Chapter 8.3, Problem 10PSA

To determine

To calculate: Coordinates of P and Q

Answer to Problem 10PSA

Coordinates of P is $(0,3)$ and Q is $(2,0)$ .

Explanation of Solution

Given information : Here, we have given that Slope of line PQ is $\frac{-3}{2}$

Formula used :

Slope of a line = $\frac{{\text{y}}_{\text{2}}{\text{-y}}_{\text{1}}}{{\text{x}}_{\text{2}}{\text{-x}}_{\text{1}}}$

Calculation: Let $(-4,9)$ be the coordinates of point T.

As, P is on y axis,

  $\Rightarrow$ Coordinates of P will be $\text{(0,y)}$ .

Similarly, Q is on x axis,

  $\Rightarrow$ Coordinates of Q will be $\text{(x,0)}$ .

Here, slope of line TP = slope of line PQ

  $\Rightarrow$ Slope of line TP = $\frac{-3}{2}$ .

  $\begin{array}{l}\frac{\text{y-9}}{\text{0-(-4)}}\text{=}\frac{\text{-3}}{\text{2}}\\ \frac{\text{y-9}}{\text{4}}\text{=}\frac{\text{-3}}{\text{2}}\\ \text{2y-18=-12}\\ \text{2y=-12+18}\\ \text{2y=6}\\ \text{y=3}\end{array}$

  $\therefore$ Coordinates of P will be $(0,3)$ .

Now, Slope of PQ = $\frac{-3}{2}$

  $\begin{array}{l}\Rightarrow \frac{0-3}{x-0}=\frac{-3}{2}\\ \Rightarrow \frac{-3}{x}=\frac{-3}{2}\\ \Rightarrow -3x=-6\\ \Rightarrow x=2\end{array}$

  $\therefore$ Coordinates will be $(2,0)$ .