Answer
424.8k+ views
Hint: We can find the vertices of the triangle using mid-point formula. Then we can check whether the triangle is right-angled. We can find the coordinates of the orthocenter.
Complete step by step answer:
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, the correct option is option A.
Note: We can plot the given points as midpoints of a triangle for a better understanding of the question. We use the distance formula to get the square of the length of the sides and then verify Pythagoras theorem to confirm it is a right-angled triangle. We can also say that the triangle is right angles by checking the coordinates of the vertices. The orthocenter of a triangle is the point at which all the altitudes of the triangle intersects with each other. An altitude is a perpendicular line from one side of a triangle to the opposite vertex. For a right-angled triangle, the altitudes meet at the right-angled vertex and hence it is the orthocenter.
Complete step by step answer:
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.
![seo images](https://www.vedantu.com/question-sets/b926fe6e-dd38-415b-8ab1-a0a7fec297874877413818910734708.png)
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, the correct option is option A.
Note: We can plot the given points as midpoints of a triangle for a better understanding of the question. We use the distance formula to get the square of the length of the sides and then verify Pythagoras theorem to confirm it is a right-angled triangle. We can also say that the triangle is right angles by checking the coordinates of the vertices. The orthocenter of a triangle is the point at which all the altitudes of the triangle intersects with each other. An altitude is a perpendicular line from one side of a triangle to the opposite vertex. For a right-angled triangle, the altitudes meet at the right-angled vertex and hence it is the orthocenter.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)