Question

ABC is an isosceles triangle such that A(-3,0) B(3,0) and AC=BC=5. Then the coordinates of C could be ​

Answer 1

Answer:

the coordinates of C(0, 4)

Step-by-step explanation:

Given that ΔABC is an isosceles

coordinates of triangle = A(-3, 0)  B(3, 0)

and AC = BC = 5

We need to find coordinates of C

let  (a, b) be the coordinates of C

from data AC = BC =5

here AC is distance between A(-3, 0) and C(a, b) = 5

        BC is distance between B(3, 0) and C(a, b) = 5

now we will find out AC and BC with C(a, b)

the distance between points (x₁, y₁) (x₂, y₂) = $\sqrt{(x_{2}- x_{1})^{2} +(y_{2} -y_{1} )^{2} }$  

distance between A(-3,0) and (a, b)

⇒ $\sqrt{(a-(-3))^{2} + (b-0)^{2} } =5$

⇒ $\sqrt{(a+3)^{2} + b^{2} } = 5$  

⇒ $(\sqrt{a^{2} +9+6a+b^{2} )} = 5$       [squaring on both sides ]

⇒ $a^{2} +b^{2}+6a+9 =25$

⇒ $a^{2} +b^{2} +6a =16$_ (1)

distance between B(3, 0) and C(a, b)

⇒ $\sqrt{(a-3)^{2} +(b-0)^{2} } =5$

⇒ $\sqrt{a^{2}+9-6a+b^{2} } =5$  

⇒ $\sqrt{a^{2}+b^{2} -6a+9 } =5$        

⇒ $(\sqrt{a^{2}+b^{2} -6a +9 } )^{2} = 5^{2}$          [squaring on both sides ]

⇒  $a^{2} +b^{2} -6a+9 =25$

⇒ $a^{2} +b^{2} -6a =16$ _(2)

do (1) -(2) ⇒  $a^{2} +b^{2} +6a -(a^{2} +b^{2} -6a ) = 16 -16$

               ⇒ $a^{2} +b^{2} +6a - a^{2}-b^{2} +6a =0$

               ⇒ $12a =0$

               ⇒ $a=0$

$a =0$ substitute in (1) ⇒ $(0)^{2}+b^{2}+6(0) =16$

                                  ⇒ $b^{2} =16$

                                  ⇒ $b=4$

the coordinates  C(a, b) = (0, 4)