Respuesta :

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.

[tex]\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 "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}[/tex]

[tex]\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(-\frac{15}{4}~~,~~\frac{7}{2} \right) \\\\\\ C=\left(-3\frac{3}{4}~~,~~3\frac{1}{2} \right)[/tex]
Ver imagen jdoe0001

Answer:

[tex]C(-\dfrac{15}{4},\dfrac{7}{2})[/tex]

Step-by-step 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 [tex]P(x_1,x_2)[/tex] and [tex]Q(x_2,y_2)[/tex], then the coordinates of that point are

[tex](\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n})[/tex]

Using section formula, the coordinates of point C are

[tex]C=(\dfrac{(1)(12)+(3)(-9)}{1+3},\dfrac{(1)(8)+(3)(2)}{1+3})[/tex]

[tex]C=(\dfrac{12-27}{4},\dfrac{8+6}{4})[/tex]

[tex]C=(\dfrac{-15}{4},\dfrac{14}{4})[/tex]

[tex]C=(-\dfrac{15}{4},\dfrac{7}{2})[/tex]

Therefore, the coordinates of point C are [tex](-\dfrac{15}{4},\dfrac{7}{2})[/tex].

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