Question

Two vertices of isosceles triangle (2,8) and (10,2) find third vertex if equal sides 10 units

Answer 1

Answer:

let the third vertex is A(x,y)

and the given vertex are of base of isosceles Triangle B(2,8) and C(10,2)

As we know that two sides of isosceles Triangle are equal

Thus we calculate AB and AC by distance formula

AB=AC

$\sqrt{( {x - 2)}^{2} + ( {y - 8)}^{2} }$ = $\sqrt{( {x - 10)}^{2} + ( {y - 2)}^{2} } \\ \\ ( {x - 2)}^{2} + ( {y - 8)}^{2}$ = $( {x - 10)}^{2} + ( {y - 2)}^{2} \\ \\ {x}^{2} - 4x + 4 + {y}^{2} - 16y + 64 = {x}^{2} - 20x + 100 + {y}^{2} - 4y + 4 \\ \\ - 4x - 16y + 20x + 4y$ = $104 - 68 \\ \\ 16x - 12x =36 \\ \\ 4x - 3y$ = $9 \\ \\ 4x = 9 + 3y \\ \\ x = \frac{9 + 3y}{4}$

put this value in

{x}^{2} - 20x + 100 + {y}^{2} - 4y + 4 = $100 \\ \\ since \: \: ab = 10 \: units \\ \\ ( { \frac{9 + 3y}{4} })^{2} - 20( \frac{9 + 3y}{4} ) + {y}^{2} - 4y + 4 = 0 \\ \\ \frac{1}{16} (81 + 54y+ 9 {y}^{2} ) - 45 - 15y + {y}^{2} - 4y + 4$ = $0 \\ \\ 81 + 54y + 9 {y}^{2} - 16(45 - 15y + {y}^{2} - 4y + 4) = 0 \\ \\ - 7 {y}^{2} + 54y + 16 \times19y + 81 - 16 \times 49 = 0 \\ \\- 7 {y}^{2}+54y+304y+81-784=0\\\\\\- 7 {y}^{2}+358y-703=0\\\\y$=$\frac{-358±\sqrt{358^2+4(7)(703)}}{-14}$

by this way we can solve the value of y by solving Quadratic equation.

thus value of x