A) If the triangle has vertices A(0,2), B(2,5) and C(-1,7)
To find the length we can use a formula, to find the length of each leg derived from the Pythagorean Theoream
[tex]d_{AB}=\sqrt{(y_2-y_1)^2+(x_2-x_2)^2}[/tex]So Plugging in the values from each Coordinate points for A (0,2) and B (2,5)
We have:
[tex]\begin{gathered} d_{AB=}\sqrt{(5-2)^2+(2-0)^2} \\ d_{AB}=\sqrt{9+4} \\ d_{AB\text{ =}}\text{ }\sqrt{13} \end{gathered}[/tex]Now let's proceed to find the distance of the leg BC
Since B (2,5) and C(-1,7). Similarly:
[tex]\begin{gathered} d_{BC\text{ = }}\sqrt{(7-5)^2+(-1-2)^2\text{ }} \\ d_{BC\text{ = }}\text{ }\sqrt{4+9} \\ d_{BC\text{ =}}\text{ }\sqrt{13} \end{gathered}[/tex]Finally, for the last leg. A(0,2) and C(-1,7)
[tex]\begin{gathered} d_{AC}=\sqrt{(7-2)^2+(-1-0)^2} \\ d_{AC=\text{ }}\sqrt{26} \end{gathered}[/tex]So we have an isosceles triangle whose legs have size above.
