Question

determine the ratio in which the line 3x+y-9=0 divides the line segment joining the point (1,3),(2,7) ????​

Answer 1

Answer:

3:4

Step-by-step explanation:

Let the point be P(x, y) and ratio be k:1

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

Since P lies on line 3x+y-9=0

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

6k+3+7k+3-9(k+1)=0 (Taking Lcm)

13k+6-9k-9=0

4k-3=0

k=3/4

Thus ratio is 3/4 :1

i.e. 3:4

Answer 2

Answer:

$K=3:4$

Step-by-step explanation:

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, the coordinates of C are

⇒$[\frac{2k+1}{k+1},\frac{7k+3}{k+1}]$

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

Therefore   $3[\frac{2k+1}{k+1}]+[\frac{7k+3}{k+1}]-9=0$

⇒ $3(2k+1)+(7k+3)-9(k+1)=0$

⇒$6k+3+7k+3-9k-9=0$

⇒$4k-3=0$

⇒$4k=3$

⇒$k=\frac{4}{3}$

⇒$K=3:4$

Hence the answer is $K=3:4$.