies \sf \: 2b + 2a = ab \: \: and \: \: a + b = 9 \\ \\ \\ \implies \sf \: 2(9 - a) + 2a = a(9 - a) \qquad\: (on \: eliminating) \\ \\ \\ \implies \sf \: {a}^{2} -9a + 18 = 0 \\ \\ \\ \implies \sf \: (a - 6)(a - 3) = 0 \\ \\ \\ \implies \sf \: a = 3,6.$

When a = 3, a + b = 9 gives b = 6. When a = 6, a + b = 9 give b = 3.

Hence, the equation of the line are

$\qquad \sf \: \frac{x}{3} + \frac{y}{6} = 1 \: \: and, \: \frac{ x }{6} + \frac{y}{3} = 1 \: \: or, \\ \\ \qquad \sf \: 2x + y = 6 \: \: and \: \: x + 2y = 6$