Question

find the ratio in which X divides the line segment joining (3,6) and (12.-3)​

Answer 1

Explanation:

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)P(x

1

,y

1

)=(3,6)andQ(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)(

m+n

m×x

2

+n×x

1

,

m+n

m×y

2

+n×y

1

)=(x,y)

Using this we get,

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

k+1

12×k+3×1

,

k+1

−3×k+6×1

)=(x,0)

By comparing y-coordinate,

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

k+1

−3×k+6×1

=0

-3k+6=0−3k+6=0

-3k = -6

k = 2

Therefore, Required ratio is 2 : 1