Answer:
Step-by-step explanation:
Let B divides AB in the ratio K:1
[tex]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} \\[/tex]
so B divides AB in the ratio 11:-36
[tex]x=\frac{-36*1+11 *\frac{-1}{2} }{11-36} \\x=\frac{83}{50}[/tex]
Answer:
[tex]\large \boxed{1.66}[/tex]
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
[tex]\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}[/tex]
(b) Find the y-intercept
Insert the coordinates of one of the points into the equation
[tex]\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}[/tex]
(c) Write the equation for the line
[tex]y = -\dfrac{20}{9}x + \dfrac{26}{9}[/tex]
2. Calculate the value of x when y = -⅘
[tex]\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 $\large \boxed{\mathbf{1.66}}$}[/tex]
The graph below shows your three collinear points.
