Question

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.

Answer 1

$\bf \left. \qquad \right.\textit{internal division of a line segment} \\\\\\ A(5,0)\qquad C(x,y)\qquad \qquad 2:5 \\\\\\ \cfrac{AB}{BC} = \cfrac{2}{5}\implies \cfrac{A}{C}=\cfrac{2}{5}\implies 5A=2C\implies 5(5,0)=2(x,y)\\\\ -------------------------------\\\\ { B=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}$

$\bf -------------------------------\\\\ B=\left(\cfrac{(5\cdot 5)+(2\cdot x)}{2+5}\quad ,\quad \cfrac{(5\cdot 0)+(2\cdot y)}{2+5}\right)=\boxed{(8,0)} \\\\\\ B=\left( \cfrac{25+2x}{7}~~,~~\cfrac{0+2y}{7} \right)=(8,0)\implies \begin{cases} \cfrac{25+2x}{7}=8\\\\ 25+2x=56\\ 2x=31\\\\ \boxed{x=\cfrac{31}{2}}\\ -------\\ \cfrac{0+2y}{7}=0\\\\ 2y=0\\ \boxed{y=0} \end{cases}$

Answer 2

Answer:

$(\frac{31}{2},0)$

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 $P(x_1,y_1)$ and $Q(x_2,y_2)$, then the coordinates of point K are

$K=(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n})$

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

Using section formula the coordinates of B are

$B=(\dfrac{(2)(a)+(5)(5)}{2+5},\dfrac{(2)(b)+(5)(0)}{2+5})$

$B=(\dfrac{2a+25}{7},\dfrac{2b}{7})$

We know that B(8,0).

$(8,0)=(\dfrac{2a+25}{7},\dfrac{2b}{7})$

On comparing both sides we get

$8=\dfrac{2a+25}{7}$

$56=2a+25$

$56-25=2a$

$31=2a$

$\frac{31}{2}=a$

Similarly,

$0=\dfrac{2b}{7}\Rightarrow 0=2b\Rightarrow b=0$

Therefore, the coordinates of C are $(\frac{31}{2},0)$.