If sides of the triangle are x, y, z and x^2+y^2+z^2 = xy+yz+zx . then it is which type of triangle

Open in App

Solution

X² + y² + z² -xy - xz - yz = 0
multiplying 2 on both sides

2x² + 2y² + 2z² - 2xy - 2xz - 2yz = 0

x² + y² - 2xy+ x² + z² - 2xz + z² + y² - 2yz = 0

(x - y)² + (x - z)² + (y - z)² = 0

so the value are in square so they can't be negative so they must be indivudually zero to satisfy the equation

so x - y = 0 ⇒ x = y
x - z = 0 ⇒ x = z
y - z = 0 ⇒ y = z

so hence x = y = z

Therefore the triangle is an equilateral triangle since sides are equal

Suggest Corrections

Similar questions

△ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} 1 & y & y^{2} 1 & z & z^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

△_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 y z & z x & x y x & y & z \end{matrix} ∣ ∣ ∣ ∣,

△_{1} + △ =

∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & y z y & y^{2} & z x z & z^{2} & x y \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (x y + y z + z x)

∣ ∣ ∣ ∣ ∣ \begin{matrix} {y z - x}^{2} & {z x - y}^{2} & x y - z^{2} {z x - y}^{2} & x y - z^{2} & {y z - x}^{2} {x y - z}^{2} & {y z - x}^{2} & {z x - y}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

(x + y + z)

∣ ∣ ∣ ∣ ∣ \begin{matrix} y z - x^{2} & z x - y^{2} & x y - z^{2} z x - y^{2} & x y - z^{2} & y z {- x}^{2} x y - z^{2} & y z - x^{2} & z x - y^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

(x + y + z)

∣ ∣ ∣ ∣ ∣ \begin{matrix} y z - x^{2} & z x - y^{2} & x y - z^{2} z x - y^{2} & x y - z^{2} & y z - x^{2} x y - z^{2} & y z - x^{2} & z x - y^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

(x + y + z)

Join BYJU'S Learning Program