[[1]] is the circumcenter of the triangle whose sides are 3x−y−5=03, x−y−5=0, x+2y−4=0, x+2y−4=0, 5x+3y+1=0

Question

[[1]] is the circumcenter of the triangle whose sides are 3x−y−5=03, x−y−5=0, x+2y−4=0, x+2y−4=0, 5x+3y+1=0

Accepted Answer

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.

https://www.vedantu.com/question-sets/c4260b20-17c8-4592-9858-0b347b2b06eb1975546094214352540.png

In the above figure, consider O as the circumcenter of △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 ⇒ y=3x−5

From (2), we get 2y=−x+4 ⇒ y=−x2+2
Therefore,

3 x - 5 = - \frac{x}{2} + 2

\Rightarrow \frac{7 x}{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≡ (2,1)
Similarly, by solving the remaining equations in pairs, we get the other two vertices.
B≡ (1,−2) and C≡(−2,3)
Now, these ABC vertices are on the circle, and we only need three points to make a circle.
The required circumcenter is the centre of the circle formed by the vertices A, B, and C.
Using A, B, and C, we will solve the circle equation.
Let (h,k) be the circumcenter.
Then the equation of the circle with radius r and centre (h,k) is given by