Question

The vertices of a triangle are $\left( -2,0 \right),\left( 2,3 \right)$and $\left( 1,-3 \right)$ then the type of the triangle is \[\]
A. Scalene\[\]
B. Equilateral\[\]
C. Isosceles \[\]
D. Right angled\[\]

Answer 1

Hint: We denote the vertices of the given triangle as $A\left( -2,0 \right),B\left( 2,3 \right)$and$C\left( 1,-3 \right)$. We find the length of the sides using the distance formula between two points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ in a plane as $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$. We recall the definitions of scalene, equilateral, isosceles and right angled triangle depending on relation among sides. We find a relation among the sides AB, BC, CA and choose the correct option.

Complete step by step answer:
Let us assume the vertices of the given triangle as $A\left( -2,0 \right),B\left( 2,3 \right)$and $C\left( 1,-3 \right)$ which means the sides are AB, BC, AC.\[\]

We know that in scalene none of the sides are equal which means $AB\ne BC\ne CA$. The equilateral triangle has all sides equal which means $AB=BC=CA$. The isosceles triangle has at least two sides equal which means $AB=BC$ or $BC=CA$ or $AB=CA$. The right angled triangle has the square of one side is the sum of squares of other two sides which means $A{{B}^{2}}+B{{C}^{2}}=C{{A}^{2}}$ or $A{{B}^{2}}+C{{A}^{2}}=B{{C}^{2}}$or$C{{A}^{2}}+B{{C}^{2}}=A{{B}^{2}}$.
We know from distance formula that the distance $d$ between points with coordinates $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ is given by,
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
Let us find the length of the sides using the distance formula. We have the side AB which has endpoints as vertices$A\left( -2,0 \right)=\left( {{x}_{1}},{{y}_{1}} \right),B\left( 2,3 \right)=\left( {{x}_{2}},{{y}_{2}} \right)$. So the length of AB is
\[AB=\sqrt{{{\left( 2-\left( -2 \right) \right)}^{2}}+{{\left( 3-0 \right)}^{2}}}=\sqrt{16+9}=\sqrt{25}=5\]
We have the side BC which has endpoints as vertices $B\left( 2,3 \right)=\left( {{x}_{1}},{{y}_{1}} \right),C\left( 1,-3 \right)=\left( {{x}_{2}},{{y}_{2}} \right)$. So the length of BC is
\[BC=\sqrt{{{\left( 1-2 \right)}^{2}}+{{\left( -3-3 \right)}^{2}}}=\sqrt{1+36}=\sqrt{37}=6.08\]
We have the side CA which has endpoints as vertices $A\left( -2,0 \right)=\left( {{x}_{2}},{{y}_{2}} \right),C\left( 1,-3 \right)=\left( {{x}_{1}},{{y}_{1}} \right)$. So the length of AC is
\[CA=\sqrt{{{\left( -2-1 \right)}^{2}}+{{\left( 0-\left( -3 \right) \right)}^{2}}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}=4.24\]
We observe that
\[AB\ne BC\ne CA\]

So, the correct answer is “Option A”.

Note: We can alternatively solve using the slopes of the sides AB, BC, CA. We find the slopes with endpoints $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ as $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. Then we find the angle between two sides with slopes say ${{m}_{1}},{{m}_{2}}$ as ${{\tan }^{-1}}\left( \dfrac{{{m}_{1}}-{{m}_{2}}}{1-{{m}_{1}}{{m}_{2}}} \right)$. We use the definitions of triangles in the options in terms of relation among angles to choose the correct option.