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

Asked by Abhisek | 11 months ago |  77

##### Solution :-

We know that equation of the line making intercepts a and b on x-and y-axis, respectively, is

$$\dfrac{x}{a}+\dfrac{y}{b}=1$$ . … (1)

Given: sum of intercepts = 9

a + b = 9

b = 9 – a

Now, substitute value of b in the above equation, we get

$$\dfrac{x}{a}+\dfrac{y}{9-a}=1$$

Given: the line passes through the point (2, 2),

So, $$\dfrac{2}{a}+\dfrac{2}{9-a}=1$$

$$\dfrac{(2(9 – a) + 2a)}{a(9 – a)}=1$$

$$\dfrac{(18 – 2a + 2a)}{a(9 – a)}$$ = 1

$$\dfrac{18}{a(9 – a)}$$ = 1

18 = a (9 – a)

18 = 9a – a2

a2 – 9a + 18 = 0

Upon factorizing, we get

a2 – 3a – 6a + 18 = 0

a (a – 3) – 6 (a – 3) = 0

(a – 3) (a – 6) = 0

a = 3 or a = 6

Let us substitute in (1),

Case 1 (a = 3):

Then b = 9 – 3 = 6

$$\dfrac{x}{3}+\dfrac{y}{6}=1$$

2x + y = 6

2x + y – 6 = 0

Case 2 (a = 6):

Then b = 9 – 6 = 3

$$\dfrac{x}{6}+\dfrac{y}{3}=1$$

x + 2y = 6

x + 2y – 6 = 0

The equation of the line is 2x + y – 6 = 0 or x + 2y – 6 = 0.

Answered by Pragya Singh | 11 months ago

### Related Questions

#### Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

#### Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices

Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

#### Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.

Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.