Question

Two vertices of a issosceles triangle are (2,0) and (2,5) find the third vertex if the length of the equal sides is 3

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Coordinate\:of\:C=(0.35,2.5)}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{Given : } \\ \implies \text{Coordinate \: of \:A=(2,0)} \\ \\ \implies \text{Coordinate \: of \: B=(2,5)} \\ \\ \implies \text{AC= BC= 3 units} \\ \\ \underline \bold{To \: Find : } \\\implies Coordinate\:of\:C=?$

$\bold{Using \: Distance \: formula : } \\ \implies \text{AB=BC} \\ \\ \implies \sqrt{(2 - x) ^{2} + (0 - y)^{2} } = \sqrt{(2 -x) ^{2} +( 5 - y)^{2} } \\ \\ \text{Squaring \: both \: side : } \\ \implies \cancel{( 2 - x)^{2}} + {y}^{2} = \cancel {{(2 - x)}^{2}} + {(5 - y)}^{2} \\ \\ \implies \cancel {y}^{2} = 25 + \cancel{y}^{2} - 10y \\ \\ \implies 25 - 10y = 0 \\ \\ \implies - 10y = - 25 \\ \\ \implies \text{y = 2.5} \\ \\ \text{Using \: Distance \: formula : } \\ \implies AC = \sqrt{( x_{2} - x_{1})^{2} + ( y_{2} - y_{1})^{2} } \\ \\ \implies 3 = \sqrt{(2 - x)^{2} + (0 - 2.5)^{2} } \\ \\ \implies 9 = (2 - x) ^{2} + 6.25 \\ \\ \implies 9 - 6.25 = ({2 - x)}^{2} \\ \\ \implies \sqrt{ 2.75} = 2 - x \\ \\ \implies 1.65 = 2 - x \\ \\ \implies x = 2 - 1.65 \\ \\ \implies \text{x =0.35 } \\ \\ \therefore \text{Coordinate \: of \: C= (0.35,2.5)}$

Answer 2

$\huge{\underline{\sf{\red{Answer}}}}$

Given:

Isosceles Triangle

Coordinates : A(2,0) B(2,5) C(x,y)

Distances of Equal Sides = 3 units

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$\star$ Formulae

• Distance Formula PQ= $\sqrt{(x_2-x_1 )^2 + (y_2-y_1)^2}$
• Quadratic Formula $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}$

Solution:

Distance AB = $\sqrt{(x_2-x_1 )^2 + (y_2-y_1)^2}$

here,$x_1=2,x_2=2,y_1=0,y_2=5$

$\leadsto$ $\sqrt{(2-2 )^2 + (5-0)^2}$

$\leadsto$ $\sqrt{0 + 5^2}$

$\implies$ AB= 5

from here we know that

AC = BC = 3

so

By applying the Distance Formula

$\sqrt{(x -2 )^2 + (y-0)^2}$ = $\sqrt{(x-2 )^2 + (y-5)^2}$

Squaring Both the Sides

$\leadsto$ $\cancel{(x-2)^2 }+ y^2 =\cancel{(x-2)^2} + (y-5)^2 =3$

$\implies$ $\cancel{y^2}= \cancel{y^2}-10y +25$

$\implies$ 10y=25

$\leadsto$ y= $\dfrac{5}{2}$

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AC = 3

$\sqrt{(x-2)^2 +y^2}=3$

Squaring Both the sides

$\implies$ $(x-2)^2 +y^2 =9$

$\implies$ $x^2 -4x +4 +y^2 =9$

putting value of y

$\leadsto$ $x^2 - 4x-5+\bigg(\dfrac{5}{2}\bigg)^2$

$\implies$ $x^2 - 4x-5+\bigg(\dfrac{25}{4}\bigg)$

$\implies$ $x^2 - 4x+\bigg(\dfrac{-20+25}{4}\bigg)$

$\implies$ $x^2 - 4x+\bigg(\dfrac{5}{4}\bigg)$

When we compare

We get

a = 1 ,b = -4 , c= 5/4

using Quadratic Formula

$\implies$ x=$\dfrac{4\pm\sqrt{4^2-4\times1\times\dfrac{5}{4}}{2\times1}}$

$\implies$ x= $\dfrac{4\pm\sqrt{16-5}}{2}$

$\implies$ $\large{\boxed{\dfrac{2\pm\sqrt{11}}{2}}}$

so the coordinates will be

$\large{\boxed{\dfrac{2\pm\sqrt{11}}{2},\dfrac{5}{2}}}$