Chapter 13.7, Problem 22PSC

(a)

To determine

To find: The coordinates of point D.

Answer to Problem 22PSC

The required coordinate is $\left(11,9\right)$ .

Explanation of Solution

Given:

The vertices of point A is $A=\left(6,12\right)$ .

The point B is $B=\left(21,3\right)$ .

The point O is $O=\left(0,0\right)$ .

The segment CD is drawn parallel to the line OB such that C lies on the line AO.

The D lies on the line AB and $C=\left(4,8\right)$ .

Calculation:

Consider the slop of OB is,

  $\begin{array}{c}m=\frac{3-0}{21-0}\\ =\frac{1}{7}\end{array}$

Consider the equation for the CD is,

  $\begin{array}{l}y-8=\frac{1}{7}\left(x-4\right)\\ 7y-x=52\end{array}$

Consider the equation for the AB is,

  $\begin{array}{l}y-3=\frac{3-12}{21-6}\left(x-21\right)\\ 5y+3x=78\end{array}$

The point at which the line $5y+3x=78$ and $7y-x=52$ intersect is point D that is,

  $\left(11,9\right)$

(b)

To determine

To find: The ratio of the area of the triangle ACD to the area AOB.

Answer to Problem 22PSC

The ratio is $1:9$ .

Explanation of Solution

Consider the length of line CD is,

  $\begin{array}{c}CD=\sqrt{{\left(11-4\right)}^2+{\left(9-8\right)}^2}\\ =5\sqrt{2}\end{array}$

Consider the length of line OB is,

  $\begin{array}{c}OB=\sqrt{{21}^2+3^2}\\ =15\sqrt{2}\end{array}$

Then, the ratio of the area of the two triangles is the ratio of square of the two sides as,

  ${\left(\frac{5\sqrt{2}}{15\sqrt{2}}\right)}^2=\frac{1}{9}$