Question

prove that (3,0) (5,5) and (-1+3) are the vertices of an isosceles triangle with one angle 90​

Answer 1

Hence proved, is right angled isosceles.

Answer 2

Answer:

Step-by-step explanation:

A(3, 0), B(5, 5), C(- 1, 3)

AB = $\sqrt{(x_{2} - x_{1} )^2 +(y_{2} -y_{1})^2 }$ = √[(5 - 3)² + (5 - 0)²] = √29

BC = √[(- 1 - 5)² + (3 - 5)²] = 2√10

AC = √[(3 - 0)² + (- 1 - 3)²] = √25 = 5

AB ≠ BC ≠ AC

ΔABC is scalene and not an isosceles.

$m_{AB}$ = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$ = $\frac{5-0}{5-3}$ = $\frac{5}{2}$

$m_{BC}$ = $\frac{3-5}{-1-5}$ = $\frac{1}{3}$

$m_{AC}$ = $\frac{3-0}{-1-3}$ = - $\frac{3}{4}$

ΔABC is an acute and not a right angled.