Question

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

Answer 1

Answer:

Step-by-step explanation:

let the points on x and y axis be (b,0) and (0,a) respectively..

given a+b= 9

let (x1,y1) be (2,2)

slope of the line is -a/b by points (0,a) and (b,0)= (a-0)/(0-b)

and by points (0,a) , (b,0) and (2,2) slope is

(a-2)/-2 =2/(2-b)

replacing b by 9-a

we get a= 3,6

when a=3, b=6 and when a=6 b=3

by slope point form

y-y1 =m(x-x1)

y-2=(-a/b)(x-2)

ax+by-2(a+b) =0

ax+by-2*9 =0

ax+by-18 =0

substitute values of a and b

a=3, b=6

equation is

3x+6y -18=0

when a=6 and b=3

equation is

6x +3y -18=0

Answer 2

Given:-

• The point (2,2) and cutting off intercepts on the axes whose sum is 9.

To find:-

• Find the equation of a line..?

Solutions:-

• The equation of a line in the intercept form is x/a + y/b = 1 ..........(i).

Here,

a and b are the intercepts on x and y axis respectively.

It is given that

=> a + b = 9

=> b = 9 - a ...........(ii).

From equation (i) and (ii), we obtain.

=> x/a + y/9 - a = 1 ............(iii).

It is given that the line passes through points (2, 2)

Therefore, equation

=> 2/a + 2/9 - a = 1

=> 2[1/a + 1/9 - a] = 1

=> 2[9 - a + a / a(9 - a)] = 1

=> 18/9a - a² = 1

=> 18 = 9a - a²

=> a² - 9a + 18 = 0

=> a² - 6a - 3a + 18 = 0

=> a(a - 6) - 3(a - 6) = 0

=> (a - 3) (a - 6) = 0

=> a = 6 or a = 3

If a = 6 and b = 9 - 6 = 3 them the equation of the line is.

=> x/6 + y/3 = 1

=> x + 2y / 6 = 1

=> x + 2y = 6

=> x + 2y - 6 = 0

If a = 3 and b = 9 - 3 = 3 them the equation of the line is.

=> x/3 + y/6 = 1

=> 2x + y / 6 = 1

=> 2x + y = 6

=> 2x + y - 6 = 0

Hence, the equation of a line is x + 2y - 6 = 0 and 2x + y - 6 = 0.