A triangle has vertices #A(1,1), B(a, 4)# and #C(6, 2)#. The triangle is isosceles with AB = BC. What is the value of #a#?

1 Answer
Nityananda
Sep 10, 2016

a = 3

Explanation:

Here AB = BC means length of AB is equals to length of BC.
Point A(1,1), B(a,4). So the distance AB = #sqrt [(1-a)^2 + (1-4)^2]#.
Point B(a,4),C(6,2). So the distance BC= #sqrt[(6-a)^2 +(2-4)^2]#
Hence, #sqrt[(1-a)^2+(1-4)^2]# = #sqrt[(6-a)^2+(2-4)^2]#
or, #(1-a)^2+(1-4)^2 = (6-a)^2 +(2-4)^2#
or, 1 - 2a + #a^2# + 9 = 36 - 12a +#a^2# + 4
or, 10a = 30
or, a = 3

Related questions
Impact of this question
1944 views around the world
You can reuse this answer
Creative Commons License