We know that the distance between the two points (x1,y1) and
(x2,y2) is
d=√(x2−x1)2+(y2−y1)2
Let the given vertices be A(−2,1), B(4,6) and C(6,−3)
We first find the distance between A(−2,1) and B(4,6) as follows:
AB=√(x2−x1)2+(y2−y1)2=√(4−(−2))2+(6−1)2
=√(4+2)2+52=√62+52=√36+25=√61
Similarly, the distance between B(4,6) and C(6,−3) is:
BC=√(x2−x1)2+(y2−y1)2=√(6−4)2+(−3−6)2
=√22+(−9)2=√4+81=√85
Now, the distance between C(6,−3) and A(−2,1) is:
CA=√(x2−x1)2+(y2−y1)2=√(6−(−2))2+(−3−1)2
=√(6+2)2+(−4)2=√82+(−4)2=√64+16
=√80
Since the perimeter P of a triangle ABC is AB+BC+CA, therefore,
P=√61+√85+√80
Hence, the perimeter of the triangle is (√61+√85+√80) units.