Question

Given points A (1, 2/3), B (x, -4/5), and C (-1/2, 4) determine the value of x such that all three points are collinear

Answer 1

Answer:

$x=\frac{83}{50}$

Step-by-step explanation:

we know that

If the three points are collinear

then

$m_A_B=m_A_C$

we have

A (1, 2/3), B (x, -4/5), and C (-1/2, 4)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope AB

we have

$A(1,\frac{2}{3}),B(x,-\frac{4}{5})$

substitute in the formula

$m_A_B=\frac{-\frac{4}{5}-\frac{2}{3}}{x-1}$

$m_A_B=\frac{\frac{-12-10}{15}}{x-1}$

$m_A_B=-\frac{22}{15(x-1)}$

step 2

Find the slope AC

we have

$A(1,\frac{2}{3}),C(-\frac{1}{2},4)$

substitute in the formula

$m_A_C=\frac{4-\frac{2}{3}}{-\frac{1}{2}-1}$

$m_A_C=\frac{\frac{10}{3}}{-\frac{3}{2}}$

$m_A_C=-\frac{20}{9}$

step 3

Equate the slopes

$m_A_B=m_A_C$

$-\frac{22}{15(x-1)}=-\frac{20}{9}$

solve for x

$15(x-1)20=22(9)$

$300x-300=198$

$300x=198+300$

$300x=498$

$x=\frac{498}{300}$

simplify

$x=\frac{83}{50}$