Question

Points A, B and C are collinear, and AB : BC = 1: 4. A is located at (-5, - 3), B is located at (-2, 0) and C is located at (I, 1), on the directedsegment AC. What are the values of x and y?A(7,9)B(10,12)C(20,18)D(-4.4,-2.4)

Answer 1

From the question

We are given the points

$A(-5,-3),B(-2,0),C(x,y)$

Where points A, B, C are collinear points

We are given that

$\bar{AB}\colon\bar{BC}=1\colon4$

Since the points are collinear then

by the ratio rule, we have

$(\frac{bx_1+ax_2}{a+b},\frac{by_1+ay_2}{a+b})=(-2,0)$

Where

a = 1, b = 4

Also

$\begin{gathered} x_1=-5,x_2=x \\ y_1=-3,y_2=y \end{gathered}$

Hence we have

$(\frac{4(-5)+1(x)}{4+1},\frac{4(-3)+y}{4+1})=(-2,0)$

Simplifying this we get

$(-\frac{20+x}{5},\frac{-12+y}{5})=(-2,0)$

This implies

$\begin{gathered} \frac{-20+x}{5}=-2 \\ -20+x=-10 \\ x=-10+20 \\ x=10 \end{gathered}$

Also we have

$\begin{gathered} \frac{-12+y}{5}=0 \\ y=12 \end{gathered}$

Therefore, x = 10, y = 12

Hence the solution is

$(10,12)$