Question

Ratio in which the line 3x + y – 9 = 0 divides the segment joining the point (1, 3) and (2, 7)​

Answer 1

Answer:

What is the question. Not able to understand

Answer 2

Answer:

Solution:

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

$= ( \frac{2k + 1}{k + 1} , \frac{7k + 3}{k + 1} )$

But C, lies on 3x + y – 9 = 0

Therefore,

$3( \frac{2k + 1}{k + 1} ) + \frac{7k + 3}{k + 1} - 9 = 0$

$\Longrightarrow$ (6k + 3) + (7k + 3) – 9k – 9 = 0

$\Longrightarrow$ k 4k – 3 = 0

$\Longrightarrow$ k = 3/4

So, the required ratio is 3 : 4 internally.