Question

In PAB,PA=PB and area of PAB= sq. units. Find the coordinates of P if coordinates of A and B are (1,2) and (3,8) respectively.

Answer 1

We have,

In $\Delta PAB,PA=PB$

Let, $P(a,b)$ be the point.

Suppose that

$A(x_1,y_1)=(1,2)$

$P(x_2,y_2)=(a,b)$

$C(x_3,y_3)=(3,8)$

\end{align}$

Then,

Given that,

$PA=PB$

$\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(x_2-x_3)}^2-{(y_2-y_3)}^2}$

$\Rightarrow \sqrt{{(a-1)}^2+{(b-2)}^2}=\sqrt{{(a-3)}^2-{(b-8)}^2}$

Squaring both side and we get,

${(a-1)}^2+{(b-2)}^2={(a-3)}^2+{(b-8)}^2$

$\Rightarrow a^2+1-2a+b^2+4-4b=a^2+9-6a+b^2+64-16b$

$\Rightarrow -2a-4b+5=-6a-16b+65$

$\Rightarrow -2a-4b+5+6a+16b-65=0$

$\Rightarrow 4a+12b-60=0$

$\Rightarrow a+3b-15=0......\left(1\right)$

But again given that,

$\text{Area}\text{of}\Delta \text{PAB=}\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$10=\frac{1}{2}[1(b-8)+a(8-2)+3(2-b)]$

$20=b-8+6a+6-3b$

$20+2=6a-2b$

$6a-2b=22$

$3a-b=11$

$3a-b-11=0......\left(2\right)$

By equation (1) and (2) to and we get,

$(a+3b=15)\times 3$

$3a-b=11$

$3a+9b=45$

$3a-b=11$

On subtracting and we get,

$10b=34$

$b=\frac{34}{10}$

$b=\frac{17}{5}$

Put the value of $b$ in equation (1) and we get,

$a+3b-15=0$

$\Rightarrow a+3\times \frac{17}{5}-15=0$

$\Rightarrow a+\frac{51}{5}-15=0$

$\Rightarrow a+\frac{51-75}{5}=0$

$\Rightarrow a-\frac{24}{5}=0$

$\Rightarrow a=\frac{24}{5}$

The value of $P(a,b)=(\frac{24}{5},\frac{17}{5})$.

Hence, this is the answer.