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# Prove that the points (0, 0), (5, 5) and (-5, 5) are the vertices of a right isosceles triangle.

Last updated date: 17th Jun 2024
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Hint: Here in this question basic properties of isosceles triangle, right triangle and distance formula will get used.
Property of isosceles triangle: - Two sides and opposite angles corresponding to them of an isosceles triangle are equal in magnitude.
Right triangle: - A triangle in which one angle is 90 degrees is termed as right triangle. With the help of Pythagoras theorem we can prove it. Below is the formula of Pythagoras theorem.
In a triangle ABC three sides are there AB, BC and AC and B is 90 degree. So Pythagoras theorem states that ${(AB)^2} + {(BC)^2} = {(AC)^2}$
Distance formula: -We will use distance formula between the two points ${x_1},{y_1}$ and ${x_2},{y_2}$ that is mentioned below: -
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ d= distance between two points.

Complete step-by-step solution:
Let points given be named as A (0, 0), B (5, 5) and C (-5, 5)
To prove: -
For isosceles triangle two sides must be equal so required conditions to satisfied are AB=BC or BC=CA or CA=AB
For the right triangle Pythagoras theorem must get satisfied.
Proof: - First of all we will find length of AB, BC and AC
Points for AB are A (0, 0) and B (5, 5)
$\Rightarrow AB = \sqrt {{{[5 - 0]}^2} + {{[5 - 0]}^2}}$ (Putting values in distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ )
$\Rightarrow AB = \sqrt {{{[5]}^2} + {{[5]}^2}}$
$\Rightarrow AB = \sqrt {25 + 25}$
$\Rightarrow AB = \sqrt {50}$ (Finding square root of 50)
$\therefore AB = 7.07$
Points for BC are B (5, 5) and C (-5, 5)
$\Rightarrow BC = \sqrt {{{[ - 5 - 5]}^2} + {{[5 - 5]}^2}}$ (Putting values in distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ )
$\Rightarrow BC = \sqrt {{{[ - 10]}^2} + {{[0]}^2}}$
$\Rightarrow BC = \sqrt {100}$ (Finding square root of 100)
$\therefore BC = 10$
Points for AC are A (0, 0) and C (-5, 5)
$\Rightarrow AC = \sqrt {{{[ - 5 - 0]}^2} + {{[ - 5 - 0]}^2}}$ (Putting values in distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ )
$\Rightarrow AC = \sqrt {{{[ - 5]}^2} + {{[ - 5]}^2}}$
$\Rightarrow AC = \sqrt {25 + 25}$
$\Rightarrow AC = \sqrt {50}$ (Finding square root of 50)
$\therefore AC = 7.07$
AB=AC=7.07
Therefore from the three conditions AB=BC or BC=CA or CA=AB one which AB=AC is satisfied so given vertices are of isosceles triangle.
Now for right angled triangle Pythagoras theorem must get satisfied
${(AB)^2} + {(BC)^2} = {(AC)^2}$
$\Rightarrow {(\sqrt {50} )^2} + {(10)^2} \ne {(\sqrt {50} )^2}$
$\therefore 50 + 100 \ne 50$
${(AB)^2} + {(AC)^2} = {(BC)^2}$
$\Rightarrow {(\sqrt {50} )^2} + {(\sqrt {50} )^2} = {(10)^2}$
$\therefore 50 + 50 = 100$
${(AC)^2} + {(BC)^2} = {(AB)^2}$
$\Rightarrow {(\sqrt {50} )^2} + {(10)^2} \ne {(\sqrt {50} )^2}$
$\therefore 50 + 100 \ne 50$

As one of the condition got satisfied i.e. ${(AB)^2} + {(AC)^2} = {(BC)^2}$ , therefore the given isosceles triangle is a right angled triangle.

Note: Students must well aware of all the properties of triangles so that they can prove these types of questions and they must apply distance formula carefully as the common mistake which is done by most of the students is that they get confused between subtraction sign instead of addition sign in distance formula