Question

The two vertices of an isosceles triangle (2,0) and (2,5). Find the third vertex when the length of the equal sides is 3.

Answer 1

Hint: The Isosceles triangle has two of its sides equal.
In order to solve the above problem we will use the concept of calculating distance using distance formula which is given as;
$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ (${x_1}{x_2}{y_1}{y_2}$ are the coordinates of x and y respectively)
We will also use the algebraic identity which is written as;
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

Complete step-by-step solution:
Let us first draw the figure which will represent two given vertices .

We will take up point 2, 0 first.
$ \Rightarrow \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 0} \right)}^2}} = 3$ (as the distance of the side is 3)
$ \Rightarrow \sqrt {{{\left( {x - 2} \right)}^2} + {y^2}} = 3$
On removing square
$ \Rightarrow {x^2} + 4 - 4x + {y^2} = 9$
Rearranging the terms
$ \Rightarrow {x^2} - 4x + {y^2} = 5$
$ \Rightarrow {y^2} = 5 - {x^2} + 4x$ ( calculated the value of y2)...........................(1)
Now, we will take up the points 2,5 are equal and 2,0 and equate them as the two distances
$ \Rightarrow \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 5} \right)}^2}} = \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 0} \right)}^2}} $ (as the distance of the side is 3)
$ \Rightarrow {\left( {x - 2} \right)^2} + \left( {y - 5} \right) = {\left( {x - 2} \right)^2} + {\left( {y - 0} \right)^2}$ (squaring both the sides)
$ \Rightarrow {x^2} + 4 - 4x + {y^2} + 25 - 10y = {x^2} + 4 - 4x + {y^2}$
$ \Rightarrow 25 - 10y = 0$ (cancel the common terms from both RHS and LHS)
$ \Rightarrow y = \dfrac{5}{2}$ ...................(2) ( one of the point of the third vertex of the triangle)
We will substitute the value of y from equation 2 in equation 1
$ \Rightarrow {\left( {\dfrac{5}{2}} \right)^2} = 5 - {x^2} + 4x$
$ \Rightarrow {x^2} - 4x + \dfrac{5}{4} = 0$ (We have rearranged the terms)................(3)
Equation 3 is a quadratic equation which is solved using formula
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ (coefficient of x2 is a, coefficient of x is b , c is the constant term)
$ \Rightarrow 4{x^2} - 16x + 5 = 0$ (We have taken LCM)
$ \Rightarrow x = \dfrac{{ - ( - 16) \pm \sqrt {{{16}^2} - 4 \times 4 \times 5} }}{{2 \times 4}}$
$ \Rightarrow x = \dfrac{{16 \pm 4\sqrt {11} }}{8}$ (we have reduced the term inside the square root)
$ \Rightarrow x = 2 \pm \dfrac{{\sqrt {11} }}{2}$

Therefore, third vertices of the triangle are $2 \pm \dfrac{{\sqrt {11} }}{2},\dfrac{5}{2}$.

Note: Triangles are of many other types such as equilateral triangle having all the sides equal and all the angles have $60^\circ$ value, scalene triangle having all the sides of different lengths, right triangle having one of the angles equal to $90^\circ$ with one side as the longest side.