Answer
The triangle $△ABC$ is shown below in the image below.
$△ABC$ is a right triangle as the square of the length of the longest side is equal to the sum of the squares of the length of two other sides. Refer to the solution below.
The area of $△ABC$ is, $100$ square units
Work Step by Step
Step 1
Plot the points $A(−2,5), B(12,3)$ and $C(10,-11)$, then connect them to form $△ABC.$ The image is given below.
Step 2
Find the length of each side using the distance formula
$d=\sqrt {({x_{1}-x_{2})}^{2}+{(y_{1}-y_{2})}^{2})}$
So,
${AB}=\sqrt{{((-2)-(12))}^{2}+{((5)-(3))}^{2}}$
$=\sqrt{(-14)^{2}+(2)^{2}}$
$=\sqrt{196 + 4}$
$=\sqrt{200}$
${AB}=10\sqrt2$
$AC=\sqrt{{(-2)-(10))}^{2}+{((5)-(-11))}^{2}}$
$=\sqrt{(-12)^{2}+(16)^{2}}$
$=\sqrt{144 + 256}$
$=\sqrt{400}$
${AC}= 20$
$BC=\sqrt{{(12-10)}^{2}+{(3-(-11))}^{2}}$
$=\sqrt{(2)^{2}+(14)^{2}}$
$=\sqrt{4 + 196}$
$=\sqrt{200}$
${BC}=10\sqrt2$
Step 3
A triangle is a right triangle if the square of the length of the longest side is equal to the sum of the square of the length of two other sides.
So here we can see,
${AC}^{2}={AB}^{2}+{BC}^{2}$
$400=200+200$
$400=400$
So,$ △ABC$ is a right triangle.
Step 4
The area $A$ of a triangle is given by the formula
$A=\frac{1}{2}bh$ [ Here, $b$ = length of the base and $h$ = height ]
Here for $△ABC$,
$b=BC=10\sqrt2$
$h=AB=10\sqrt2$
Therefore, the area of $△ABC$ is,
$A=\frac{1}{2}.10\sqrt 2.\cdot10 \sqrt2$
$A=\frac{1}{2} \cdot 100(2)$
$A=100$ square units
