Question

If zi, z2, z3 are the vertices of the equilateral triangle and the zo be its orthocentre, such
that z12+z22+z32 = K202, then K equals​

Answer 1

SOLUTION

GIVEN

$\sf{z_1, z_2,z_3 \: are \: vertices \: \: of \: the \: equilateral \: triangle }$

$\sf{z_0 \: \: is \: orthocentre }$

$\sf{ {z_1}^{2} + {z_2}^{2} + {z_3}^{2} = k {z_0}^{2} }$

TO DETERMINE

The value of k

EVALUATION

Here it is given that

$\sf{z_1, z_2,z_3 \: are \: vertices \: \: of \: the \: equilateral \: triangle }$

$\sf{z_0 \: \: is \: orthocentre }$

Thus we get

$\displaystyle \sf{z_0 = \frac{z_1 + z_2 + z_3}{3} }$

$\displaystyle \sf{ \implies \: 3z_0 = z_1 + z_2 + z_3} \: \: \: - - - (1)$

Since the triangle is equilateral triangle

Therefore we get

$\sf{ {z_1}^{2} + {z_2}^{2} + {z_3}^{2} = z_1z_2 +z_2 z_3 +z_1 z_3} \: \: - - - (2)$

$\sf{ \implies \: {z_1}^{2} + {z_2}^{2} + {z_3}^{2} + 2(z_1z_2 +z_2 z_3 +z_1 z_3) = 3(z_1z_2 +z_2 z_3 +z_1 z_3)}$

$\sf{ \implies \: {({z_1} + {z_2} + {z_3})}^{2} = 3(z_1z_2 +z_2 z_3 +z_1 z_3)}$

$\sf{ \implies \: {(3{z_0} )}^{2} = 3(z_1z_2 +z_2 z_3 +z_1 z_3)} \: \: \: ( \: using \: 1)$

$\sf{ \implies \: 3(z_1z_2 +z_2 z_3 +z_1 z_3) = 9{{z_0}}^{2}} \:$

$\sf{ \implies \: (z_1z_2 +z_2 z_3 +z_1 z_3) = 3{{z_0}}^{2}} \:$

From Equation 2 we get

$\sf{ {z_1}^{2} + {z_2}^{2} + {z_3}^{2} = 3{z_0}^{2} }$

Now it is given that

$\sf{ {z_1}^{2} + {z_2}^{2} + {z_3}^{2} = k {z_0}^{2} }$

Comparing we get k = 3

FINAL ANSWER

Hence the required value of k = 3

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. show that sinix=isinhx

https://brainly.in/question/11810908

2. if a+ib/c+id is purely real complex number then prove that ad=bc

https://brainly.in/question/25744720