Question

find the ratio in which x-axis divides the line segment joining the points 3,6 and 12,-3

Answer 1

take the x axis to be (x,0)...nos take the 0 and then calculate the ratio by section formula

Answer 2

Answer:

Required ratio is 2 : 1

Step-by-step explanation:

Let say given points are $P(x_1,y_1)=(3,6)\:\:and\:\:Q(x_2,y_2)=(12,-3)$

We have to find: Ratio in which x-axis divides the line segment joining given points.

We know that point on x-axis is written as ( x , 0 ) that is y-coordinate is always 0 on x-axis.

let the ratio is k : 1

Now we use section formula to find value of k in the  ratio.

$(\frac{m\times x_2+n\times x_1}{m+n},\frac{m\times y_2+n\times y_1}{m+n})=(x,y)$

Using this we get,

$(\frac{12\times k+3\times1}{k+1},\frac{-3\times k+6\times1}{k+1})=(x,0)$

By comparing y-coordinate,

$\frac{-3\times k+6\times1}{k+1}=0$

$-3k+6=0$

-3k = -6

k = 2

Therefore, Required ratio is 2 : 1