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:

$\large \boxed{1.66}$

Step-by-step explanation:

1. Calculate the equation of the straight line joining A and C.

The equation for a straight line is

y = mx + b

where m is the slope of the line and b is the y-intercept.

The line passes through the points (-½, 4) and (1, ⅔)

(a) Calculate the slope of the line

$\begin{array}{rcl}m & = & \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\ & = & \dfrac{\frac{2}{3} - 4}{1 - (-\frac{1}{2})}\\\\& = & \dfrac{-\frac{10}{3}}{\frac{3}{2}}\\\\& = & \dfrac{-10}{3}\times{\dfrac{2}{3}}\\\\& = & \dfrac{-20}{9}\\\\\end{array}$

(b) Find the y-intercept

Insert the coordinates of one of the points into the equation

$\begin{array}{rcl}y & = & mx + b\\4 & = & \dfrac{-20}{9}\left(-\dfrac{1}{2}\right) + b \\\\4 & = & \dfrac{10}{9} + b\\\\b & = & \dfrac{36}{9} - \dfrac{10}{9}\\\\b & = & \dfrac{26}{9}\\\\\end{array}$

(c) Write the equation for the line

$y = -\dfrac{20}{9}x + \dfrac{26}{9}$

2. Calculate the value of x when y = -⅘

$\begin{array}{rcl}y & = & -\dfrac{20}{9}x + \dfrac{26}{9}\\\\-\dfrac{4}{5} & = & -\dfrac{20}{9}x+ \dfrac{26}{9}\\\\36 & = & 100x -130\\100x & = & 166\\x & = & 1.66\\\end{array}\\\text{The value of x is \large \boxed{\mathbf{1.66}} }$

The graph below shows your three collinear points.

Answer 2

Answer:

Step-by-step explanation:

Let B divides AB in the ratio K:1

$x=\frac{nx1+mx2}{m+n} \\y=\frac{ny1+my2}{m+n} \\\frac{-4}{5} =\frac{1*\frac{2}{3}+k*4}{k+1} \\-4k-4=\frac{10}{3} +20k\\-12 k-12=10+60k\\72k=-22\\36k=-11\\k=-\frac{11}{36}$

so B divides AB in the ratio 11:-36

$x=\frac{-36*1+11 *\frac{-1}{2} }{11-36} \\x=\frac{83}{50}$