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Question

Answers

A.$\left( {0,0} \right)$

B.$\left( {0,24} \right)$

C.$\left( {10,0} \right)$

D.None of these

Answer

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Given the mid points of the sides of a triangle are $\left( {5,0} \right)$,$\left( {5,12} \right)$ and $\left( {0,12} \right)$.

We can plot the given points as midpoints of a triangle.

Let A$({x_1},{y_1})$, B$({x_2},{y_2})$and C$({x_3},{y_3})$ be the vertices of the triangle. We can use mid-point formula to find the vertices of a triangle. According to midpoint formula, the coordinates of the midpoint of the line connecting the points $({x_1},{y_1})$ and $({x_2},{y_2})$is $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$.

Then by mid-point formula for x coordinates of AB, we get,

$\dfrac{{{x_1} + {x_2}}}{2} = 0$

${x_1} + {x_2} = 0$

\[ \Rightarrow {x_1} = - {x_2}\] … (1)

Using mid-point formula for x coordinates of BC, we get,

$\dfrac{{{x_2} + {x_3}}}{2} = 5$

${x_2} + {x_3} = 10$

$ \Rightarrow {x_3} = 10 - {x_2}$ … (2)

Using mid-point formula for x coordinates of AC, we get

$\dfrac{{{x_1} + {x_3}}}{2} = 5$

$ \Rightarrow {x_1} + {x_3} = 10$ … (3)

Substituting (1) and (2) in (3), we get,

$

- {x_2} + 10 - {x_2} = 10 \\

\Rightarrow - 2{x_2} = 0 \\

\Rightarrow {x_2} = 0 \\

$

Substituting ${x_2} = 0$ in (1) and (2), we get,

\[{x_1} = - {x_2} = 0\]

${x_3} = 10 - {x_2} = 10 - 0 = 10$

Now we have the x coordinates of points A, B, and C.

Now we can take mid-point formula for y coordinates of AB,

$\dfrac{{{y_1} + {y_2}}}{2} = 12$

${y_1} + {y_2} = 24$

\[ \Rightarrow {y_1} = 24 - {y_2}\] … (4)

Using mid-point formula for y coordinates of BC, we get

$\dfrac{{{y_2} + {y_3}}}{2} = 0$

${y_2} + {y_3} = 0$

\[ \Rightarrow {y_3} = - {y_2}\] … (5)

Using mid-point formula for y coordinates of AC, we get

$\dfrac{{{y_1} + {y_3}}}{2} = 12$

$ \Rightarrow {y_1} + {y_3} = 24$ … (6)

Substituting (4) and (5) in (6), we get,

$

{y_1} + {y_3} = 24 - {y_2} - {y_2} = 24 \\

\Rightarrow - 2{y_2} = 0 \\

\Rightarrow {y_2} = 0 \\

$

Substituting ${y_2} = 0$ in (4) and (5), we get,

\[{y_1} = 24 - {y_2} = 24 - 0 = 24\]

\[{y_3} = - {y_2} = 0\]

Now we have the vertices of the triangle as A$(0,24)$, B$(0,0)$ and C$(10,0)$

Using distance formula,

\[

A{B^2} = {\left( {{x_2} - {x_1}} \right)^2} + {\left( {{y_2} - {y_1}} \right)^2} \\

= {\left( {0 - 0} \right)^2} + {\left( {24 - 0} \right)^2} \\

= {24^2} \\

= 576 \\

\]

\[

B{C^2} = {\left( {10 - 0} \right)^2} + {\left( {0 - 0} \right)^2} \\

= {10^2} \\

= 100 \\

\]

$

A{C^2} = {\left( {10 - 0} \right)^2} + {\left( {0 - 24} \right)^2} \\

= {10^2} + {24^2} \\

= 100 + 576 \\

= 676 \\

$

$

A{B^2} + B{C^2} \\

= 576 + 100 \\

= 676 \\

= A{C^2} \\

$

As $A{B^2} + B{C^2} = A{C^2}$, by Pythagoras theorem, ABC is a right-angled triangle right angled at B.

The orthocenter of a right-angled triangle is the vertex on the right angle. So, the orthocenter of the triangle ABC is B $(0,0)$.

So,