Answer:
- AB = AC = 5 units
- BC = 6 units
- No it is not a equilateral triangle
Step-by-step explanation:
Some coordinates of a triangle is given to us. And we need to find the length of AB , BC and AC . The given coordinates are ,
[tex]\rm\implies Coordinates = A(-1,3) , B(2,-1) \ and \ C(-4,-1) [/tex]
We can use the Distance Formula to find the length of sides of the triangle .
Finding length of AB :-
[tex]\rm\implies Distance =\sqrt{ ( x_2-x_1)^2+(y_2-y_1)^2} \\\\\rm\implies \overline{AB}= \sqrt{ (-1-2)^2+(-1-3)^2} \\\\\rm\implies \overline{AB}= \sqrt{ (-3)^2+(-4)^2} \\\\\rm\implies \overline{AB}= \sqrt{ 9 + 16} \\\\\rm\implies \overline{AB}= \sqrt{25} \\\\\rm\implies\boxed{\red{\rm \overline{AB}= 5 \ units}}[/tex]
Finding length of AC :-
[tex]\rm\implies Distance =\sqrt{ ( x_2-x_1)^2+(y_2-y_1)^2} \\\\\rm\implies \overline{AC}= \sqrt{ (-1+4)^2+(-1-3)^2 }\\\\\rm\implies \overline{AC}= \sqrt{ (-3)^2+(-4)^2} \\\\\rm\implies \overline{AC}= \sqrt{ 9 + 16} \\\\\rm\implies \overline{AC}= \sqrt{25} \\\\\rm\implies\boxed{\red{\rm \overline{AC}= 5 \ units}}[/tex]
Finding length of BC :-
[tex]\rm\implies Distance =\sqrt{ ( x_2-x_1)^2+(y_2-y_1)^2} \\\\\rm\implies \overline{BC}= \sqrt{ (-4-2)^2+(-1-1)^2} \\\\\rm\implies \overline{BC}= \sqrt{ 6^2+0^2} \\\\\rm\implies \overline{BC}= \sqrt{ 36+0} \\\\\rm\implies \overline{BC}= \sqrt{36} \\\\\rm\implies\boxed{\red{\rm \overline{BC}= 6 \ units}}[/tex]
Is it a Equilateral triangle ?
No the given triangle is not a equilateral triangle since the measure of all sides is not same.
