Chapter 5.5, Problem 26PSB

a

To determine

Tofind: the coordinates of M

Answer to Problem 26PSB

Mid-point of $\overline{AB}$ is, $\left(M\right)=\left(-2,0\right)$

Explanation of Solution

Given information:

  

Y13y $\text{In }\Delta ABC,A\left(-6,2\right),B\left(2,-2\right),C\left(2,8\right)$ and

  $\overline{CM}$ is the median.

Formula used:

Mid-point of $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Calculation:

Mid-point of $\overline{AB}$ is $\left(M\right)=\left(\frac{\left(-6\right)+2}{2},\frac{2+\left(-2\right)}{2}\right)=\left(\frac{-4}{2},0\right)=\left(-2,0\right)$

b

To determine

To find: whether $\overline{CM}$ is an altitude or not.

Answer to Problem 26PSB

Yes, $\overline{CM}$ is altitude.

Explanation of Solution

Given information:

  

  $\text{In }\Delta ABC,A\left(-6,2\right),B\left(2,-2\right),C\left(2,8\right),\text{}M\left(-2,0\right)$ and $\overline{CM}$ is the median.

To check whether $\overline{CM}$ altitude, it is enough to check that the product of the slopes $\left(\overline{CM}\text{ and }\overline{AB}\right)$ is $-1$ .

Slope of $\overline{CM}=\frac{0-8}{-2-2}=\frac{-8}{-4}=2$

Slope of $\overline{AB}=\frac{-2-2}{2-\left(-6\right)}=\frac{-4}{8}=-\frac{1}{2}$

  $\left(\text{Slope of }\overline{CM}\right)\times \left(\text{Slope of }\overline{AB}\right)=2\times \left(-\frac{1}{2}\right)=-1$

Hence, $\overline{CM}$ altitude

c

To determine

To find:the type of the triangle $\Delta ABC$

Answer to Problem 26PSB

  $\Delta ABC$ is an isosceles triangle.

Explanation of Solution

Given information:

  $\text{In }\Delta ABC,A\left(-6,2\right),B\left(2,-2\right),C\left(2,8\right),\text{}M\left(-2,0\right)$ and $\overline{CM}$ is the median.

  

Formula used:

• Distance between $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$
• If length of two sides in a triangle $\Delta ABC$ are equal, then $\Delta ABC$ is isosceles triangle.
• If length of all the sides are equal in a triangle $\Delta ABC$ are equal, then $\Delta ABC$ is equilateral triangle.

Calculation:

  $\begin{array}{l}AB=\sqrt{{\left(2-\left(-6\right)\right)}^2+{\left(\left(-2\right)-2\right)}^2}=\sqrt{{\left(2+6\right)}^2+{\left(-2-2\right)}^2}=\sqrt{{\left(8\right)}^2+{\left(-4\right)}^2}=\sqrt{64+16}=\sqrt{80}=4\sqrt{5}\\ BC=\sqrt{{\left(2-2\right)}^2+{\left(8-\left(-2\right)\right)}^2}=\sqrt{{\left(0\right)}^2+{\left(8+2\right)}^2}=\sqrt{0+{10}^2}=\sqrt{100}=10\\ CA=\sqrt{{\left(\left(-6\right)-2\right)}^2+{\left(2-8\right)}^2}=\sqrt{{\left(-8\right)}^2+{\left(-6\right)}^2}=\sqrt{64+36}=\sqrt{100}=10\end{array}$ Observe from these values, only $BC\text{ and }CA$ are equal in lengths.

Hence, $\Delta ABC$ is an isosceles triangle.