Question

prove that the points (- 6 - 8) (- 16 12) and( - 26 - 18) are the vertices of an isosceles right angled triangle​

Answer 1

The answer is "True".

Step-by-step explanation:

Given points:

`A= (-6,-8)\\B=(-16, 12)\\C=(-26, -18)\\`

Condition of the right-angle triangle is:
`\bold{AC^2= AB^2+BC^2}`

Formula to find length:

`Length= √((x_2-x_1)^2+(y_2-y_1)^2)`

`AC = (-6,-8)(-26,-18)\\\\x_1= -6 \ \ y_1=-8 \ \ \ and \ \ \ x_2=-26 \ \ y_2=-18\\`

`\Rightarrow AC= √((-26-(-6))^2+(-18-(-8))^2)\\\\\Rightarrow AC= √((-26+6))^2+(-18+8))^2)\\\\\Rightarrow AC= √((-20))^2+(-10))^2)\\\\\Rightarrow AC= √(400+100)\\\\\Rightarrow AC= √(500)\\\\`

`AB = (-6,-8)(-16,12)\\\\x_1= -6 \ \ y_1=-8 \ \ \ and \ \ \ x_2=-16 \ \ y_2=12\\`

`\Rightarrow AB= √((-16-(-6))^2+(12-(-8))^2)\\\\\Rightarrow AB= √((-16+6))^2+(12+8))^2)\\\\\Rightarrow AB= √((-10))^2+(20))^2)\\\\\Rightarrow AB= √(100+400)\\\\\Rightarrow AB= √(500)\\\\`

`BC = (-16,12)(-26,-18)\\\\x_1= -16 \ \ y_1=12 \ \ \ and \ \ \ x_2=-26 \ \ y_2=-18\\`

`\Rightarrow BC= √((-26-(-16))^2+(-18-12)^2)\\\\\Rightarrow BC= √((-26+16))^2+(-30)^2)\\\\\Rightarrow BC= √((-10)^2+(-30)^2)\\\\\Rightarrow BC= √(100+900)\\\\\Rightarrow BC= √(1000)\\\\`

`\Rightarrow {AC^2= AB^2+BC^2}\\\\\Rightarrow (√(1000))^2= (√(500+500))^2\\\\\Rightarrow 1000=1000\\`

The angle is right-angled triangle