Question

The points $A\left( {12,8} \right)$, $B\left( { - 2,6} \right)$ and $C\left( {6,0} \right)$ are vertices of
A. Right-angled triangle
B. Isosceles triangle
C. Equilateral triangle
D. None of these

Answer 1

We need to find the distance between the points. This distance is nothing but the length of that side of that triangle. These lengths will help us in identifying the type of triangle. To find the distance between the points using the distance formula.
Formula used:
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$

Complete step by step answer:

Given vertices of that triangle are $A\left( {12,8} \right),B\left( { - 2,6} \right) and C\left( {6,0} \right)$.
Now let’s find the distance between them one by one.
That is distance AB, BC and CA.
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$

d(AB) = \sqrt {{{\left( { - 2 - 12} \right)}^2} + {{\left( {6 - 8} \right)}^2}} \\\ \Rightarrow \sqrt {{{( - 14)}^2} + {{( - 2)}^2}} \\\ \Rightarrow \sqrt {196 + 4} \\\ \Rightarrow \sqrt {200} \\\ \Rightarrow 10\sqrt 2 \\\ \end{gathered} $$ Then, $$\begin{gathered} d(BC) = \sqrt {{{\left( {6 - ( - 2)} \right)}^2} + {{\left( {0 - 6} \right)}^2}} \\\ \Rightarrow \sqrt {{{(8)}^2} + {{( - 6)}^2}} \\\ \Rightarrow \sqrt {64 + 36} \\\ \Rightarrow \sqrt {100} \\\ \Rightarrow 10 \\\ \end{gathered} $$ And at last, $$\begin{gathered} d(CA) = \sqrt {{{\left( {6 - 12} \right)}^2} + {{(0 - 8)}^2}} \\\ \Rightarrow \sqrt {{{( - 6)}^2} + {{( - 8)}^2}} \\\ \Rightarrow \sqrt {36 + 64} \\\ \Rightarrow \sqrt {100} \\\ \Rightarrow 10 \\\ \end{gathered} $$ Since two lengths are same that must be an isosceles triangle. Thus, option b is correct.  **Note:** Students should find all the distances or we can say lengths of sides of the triangle. They should not judge the type of triangle just by finding any two sides of the triangle. In problems like these if you get all the three distances equal then it is an equilateral triangle. To find whether it is a right-angled triangle using the Pythagoras theorem. $${(hypt)^2} = {(base)^2} + {(height)^2}$$. One side will be your hypotenuse and the remaining two will form base and height. **Additional information:** An equilateral triangle is a triangle having all sides equal. Each angle of the equilateral triangle is of $${60^ \circ }$$. An isosceles triangle is a triangle having any two sides of the same length. The base angles of the triangle are the same. The right-angled triangle is having one of the angles $${90^ \circ }$$ and the remaining two are acute angles.