Answer
423.9k+ views
Hint: Solve the given equations pairwise to obtain the vertices of the triangle. Substitute the coordinates of the vertices in the general equation of a circle ${(x - h)^2} + {(y - k)^2} = {r^2}$ with $(h,k)$ as its centre. Solve the equations thus obtained to get $(h,k)$ which is the required answer.
Complete step by step solution:
We are given the equation of the lines which form the sides of the triangle.
They are $3x - y - 5 = 0$, $x + 2y - 4 = 0$, $5x + 3y + 1 = 0$.
We need to find the circumcenter of the triangle.
In the above figure, consider O as the circumcenter of $\vartriangle ABC$.
Consider the system of linear equations
$3x - y - 5 = 0......(1)$
$x + 2y - 4 = 0.....(2)$
$5x + 3y + 1 = 0....(3)$
As can be understood from the figure, every pair of lines has a point of intersection and these 3 points of intersection form the vertices of the triangle.
From (1), we get \[3x - 5 = y \Rightarrow y = 3x - 5\]
From (2), we get $2y = - x + 4 \Rightarrow y = - \dfrac{x}{2} + 2$
Therefore,
\[
3x - 5 = - \dfrac{x}{2} + 2 \\
\Rightarrow \dfrac{{7x}}{2} = 7 \\
\Rightarrow x = 2 \\
\]
Substituting\[x = 2\]in\[y = 3x - 5\], we get\[y = 3(2) - 5 = 1\]
Thus, we get one of the vertices of the triangle. Call this vertex as A.
$A \equiv (2,1)$
Similarly, by solving the remaining equations in pairs, we get the other two vertices.
$B \equiv (1, - 2)$ and $C \equiv ( - 2,3)$
Now, these vertices of $\vartriangle ABC$lie on the circle and we need only 3 points to construct a circle.
The centre of the circle formed by the vertices A, B, C is the required circumcenter.
We will construct the equation of the circle using A, B, and C.
Let $(h,k)$be the circumcenter.
Then the equation of the circle with radius r and centre $(h,k)$ is given by ${(x - h)^2} + {(y - k)^2} = {r^2}....(4)$
As the points A, B, and C lie on the circle, they will satisfy equation (4).
$A \equiv (2,1) \Rightarrow {(2 - h)^2} + {(1 - k)^2} = {r^2}.....(5)$
$B \equiv (1, - 2) \Rightarrow {(1 - h)^2} + {( - 2 - k)^2} = {r^2}.....(6)$
$C \equiv ( - 2,3) \Rightarrow {( - 2 - h)^2} + {(3 - k)^2} = {r^2}....(7)$
Comparing the equations (5) and (6), we get
\[
{(2 - h)^2} + {(1 - k)^2} = {(1 - h)^2} + {( - 2 - k)^2} \\
\Rightarrow 4 - 4h + {h^2} + 1 - 2k + {k^2} = 1 - 2h + {h^2} + 4 + 4k + {k^2} \\
\Rightarrow - 2h - 6k = 0 \\
\Rightarrow 2h + 6k = 0....(8) \\
\]
Similarly, from (6) and (7), we get
$3h - 5k = - 4.....(9)$
Solving (8) and (9), we get $(h,k) \equiv (\dfrac{{ - 6}}{7},\dfrac{2}{7})$
Hence, the circumcenter is $(\dfrac{{ - 6}}{7},\dfrac{2}{7})$.
Note: In the given $\vartriangle ABC$, the line passing through the circumcenter and the vertex A is the perpendicular bisector of its opposite side BC. This holds true for all the vertices.
In fact, the circumcenter of a triangle is defined as the point of intersection of the perpendicular bisectors of the sides of the triangle.
Complete step by step solution:
We are given the equation of the lines which form the sides of the triangle.
They are $3x - y - 5 = 0$, $x + 2y - 4 = 0$, $5x + 3y + 1 = 0$.
We need to find the circumcenter of the triangle.
![seo images](https://www.vedantu.com/question-sets/c4260b20-17c8-4592-9858-0b347b2b06eb1975546094214352540.png)
In the above figure, consider O as the circumcenter of $\vartriangle ABC$.
Consider the system of linear equations
$3x - y - 5 = 0......(1)$
$x + 2y - 4 = 0.....(2)$
$5x + 3y + 1 = 0....(3)$
As can be understood from the figure, every pair of lines has a point of intersection and these 3 points of intersection form the vertices of the triangle.
From (1), we get \[3x - 5 = y \Rightarrow y = 3x - 5\]
From (2), we get $2y = - x + 4 \Rightarrow y = - \dfrac{x}{2} + 2$
Therefore,
\[
3x - 5 = - \dfrac{x}{2} + 2 \\
\Rightarrow \dfrac{{7x}}{2} = 7 \\
\Rightarrow x = 2 \\
\]
Substituting\[x = 2\]in\[y = 3x - 5\], we get\[y = 3(2) - 5 = 1\]
Thus, we get one of the vertices of the triangle. Call this vertex as A.
$A \equiv (2,1)$
Similarly, by solving the remaining equations in pairs, we get the other two vertices.
$B \equiv (1, - 2)$ and $C \equiv ( - 2,3)$
Now, these vertices of $\vartriangle ABC$lie on the circle and we need only 3 points to construct a circle.
The centre of the circle formed by the vertices A, B, C is the required circumcenter.
We will construct the equation of the circle using A, B, and C.
Let $(h,k)$be the circumcenter.
Then the equation of the circle with radius r and centre $(h,k)$ is given by ${(x - h)^2} + {(y - k)^2} = {r^2}....(4)$
As the points A, B, and C lie on the circle, they will satisfy equation (4).
$A \equiv (2,1) \Rightarrow {(2 - h)^2} + {(1 - k)^2} = {r^2}.....(5)$
$B \equiv (1, - 2) \Rightarrow {(1 - h)^2} + {( - 2 - k)^2} = {r^2}.....(6)$
$C \equiv ( - 2,3) \Rightarrow {( - 2 - h)^2} + {(3 - k)^2} = {r^2}....(7)$
Comparing the equations (5) and (6), we get
\[
{(2 - h)^2} + {(1 - k)^2} = {(1 - h)^2} + {( - 2 - k)^2} \\
\Rightarrow 4 - 4h + {h^2} + 1 - 2k + {k^2} = 1 - 2h + {h^2} + 4 + 4k + {k^2} \\
\Rightarrow - 2h - 6k = 0 \\
\Rightarrow 2h + 6k = 0....(8) \\
\]
Similarly, from (6) and (7), we get
$3h - 5k = - 4.....(9)$
Solving (8) and (9), we get $(h,k) \equiv (\dfrac{{ - 6}}{7},\dfrac{2}{7})$
Hence, the circumcenter is $(\dfrac{{ - 6}}{7},\dfrac{2}{7})$.
Note: In the given $\vartriangle ABC$, the line passing through the circumcenter and the vertex A is the perpendicular bisector of its opposite side BC. This holds true for all the vertices.
In fact, the circumcenter of a triangle is defined as the point of intersection of the perpendicular bisectors of the sides of the triangle.
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)