In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

Asked by Abhisek | 1 year ago |  114

##### Solution :-

The equation of the line joining the points (-1,1) and (5,7) is,

$$y-1=\dfrac{7-1}{5+1}(x+1)$$

$$y-1=\dfrac{6}{6}(x+1)$$

x - y + 2 = 0............(1)

The equation of the given line is x + y -4 = 0..............(2) .

The points of intersection of line (1) and (2) is x =1 and y = 3.

Let point (1,3) divide the line segment joining (-1,1) and (5,7) in the ratio 1: k .

Then, by section formula,

$$( 1,3)=(\dfrac{-k+5}{1+k},\dfrac{k+7}{1+k})$$

$$\dfrac{-k+5}{1+k}=1,$$$$\dfrac{k+7}{1+k}=3$$

$$\dfrac{-k+5}{1+k}=1$$

By cross multiplicatio

= -k+5=1+k

= 2k =4

k =2

Therefore, the line joining the points (-1,1) and (5,7) is divided by line x + y = 4 in the ratio 1: 2.

Answered by Pragya Singh | 1 year 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.