Ibrahim Rosales

Answered

2022-07-20

Point $A$ has coordinates $({x}_{1},{y}_{1})$, point $B$ has coordinates $({x}_{2},{y}_{2})$ andpoint $M$ has coordinates $(\frac{{x}_{1}+{x}_{1}}{2},\frac{{y}_{1}+{y}_{1}}{2})$

Prove that $AM=\frac{1}{2}AB$

Prove that $AM=\frac{1}{2}AB$

Answer & Explanation

decoratesuw

Expert

2022-07-21Added 11 answers

The points $A({x}_{1},{y}_{1}),B({x}_{2},{y}_{2}),M({x}_{m}=\frac{{x}_{1}+{x}_{2}}{2},{y}_{m}=\frac{{y}_{1}+{y}_{2}}{2})$ are given.

First note that the points $A$, $M$ and $B$ are colinear (i.e. lie on the same line):

$\text{slope of AM}=\frac{{y}_{m}-{y}_{1}}{{x}_{m}-{x}_{1}}=\frac{\frac{{y}_{1}+{y}_{2}}{2}-{y}_{1}}{\frac{{x}_{1}+{x}_{2}}{2}-{x}_{1}}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\text{slope of AB}.$

The triangles $\mathrm{\Delta}AMD$ and $\mathrm{\Delta}ABC$ are similar, because their corresponding angles are equal. Hence:

$\frac{AM}{AB}=\frac{AD}{AC}=\frac{AD}{2AD}=\frac{1}{2}\Rightarrow AM=\frac{1}{2}AB.$

Taniya Burns

Expert

2022-07-22Added 4 answers

$\begin{array}{rl}A{M}^{2}& ={(\frac{{x}_{1}+{x}_{2}}{2}-{x}_{1})}^{2}+{(\frac{{y}_{1}+{y}_{2}}{2}-{y}_{1})}^{2}\\ & ={\left(\frac{-{x}_{1}+{x}_{2}}{2}\right)}^{2}+{\left(\frac{-{y}_{1}+{y}_{2}}{2}\right)}^{2}\\ & =\frac{1}{4}{\textstyle (}(-{x}_{1}+{x}_{2}{)}^{2}+(-{y}_{1}+{y}_{2}{)}^{2}{\textstyle )}=\frac{1}{4}A{B}^{2},\end{array}$

from which it follows that $AM=\frac{1}{2}AB$.

from which it follows that $AM=\frac{1}{2}AB$.

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