If the coordinates of two points are A(3,4), B(5,−2), and a point P (x,5) is such that PA=PB, then find the area of  ΔPAB. 

 

 

 

 

 

Given that,  

The coordinates of two points are A(3,4), B(5,−2) 

$\begin{array}{r}P(x,5)\text{ is a point }\\ PA=PB \\ \Rightarrow P{A}^2=P{B}^2 \end{array}$

We know that the distance between 𝐴(𝑥₁, 𝑦₁) and 𝐵(𝑥₂, 𝑦₂) is given by

$AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\text{ units }$

The distance between 𝑃(𝑥, 5) and 𝐴(3, 4), 𝑃(𝑥, 5) and 𝐵(5, − 2) is

$\begin{array}{l}\Rightarrow P{A}^2=P{B}^2\\ \Rightarrow (x-3{)}^2+(5-4{)}^2=(x-5{)}^2+(5+2{)}^2\\ \Rightarrow x^2-6x+9+1=x^2-10x+25+49\\ \Rightarrow -6x+10=-10x+74\\ \Rightarrow 4x=64\\ \Rightarrow x=16\\ \Rightarrow P(x,5)=P(16,5)\end{array}$

$\text{ Area of triangle }=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|\text{ sq units }$

$A(3,4),B(5,-2),P(16,5)$

$\mathrm{area} \left(\mathrm{\Delta }PAB\right)=\frac{1}{2}\left|3\right(-2-5)+5(5-16)+16(4+2\left)\right|$

$\begin{array}{l}\text{=}\frac{\text{1}}{\text{2}}\text{|}\text{3}\text{(}\text{−}\text{7}\text{)}\text{+}\text{5}\text{(}\text{1}\text{)}\text{+}\text{16}\text{(}\text{6}\text{)}\text{|}\\ \text{=}\frac{\text{1}}{\text{2}}\text{|}\text{−}\text{21}\text{+}\text{5}\text{+}\text{96}\text{|}\\ \text{=}\frac{\text{1}}{\text{2}}\text{|}\text{80}\text{|}\\ \text{=}\text{40}\text{ sq units }\end{array}$

Hence, the area of ∆𝑃𝐴𝐵 = 40 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠