The approach will leave its marks

Geometry Level 4

$\begin{cases} (x_1-x_2)^2+(y_1-y_2)^2=9 \\ (x_2-x_3)^2+(y_2-y_3)^2=16 \\ (x_1-x_3)^2+(y_1-y_3)^2=25 \end{cases}$

If above 3 conditions hold simultaneously,

then find the value of $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2$

Try the set : Practice Problems for JEE-Mains and Give a final Boost to your JEE_preparation. All the best!

The answer is 144.

2 solutions

Sandeep Bhardwaj
Mar 31, 2015

Let $A \equiv (x_1,y_1) \quad , B \equiv (x_2,y_2) \quad and \quad C \equiv (x_3,y_3)$

$\therefore \quad AB=3 \quad , BC=4 \quad and \quad CA=5$

$\because \text{area of} \ \ \Delta ABC = \dfrac{1}{2} \cdot \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \dfrac{1}{2} \times 3 \times 4$

$(\Delta ABC \ \text{is right angled.})$

$\implies \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2=(12)^2=\Large \boxed{144}$

enjoy !

Good one, did exactly the same . (:

Aniket Verma - 6 years, 2 months ago

did the same thing :)

Krishna Ramesh - 6 years, 1 month ago

Heer Shah
Apr 7, 2015

put x1=3 x2=0 x3=0 y1=y2=0 and y3=4

The approach will leave its marks

Try the set : Practice Problems for JEE-Mains and Give a final Boost to your JEE_preparation. All the best!

The answer is 144.

2 solutions

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