The points A(4,7), B(p,3), and C(7,3) are the vertices of a right triangle, right-angled at B. Find the values of p.

Given A(4,7), B(p,3), and C(7,3)
Distance Formula = √(x₂-x₁)² + (y₂-y₁)²
AB = √(p-4)² + (3-7)² = √(p-4)² + 16
BC = √(7-p)² + (3-3)² = √(7-p)²
AC = √(7-4)² + (3-7)² = √(3)² + (4)² = 5
Triangle ABC is right-angled at B, hence (AC)² = (AB)² + (BC)²
∴ (√(p-4)² + 16)² + (√(7-p)²)² = 25
⇒ p² - 8p + 16 + 16 + 49 - 14p + p² = 25
⇒ 2p² - 22p + 81 = 25
⇒ 2p² - 22p + 56 = 0
⇒ p² - 11p + 28 = 0
⇒ p² - 7p - 4p + 28 = 0
⇒ p(p - 7) - 4(p - 7) = 0
Or (p - 7)(p - 4) = 0
If p - 7 = 0 then p = 7, which is not possible as B and C will be the same.
If p - 4 = 0 then p = 4
Hence, p = 4.