a triangle had vertices A(1,1), B(a,4), C(6,2). The triangle is an isosceles with AB=BC. Find the value of a.?

Answer 1

AB² = (a - 1)² + 3²
AB² = a² - 2a + 1 + 9
AB² = a² - 2a + 10

BC ² = (a - 6) ² + 2 ²
BC ² = a ² - 12 a + 36 + 4
BC ² = a ² - 12 a + 40

- 2 a + 10 = - 12 a + 40
10 a = 30
a = 3

Answer 2

AB = BC
→ (a-1)² + (4-1)² = (a-6)² + (4-2)²
→ a² - 2a + 10 = a² - 12a + 40
→ 10a = 30
→ a = 3

Answer 3

(a-1)²+9=(a-6)² +4
a² -2a +1 +9 = a² -12a+36 +4
10a=30 ,then a = 3