Question

determine the ratio in which line
3x+y-9=0 divides the line segment joining points (1,3) and (2,7)?
Please answer as soon as possible. ​

Answer 1

Answer:

Let the point (x,y) divides AB in the ratio k:1.

By section formula,

x = (2k+1)/(k+1) , y = (7k+3)/(k+1)

Now, this point also lies on the given line so it would satisfy the equation 3x+y-9=0

Therefore,

3(2k+1) + (7k+3) -9(k+1) = 0

k = 3/4

Hence, the required ratio is 3:4.

Answer 2

Answer:

Suppose the line 3x+y−9=0 divides  the line segment joining A(1,3) and B(2,7) in the ratio k:1 at point C. Then coordinates of C are

(k+12k+1,k+17k+3)

But, C lies on $$3x+y-9=0$4. Therefore,

3(k+12k+1)+k+17k+3−9=0

k=43

So, the required ratio is 3 : 4 internally.