Number 8 please, What are the coordinates of point C below segment AC that is partitioned at point B in a ratio of 2 to 5.

[tex]\bf \left. \qquad \right.\textit{internal division of a line segment} \\\\\\ A(5,0)\qquad C(x,y)\qquad \qquad 2:5 \\\\\\ \cfrac{AB}{BC} = \cfrac{2}{5}\implies \cfrac{A}{C}=\cfrac{2}{5}\implies 5A=2C\implies 5(5,0)=2(x,y)\\\\ -------------------------------\\\\ { B=\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 -------------------------------\\\\ B=\left(\cfrac{(5\cdot 5)+(2\cdot x)}{2+5}\quad ,\quad \cfrac{(5\cdot 0)+(2\cdot y)}{2+5}\right)=\boxed{(8,0)} \\\\\\ B=\left( \cfrac{25+2x}{7}~~,~~\cfrac{0+2y}{7} \right)=(8,0)\implies \begin{cases} \cfrac{25+2x}{7}=8\\\\ 25+2x=56\\ 2x=31\\\\ \boxed{x=\cfrac{31}{2}}\\ -------\\ \cfrac{0+2y}{7}=0\\\\ 2y=0\\ \boxed{y=0} \end{cases}[/tex]

erinna erinna · Answer 2 · 2019-12-19T14:49:57+01:00

Answer:

[tex](\frac{31}{2},0)[/tex]

Step-by-step explanation:

From the given figure it is clear that the coordinates of points are A(5,0) and B(8,0).

Let the coordinates of C are (a,b).

Section formula:

If a point K divides a line segment PQ in m:n and end point of segment are [tex]P(x_1,y_1)[/tex] and [tex]Q(x_2,y_2)[/tex], then the coordinates of point K are

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

It is given that point B divides the line AC in 2:5.

Using section formula the coordinates of B are

[tex]B=(\dfrac{(2)(a)+(5)(5)}{2+5},\dfrac{(2)(b)+(5)(0)}{2+5})[/tex]

[tex]B=(\dfrac{2a+25}{7},\dfrac{2b}{7})[/tex]

We know that B(8,0).

[tex](8,0)=(\dfrac{2a+25}{7},\dfrac{2b}{7})[/tex]

On comparing both sides we get

[tex]8=\dfrac{2a+25}{7}[/tex]

[tex]56=2a+25[/tex]

[tex]56-25=2a[/tex]

[tex]31=2a[/tex]

[tex]\frac{31}{2}=a[/tex]

Similarly,

[tex]0=\dfrac{2b}{7}\Rightarrow 0=2b\Rightarrow b=0[/tex]

Therefore, the coordinates of C are [tex](\frac{31}{2},0)[/tex].