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.
[4 MARKS]

Answer 1

Formula: 1 Mark
Concept of right-angled triangles: 1 Mark
Steps: 1 Mark
Answer: 1 Mark

Let A(-3, 0), B(1, -3) and C(4, 1) be the given points. Then,

Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

$AB=\sqrt{(1-{(-3))}^2+{(-3-0)}^2}=\sqrt{16+9}=5units$.

$BC=\sqrt{{(4-1)}^2+{(1+3)}^2}=\sqrt{9+16}=5units$.

$AC=\sqrt{{(4+3)}^2+{(1-0)}^2}=\sqrt{49+1}=5\sqrt{2}$ units.

Clearly, AB = BC. Therefore, $\Delta ABC$ is isosceles.

Also,

${AB}^2+{BC}^2=25+25=({5\sqrt{2})}^2={CA}^2$

$\Rightarrow \Delta ABC$ is right- angled at B because it is satisfying pythagoras theorem.

Thus, $\Delta ABC$ is right- angled isosceles triangle.

Now, Area of $\Delta ABC=\frac{1}{2}(Base\times Height)$

$=\frac{1}{2}(AB\times BC)$

$\Rightarrow Areaof\Delta ABC=(\frac{1}{2}\times AB\times BC)sq.units$

$=\frac{25}{2}sq.units$.