Question

Find the ratio in which the line 3x+y-9=0 divides the line segment joining the points A(1,3) and B(2,7)​

plz... answer it

Answer 1

let ratio be k:1

then,B y section formula

x=2k+1/k+1.

y=7k+3/k+1.

put in the equation ,

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

=6k+3+7k+3-9(k+1) = 0 × (k+1)

=6k+3+7k+3-9k-9 =0

=4k-5=0

4k=5

k=5:4

Answer 2

have the equation for first line,

3x+ y-9= 0

3x+y= 9

Now, about the second line.

The line segment joins the points A(1,3) and B(2,7)

So, let's use the Two point formula to find the equation of second line. According to the formula,

x- x1/x2-x1= y-y1/y2-y1

LetA(x1,y1) and B(x2,y2)

x-1/2–1= y- 3/7–3

x-1/1= y-3/4

4x-4= y-3

So, 4x-y= 1

This is the equation of second line. Now, we have to solve them simultaneously to find the coordinates of point of intersection. By solving it , as we solve the linear equation in two variable, we get the coordinates as C(10/7, 33/7).Let the ratio be m:n

By using the formula for internal division,

X=mx2-nx1 /m+n

10/7= 2m-n/ m+n

10(m+n)= 7(2m-n)

10m+10n= 14m-7n

10n+7n= 14m-10m

17n= 4m

m/n=17/4

So, the ratio is 17:4.