Question

Prove that the points $(-3,0),(1,-3)$ and $(4,1)$ are the vertices of an isosceles right-angled triangle. Find the area of this triangle

Answer 1

Let A(-3,0), B(1,-3) and C(4,1) by the given points. Then,
$AB=\sqrt{{\{1-(-3)\}}^2+{(-3-0)}^2}=\sqrt{4^2+{(-3)}^2}=\sqrt{16+9}=5units$
$BC=\sqrt{{(4-1)}^2+{(1+3)}^2}=\sqrt{19+16}=5units$
and, $CA=\sqrt{{(4+3)}^2+{(1-0)}^2}=\sqrt{49+1}=5\sqrt{2}unit$
Clearly, AB=BC. Therefore, $△$ABC is isosceles.
Also, ${AB}^2+{BC}^2=25+25={\left(5\sqrt{2}\right)}^2={CA}^2$
$\Rightarrow$ $△$ ABC is right-angled at B.
Thus, $△$ ABC is right-angled isosceles triangle.
Now, Area of $△$ ABC $=\frac{1}{2}(Base\times Height)=\frac{1}{2}(AB\times BC)$
$\Rightarrow$ Area of $△$ ABC=$(\frac{1}{2}\times 5\times 5)$sq. Units=$\frac{25}{2}$sq. Units