Question

Given line segment AB with endpoints A(-9, 2) and B(12, 8) what are the coordinated of point c that is partitioned one third from A to B. PLEASE HELP.

Answer 1

check the picture below.

so, from A to B cut at 1/3, simply means, splitting the AB segment into 1+3 equal pieces, and from those four, AB takes 1 piece, and CB takes the other 3 pieces.


`\bf \left. \qquad \right.\textit{internal division of a line segment} \\\\\\ A(-9,2)\qquad B(12,8)\qquad \qquad 1:3 \\\\\\ \cfrac{AC}{CB} = \cfrac{1}{3}\implies \cfrac{A}{B} = \cfrac{1}{3}\implies 3A=1B\implies 3(-9,2)=1(12,8)\\\\ -------------------------------\\\\ { C=\left(\cfrac{\textit{sum of`


`\bf -------------------------------\\\\ C=\left(\cfrac{(3\cdot -9)+(1\cdot 12)}{1+3}\quad ,\quad \cfrac{(3\cdot 2)+(1\cdot 8)}{1+3}\right) \\\\\\ C=\left(\cfrac{-27+12}{4}~~,~~\cfrac{6+8}{4} \right)\implies C=\left(-(15)/(4)~~,~~(7)/(2) \right) \\\\\\ C=\left(-3(3)/(4)~~,~~3(1)/(2) \right)`

Answer 2

`C(-(15)/(4),(7)/(2))`

Explanation:

The given endpoints of the line segment AB are A(-9, 2) and B(12, 8).

It is given that point C partitioned one third from A to B. It means point C divide the line segment AB in 1:3.

Section formula:

If a point divides a line segment PQ in m:n, where endpoints are
`P(x_1,x_2)` and
`Q(x_2,y_2)`, then the coordinates of that point are

`((mx_2+nx_1)/(m+n),(my_2+ny_1)/(m+n))`

Using section formula, the coordinates of point C are

`C=(((1)(12)+(3)(-9))/(1+3),((1)(8)+(3)(2))/(1+3))`

`C=((12-27)/(4),(8+6)/(4))`

`C=((-15)/(4),(14)/(4))`

`C=(-(15)/(4),(7)/(2))`

Therefore, the coordinates of point C are
`(-(15)/(4),(7)/(2))`.