First, we need to remember that the distance between two points (x1, y1) and (x2, y2) can be calculated with √[ (x1 - x2)^2 + (y1 - y2)^2 ].
Thus, we apply this formula to measure the lengths of AB, BC, and AC in ∆ABC.
AB = √[ (1 - -2)^2 + (7 - 2)^2 ] = √25 = 5 units
BC = √[ (-2 - 4)^2 + (2 - 2)^2 ] = √36 = 6 units
CA = √[ (4 - 1)^2 + (2 - 7)^2 ] = √25 = 5 units
From this, we can clearly see that BC is the longest side of ∆ABC with a length of √36 = 6 units. Thus, the answer is B: 6.
Since ∆ABC sides 5, 5, and 6. That makes it an isosceles triangle. Which makes the right answer to be B: isosceles.
Now, if we form a new triangle, ∆ABD, with D at (1, 2), we have the following lengths:
AB = 5 units
BD = √[ (-2 - 1)^2 + (2 - 2)^2 ] = √9 = 3 units
AD = √[ (1 - 1)^2 + (7 - 2)^2 ] = √25 = 5 units
Similarly, since ∆ABD has sides with lengths of 5, 3, and 5. This means it is isosceles. The answer for this item is B: isosceles.
We have shown above that AD is 5 units. Thus, answer is B: 5.
