Question

an isosceles triangle is such that AC=BC and has vertices A=(3,4), B=(7,4) and C=(5,8) a) Calculate the length of AC B) the line of symmetry of the triangle meets the line AB at M what are the coordinates of M​

Answer 1

Explanation:

A( 3 , 4) & C(5 , 8)

Distance =
`\sqrt{(x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2)}`

`AC = \sqrt{(5-3)^(2)+(8-4)^(2)}\\\\=\sqrt{(2)^(2)+(4)^(2)}\\\\=√(4+16)\\\\=√(20)\\\\= 4.47 units`

M is the midpoint of AB

A(3,4) &B(7,4)

`M((x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2))\\\\M((3+7)/(2),(4+4)/(2))\\\\M((10)/(2),(8)/(2))`

M(5,4)