Question

find the ratio in which axes divide the line joining points (2,5) and (1,9)​

Answer 1

therefore point c divides the line segment AB in to -5:9.

Answer 2

Given : points (2,5) and (1,9)​

To find : the ratio in which x axis divide the lone segment

Solution :

since , the axis is x axis therefore the point is (x,0)

i.e y coordinate will be zero

now

let the point A(2,5)

B (1,9) C = (x,0)

and let the ratio be k:1

now , using the section formula of coordinate geometry

we have

$\frac{m_1x_2+m_2x_1}{m_1+m_2}, \frac{m_1y_2+m_2y_1}{m_1+m_2}\\\\\frac{k\times 1+1\times 2}{k+1}, \frac{k\times 9+1\times 5}{k+1}\\\\\frac{k+2}{k+1},\frac{9k+5}{k+1}$

since , the y coordinate will be 0 thus

$\frac{9k+5}{k+1} = 0 \\\\9k +5 =0\\\\9k = -5\\\\k = \frac{-5}{9}$

neglecting the negative sign

hence  , the ratio is 5:9