Chapter 4.3, Problem 29E

a.

To determine

To check: Whether the given quadrilateral is a parallelogram.

Answer to Problem 29E

Given quadrilateral is a parallelogram.

Explanation of Solution

Given information: We are given the vertices of a quadrilateral A $(2,2)$ , B $(6,4)$ , C $(8,10)$ , and D $(4,8)$ .

Calculation: Slope of AB $=\frac{4-2}{6-2}=\frac{2}{4}=\frac{1}{2}$ .

Slope of BC $=\frac{10-4}{8-6}=\frac{6}{2}=3$ .

Slope of CD $=\frac{8-10}{4-8}=\frac{-2}{-4}=\frac{1}{2}$ .

Slope of DA $=\frac{8-2}{4-2}=\frac{6}{2}=3$ .

Now slope of AB $=$ slope of CD.

Slope of BC $=$ slope of DA.

Therefore AB is parallel to CD and BC is parallel to DA.

Therefore ABCD is a parallelogram.

  $=$

As the required line is parallel to given line so slope is equal.

So slope of required line is $=3$ .

Equation of required line is:

  $\begin{array}{l}y+2=3(x-3)\\ y+2=3x-9\\ 3x-y=9+2\\ 3x-y=11\end{array}$

Thus Equation of required line is $3x-y=11$ .

b.

To determine

To write: An equation of the line that passes through the given point and is perpendicular to given line.

Answer to Problem 29E

Equation of required line is $x+3y=-3$ .

Explanation of Solution

Given information: Given point is $(3,-2)$ and a line is passing through the points $(1-4)\mathrm{and}(2,-1)$ .

Calculation: As line is passing through the points $(1,-4)\mathrm{and}(2,-1)$ .

Therefore slope of line is $=\frac{-1+4}{2-1}=3$ .

Let Slope of required line is $=m_2$ .

As the required line is perpendicular to given line.

  $\begin{array}{l}{m}_1.m_2=-1\\ 3.m_2=-1\\ m_2=\frac{-1}{3}\end{array}$

Equation of required line is:

  $\begin{array}{l}y+2=\frac{-1}{3}(x-3)\\ 3y+6=-x+3\\ x+3y=-6+3\\ x+3y=-3\end{array}$

Thus Equation of required line is $x+3y=-3$ .