Chapter 4.3, Problem 54PFA

a.

To determine

To write:The equation for the line containing the segment shown.

Answer to Problem 54PFA

The equation is $3x+5y-15=0$ .

Explanation of Solution

Given:

The line segment in the graph represents one side of a square.

Calculation:

From the graph it is seen that the line segment is passing through the points (0,-3) and (-5,0).

So the equation of the line passing through these two points can be found out using two-point form, which is given as,

  $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

So, plugging in the values as $x_1=0,y_1=-3,x_2=-5,y_2=0$ into the above equation we get,

  $\begin{array}{l}\frac{y+3}{0+3}=\frac{x-0}{-5-0}\\ \therefore \frac{y+3}{3}=\frac{x}{-5}\\ \therefore -5y-15=3x\\ \therefore 3x+5y+15=0\dots (2)\end{array}$

Conclusion:

Therefore equation 2 is the equation of the line containing the segment.

b.

To determine

To identify:What must be true about the sides adjacent and opposite the segment to form a square?

Answer to Problem 54PFA

The slope of the side opposite to the segment should be $\frac{-3}{5}$ .

And the slopes of the sides adjacent to the segment should be $\frac{5}{3}$ .

Explanation of Solution

Given:

Equation of the line segment is $3x+5y+15=0$ .

Calculation:

The segment adjacent to the given segment must have perpendicular slope.

Now, slope of the given segment is,

  $\begin{array}{l}m=\frac{-A}{B}\\ \therefore m=\frac{-3}{5}\dots (1)\end{array}$

Slope of the opposite side of the square will be same as that of the above segment i.e. $m=\frac{-3}{5}$

Now, the slope of the adjacent sides will be perpendicular to the above segment. So,

  $\begin{array}{l}{m}_1=\frac{-1}{m}\\ \therefore m_1=\frac{5}{3}\dots (2)\end{array}$

Conclusion:

c.

To determine

To write:The equations of the lines containing the sides of the square adjacent to the segment.

Answer to Problem 54PFA

1. $5x-3y-9=0$

2. $5x-3y+25=0$

Explanation of Solution

Given:

Slopes of the lines containing the sides of the square adjacent to the segment has been found above as $m_1=\frac{5}{3}$ .

Also these lines pass through the points (0,-3) and (-5,0).

Calculation:

Therefore equation of the line passing through the point (0,-3) and having slope $m_1=\frac{5}{3}$ will be,

  $y-y_1=m_1(x-x_1)$

  $\begin{array}{l}\therefore y+3=\frac{5}{3}(x-0)\\ \therefore 3y+9=5x\end{array}$

  $\therefore 5x-3y-9=0\dots (3)$

Andequation of the line passing through the point (-5,0) and having slope $m_1=\frac{5}{3}$ will be,

  $\begin{array}{l}y-0=m_1(x+5)\\ \therefore y=\frac{5}{3}(x+5)\\ \therefore 3y=5x+25\end{array}$

  $\therefore 5x-3y+25=0\dots (4)$

Conclusion:

d.

To determine

To Find:The equation of the line containing the opposite segment passing through the vertex (3,2).

Answer to Problem 54PFA

The equation is $3x+5y-19=0$ .

Explanation of Solution

Given:

The vertex (3,2) lies on the opposite segment.

And the slope of opposite segment we found above as $m=\frac{-3}{5}$ .

Calculation:

Therefore, the equation of the line passing through the segment having slope $m=\frac{-3}{5}$ and (3,2) will be,

  $y-y_1=m_1(x-x_1)$

  $\begin{array}{l}\therefore y-2=\frac{-3}{5}(x-3)\\ \therefore 5y-10=-3x+9\end{array}$

  $\therefore 3x+5y-19=0$ …(5)

e.

To determine

To find: The 4 vertices of the square.

Answer to Problem 54PFA

The four vertices of the square are (0,-3), (-5,0), (3,2) and (-2,5).

Explanation of Solution

Given:

The three vertices are found above as (0,-3), (-5,0) and (3,2).

Calculation:

The fourth vertex should be the point of intersection of the sides with equations $5x-3y+25=0$ and $3x+5y-19=0$

So solve equations $5x-3y=-25$ and $3x+5y=19$ simultaneously to find the point of intersection of the two lines.

  $5x-3y=-25$ …(A)

  $3x+5y=19$ …(B)

So, multiplying equation A by 5 and equation B by 3 and then adding them, we get

  $\begin{array}{l}34x=-68\\ \therefore x=\frac{-68}{34}\\ \therefore x=-2\end{array}$

Plugging $x=-2$ in equation B, we find the value of y as follows:

  $\begin{array}{l}3(-2)+5y=19\\ \therefore -6+5y=19\\ \therefore 5y=19+6\\ \therefore 5y=25\\ \therefore y=\frac{25}{5}\end{array}$

  $\Rightarrow y=5$

Therefore, the fourth vertex of the square is (-2,5)

Conclusion:

f.

To determine

To find:The quadrilateral figure is a square or not.

Answer to Problem 54PFA

The quadrilateral figure is a square.

Explanation of Solution

Given:

The four vertices of the square are given as (0,-3), (-5,0), (3,2) and (-2,5).

Calculation:

Therefore, one diagonal of the square will pass through the vertices (0,-3) and (-2,5).

And the other diagonal will pass through the vertices (-5,0) and (3,2).

Find slope of the diagonal passing through the points (0,-3) and (-2,5) by using,

  $\begin{array}{l}{m}_3=\frac{y_2-y_1}{x_2-x_1}\\ \therefore m_3=\frac{5+3}{-2-0}\\ \therefore m_3=\frac{8}{-2}\\ \therefore m_3=-4\end{array}$

Find slope of the diagonal passing through the points (-5,0) and (3,2) by using,

  $\begin{array}{l}{m}_4=\frac{y_2-y_1}{x_2-x_1}\\ \therefore m_4=\frac{2-0}{3+5}\\ \therefore m_4=\frac{2}{8}\\ \therefore m_4=\frac{1}{4}\end{array}$

Two lines are perpendicular, then the product of their slopes is -1.

And from above,

  $\begin{array}{l}{m}_3\times m_4=-(4)(\frac{1}{4})\\ \therefore m_3\times m_4=-1\end{array}$

Conclusion:

Now, since the two diagonals of the quadrilateral are perpendicular, also its adjacent sides are perpendicular to each other, the quadrilateral is a square.