∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices.
the measure of the longest side of ∆abc is
a)5
b)6
c)29^(1/2)
d)34^(1/2)
units. ∆abc is triangle.
a)equilateral
b)isosceles
c)right
d)scalene
if ∆abd is formed with the point d(1, 2) as its third vertex, then ∆abd is triangle
a)equilateral
b)isosceles
c)right
d)scalene
the length of side ad is
a)3
b)5
c)8
d)10 units.

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.