Chapter 13.7, Problem 12PSA

(a)

To determine

To find:The coordinate of K.

Answer to Problem 12PSA

The coordinate of Kis $\left(13,9\right)$ .

Explanation of Solution

Let the coordinate of K be $\left(x,y\right)$ .

HJ and MK are parallel. So their slope are equal.

  $\begin{array}{c}\frac{4-2}{14-4}=\frac{y-7}{x-3}\\ \frac{1}{5}=\frac{y-7}{x-3}\\ x-3=5y-35\\ x=5y-32\end{array}$

HJ and KJ are perpendicular. So their slopes are reciprocal to each other.

  $\begin{array}{c}\frac{y-4}{x-14}=-\frac{1}{1/5}\\ y-4=-5x+70\\ y=-5x+74\end{array}$

On solving the above two equations.

  $\begin{array}{l}x=13\\ y=9\end{array}$

Therefore, the coordinate of K is $\left(13,9\right)$ .

(b)

To determine

To find: The area of rectange.

Answer to Problem 12PSA

The area of rectangle is $52\text{sq}\text{.units}$ .

Explanation of Solution

The height of rectangle is calculated as,

  $\begin{array}{c}h=\sqrt{{\left(3-4\right)}^2+{\left(7-2\right)}^2}\\ =\sqrt{26}\end{array}$

The base of rectangle is calculated as,

  $\begin{array}{c}b=\sqrt{{\left(14-4\right)}^2+{\left(4-2\right)}^2}\\ =\sqrt{100+4}\\ =\sqrt{104}\end{array}$

The area of the rectangle is calculated as,

  $\begin{array}{c}A=bh\\ =\sqrt{104}\sqrt{26}\\ =52\text{sq}\text{.units}\end{array}$

Therefore, the area of rectangle is $52\text{sq}\text{.units}$ .