Chapter 3.1, Problem 72PPS

To determine

To calculate: points at which the diagonals of the quadrilateral intersect

Answer to Problem 72PPS

point of intersection is $(6,3.5)$

Explanation of Solution

Given information: The following graph is given

  

Formula used: Equation of a line joining points $\left(x_1,y_1\right),\left(x_2,y_2\right)$ is given by $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Calculation:

  

First find the equations of both the lines shown in the graph.

Take $\left(x_1,y_1\right)=(8,2),\left(x_2,y_2\right)=(0,8)$

Equation is calculated as follows

  $\begin{array}{l}y-2=\frac{8-2}{0-8}(x-8)\\ \Rightarrow y-2=\frac{-3}{4}(x-8)\\ \Rightarrow 4y-8=-3x+24\\ \Rightarrow 3x+4y=32(i)\end{array}$

Take $\left(x_1,y_1\right)=(2,2),\left(x_2,y_2\right)=(10,5)$

Equation is calculated as follows

  $\begin{array}{l}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\ y-2=\frac{5-2}{10-2}(x-2)\\ \Rightarrow y-2=\frac{3}{8}(x-2)\\ \Rightarrow 8y-16=3x-6\\ \Rightarrow 3x-8y=-10(ii)\end{array}$

Now find point of intersection of equations (i) and (ii).

Subtract (i) from (ii)

  $\begin{array}{l}(3x-8y)-(3x+4y)=-10-32\\ -12y=-42\\ y=\frac{42}{12}=3.5\end{array}$

Put $y=3.5$ in (i)

  $\begin{array}{l}3x+4(3.5)=32\\ \Rightarrow 3x+14=32\\ \Rightarrow 3x=32-14\\ \Rightarrow 3x=18\\ \Rightarrow x=6\end{array}$

So, point of intersection is $(6,3.5)$