Answer:
The triangle ABC is an isosceles right triangle
Step-by-step explanation:
we have
The coordinates of triangle ABC are
A (0, 2), B (2, 5), and C (−1, 7)
we know that
An isosceles triangle has two equal sides and two equal internal angles
The formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
step 1
Find the distance AB
substitute in the formula
[tex]d=\sqrt{(5-2)^{2}+(2-0)^{2}}[/tex]
[tex]d=\sqrt{(3)^{2}+(2)^{2}}[/tex]
[tex]dAB=\sqrt{13}\ units[/tex]
step 2
Find the distance BC
substitute in the formula
[tex]d=\sqrt{(7-5)^{2}+(-1-2)^{2}}[/tex]
[tex]d=\sqrt{(2)^{2}+(-3)^{2}}[/tex]
[tex]dBC=\sqrt{13}\ units[/tex]
step 3
Find the distance AC
substitute in the formula
[tex]d=\sqrt{(7-2)^{2}+(-1-0)^{2}}[/tex]
[tex]d=\sqrt{(5)^{2}+(-1)^{2}}[/tex]
[tex]dAC=\sqrt{26}\ units[/tex]
step 4
Compare the length sides
[tex]dAB=\sqrt{13}\ units[/tex]
[tex]dBC=\sqrt{13}\ units[/tex]
[tex]dAC=\sqrt{26}\ units[/tex]
[tex]dAB=dBC[/tex]
therefore
Is an isosceles triangle
Applying the Pythagoras Theorem
[tex](AC)^{2} =(AB)^{2}+(BC)^{2}[/tex]
substitute
[tex](\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}[/tex]
[tex]26=13+13[/tex]
[tex]26=26[/tex] -----> is true
therefore
Is an isosceles right triangle
