Question

The midpoint of a segment can be found using the formulas for a directed line segment, x =C DaQ+bX2-X1) + X, and46]«y =aa + b-6/2 – Yı) + y1. When using these formulas to find a midpoint, which is true?O a = 1 and b= 2a = 2 and b = 1a = 1 and a + b = 2a = 2 and a + b = 2NextSubmitSave and ExitMark this and return

Answer 1

In general, the formula to find the midpoint of a line segment which ends have the coordinates (x₁,y₁) and (x₂,y₂) is given by

$M=(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})$

Working with the formula we are given, for the x coordinate of the midpoint we have:

$x=(\frac{a}{a+b})(x_2-x_1)+x_1=\frac{a(x_2-x_1)}{a+b}+\frac{(a+b)x_1}{a+b}$

In order for this to be like the previous formula, we have the following equation:

$\frac{a(x_2-x_1)}{a+b}+\frac{(a+b)x_1}{a+b}=\frac{x_2+x_1}{2}$

From here, we see that a+b must be equal to 2, so:

$\frac{a(x_2-x_1)}{2}+\frac{2x_1_{}}{2}=\frac{ax_2-ax_1}{2}+\frac{2x_1}{2}=\frac{ax_2-ax_1+2x_1}{2}=\frac{x_2+x_1}{2}$

In order for this last equation to be true, a must equal 1:

$\frac{x_2-x_1+2x_1}{2}=\frac{x_2+x_1}{2}$

Let's verify this with the formula for the y coordinate of the midpoint:

$(\frac{1}{2})(y_2-y_1)+y_1=\frac{y_2-y_1}{2}+y_1=\frac{y_2-y_1}{2}+\frac{2y_1}{2}=\frac{y_2-y_1+2y_1}{2}=\frac{y_2+y_1}{2}$

Since using our previous deduction leads us to the correct formula for the y coordinate of the midpoint as well, we can conclude that a=1 and a+b=2.