Question

ABC is an isosceles triangle with coordinates of B and C as B( -4,0) and C(4,0) and
AB=AC=5 units. Find the coordinates of the vertex A.
PLEASE DON'T SPAM OR I WILL REPORT YOU

Answer 1

Answer:

A(xy),B(-4,0),C(40)

Step-by-step explanation:

$ab = \sqrt{(x + 4)^{2} + {(y - 0)}^{2} } \\ ac = \sqrt{ {(x - 4)}^{2} + (y - 0) ^{2} } \\ but \: ab = ac \\ {(x + 4)}^{2} + {y}^{2} = {(x - 4)}^{2} + {y}^{2} \\ {x}^{2} + 8x + 16 = {x}^{2} - 8x + 16 \\ 16x = 0 \\ x = 0 \\ ac ^{2} = {x}^{2} - 8x + 16 + y^{2} \\ 25 = 16 + {y }^{2} \\ y = 3 \\a (x.y) = (0.3)$

Answer 2

Answer:

(0,±3)

Step-by-step explanation:

Let the coordinate of vertex A of triangle ABC be (x,y)

We know, AB=5

Hence,

$5=\sqrt{(x+4)^2 + (y-0)^2$

$25= x^2 + 16 + 8x + y^2$

$x^2 + y^2 + 8x - 9 =0$ ........ (i)

Now, AC = 5

Hence,

$5=\sqrt{(x-4)^2 + (y-0)^2}$

$25 = x^2 + 16 - 8x + y^2$

$x^2 + y^2 - 8x - 9 = 0$ .......... (ii)

From (i) and (ii)

We get, 16x=0

Hence, x = 0

Now, putting the value of x into equation (i),

We get,

$y^2 = 9$

$y=\sqrt{9}$

Hence y = ±3

Hence, coordinates of vertex A are (0,±3)

Hope This Helps :)