Chapter 4.2, Problem 11CYU

a.

To determine

To write: The equation in slope-point form of the line containing $\overline{GH}$ .

Answer to Problem 11CYU

The equation is $y-7=\frac{-6}{7}(x+3)$ .

Explanation of Solution

Given:

The graph of the triangle FGH is given as:

  

Formula used:

The equation of slope of a line is given by,

  $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

From the graph the coordinates of the points F, G and H.

They are as follows:

F(-3,1), G(-3,7) and H(4,1).

Now find out slope of the equation of the line containing $\overline{GH}$ by using the two-point form, which is given by,

  $m=\frac{y_2-y_1}{x_2-x_1}$ …(1)

Therefore, plugging in the values of $x_1=-3,y_1=7,x_2=4,y_2=1$ into equation 1.

  $\begin{array}{l}\therefore m=\frac{1-7}{4+3}\\ \therefore m=\frac{-6}{7}\end{array}$

So, equation of the line in slope-point form can be found as,

  $y-y_1=m(x-x_1)\dots (2)$

So, plug in the values of $m,x_1,y_1$ in equation 2 to get,

  $\begin{array}{l}y-7=m(x+3)\\ \therefore y-7=\frac{-6}{7}(x+3)\end{array}$

b.

To determine

To find: The Standard form of the line containing $\overline{GH}$ .

Answer to Problem 11CYU

The equation is $6x+7y-31=0$ .

Explanation of Solution

Given:

The slope-point form of the line containing $\overline{GH}$ is $y-7=\frac{-6}{7}(x+3)$ .

Calculation:

Simplify the slope-point form of $\overline{GH}$ to get the equation in standard form, which is given by,

  $Ax+By+C=0$

  $\begin{array}{l}y-7=\frac{-6}{7}(x+3)\\ \therefore 7y-49=-6x-18\\ \therefore 6x+7y-49+18=0\end{array}$

Thus, the equation is $\therefore 6x+7y-31=0$

(3)

c.

To determine

To find: The slope-intercept form of the line containing $\overline{GH}$ .

Answer to Problem 11CYU

The equation is $y=-\frac{6}{7}x+\frac{31}{7}$ .

Explanation of Solution

Given:

The slope-point form of $\overline{GH}$ from equation (3) is given as $6x+7y-31=0$ .

Calculation:

Simplify equation 2 to get the slope-intercept form of the line in the form of $y=mx+c.$

  $\begin{array}{l}\therefore 6x+7y-31=0\\ \therefore 7y=-6x+31\end{array}$

Thus, the equation is $\therefore y=-\frac{6}{7}x+\frac{31}{7}$

(4)

d.

To determine

To find: The point-slope form of the line containing $\overline{FH}$ .

Answer to Problem 11CYU

The point slope form is $y=1$ .

Explanation of Solution

Given:

The co-ordinates of the points are F(-3,1) and H(4,1).

Calculation:

The slope of the line containing $\overline{FH}$ using,

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ \therefore m=\frac{1-1}{4+3}\end{array}$

  $\therefore m=0$

Therefore, point-slope form of the equation will be,

  $\begin{array}{l}y-y_1=m(x-x_1)\\ \therefore y-1=0(x+3)\\ \therefore y-1=0\end{array}$

The point slope form is $y=1$ (5)