Question

Need help! Find the point, Q, along the directed line segment BA that divides BA into the ratio 2:3.

Answer 1

see explanation

Explanation:

Using the section formula

`x_(Q)` =
`(2(-4)+3(2))/(2+3)` =
`(-8+6)/(5)` = -
`(2)/(5)`

`y_(Q)` =
`(2(7)+3(-3))/(2+3)` =
`(14-9)/(5)` = 1

Hence Q = (-
`(2)/(5)`, 1 )

Answer 2

`(-(2)/(5),1)`

Explanation:

The problem is based on Section Formula which is given as under

`x=(mx_2+nx_1)/(m+n)`

`y=(my_2+ny_1)/(m+n)`

Where the m and n are the ratio in which the point
`P(x,y) bisect the line joining [tex]A(x_1,y_1)` and
`B(x_2,y_2)` internally

Now from the attached image we can guess the values of
`(x_1,y_1)` and
`(x_2,y_2)` and the values of m and n are 3 and 2 respectively

Substituting the values and simplifying we get

`x=(3 * 2+2 * -4)/(3+2)`

`x=(6-8)/(5)`

`x=(-2)/(5)`

`x=-(2)/(5)`

Now solving for y coordinate

`y=(3 * -3+2 * 7)/(3+2)`

`y=(-9+14)/(3+2)`

`y=(5)/(5)`

`y=1`

Hence the coordinates of point p which divides the line BA in ration 3:2 will be

`(-(2)/(5),1)`