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Find the Circumcenter of the Triangle Whose Vertices Are (-2, -3), (-1, 0), (7, -6). - CBSE Class 10 - Mathematics

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Question

Find the circumcenter of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6).

Solution

The circumference of a triangle is equidistance from the vertices of a triangle.

Let A(-2, -3), B(-1, 0) and C(7, -6) vertices of the given triangle and let P(x,y) be the circumference of this triangle, Then

PA = PB = PC

Now, PA = PB

`=> sqrt((-2-x)^2 + (-3 -y)^2) = sqrt((-1  - x)^2 + (0 - y)^2`

`=> 4 + x^2 + 4x + 9 + y^2 + 6y = 1 + x^2 + 2xz + y^2`

`=> 4 + x^2 + 4x + 9 + y^2 + 6y - 1 - x^2 - 2x - y^2 = 0`

`=> 2x + 6y + 12 = 0

=> 2(x + 3y + 6) = 0

=> x + 3y + 6 = 0 ... eq (1)

And PB = PC

`=> sqrt((-1-x)^2 + (0 - y)^2) = sqrt((7- x)^2 + (-6 - y)^2)`

Squaring both the sides

`=> (-1 - x)^2 + y^2 = (7 - x)^2 + (-6 -y)^2`

`=> 1 + x^2 + 2x + y^2 = (7 - x)^2 + (-6 - y)^2`

`=> 1 + x^2 + 2x + y^2 - 49 - x^2 + 14x - 36 - y^2 - 12y`

=> 16x - 12y - 84 = 0

`=> 4(4x - 3y - 21) = 0`

=> 4x - 3y - 21 = 0 ......eq(2)

Adding eq(1) and eq(2)

`=> x + 3y + 6 + 4x - 3y - 21 = 0` 

=> x + 3y + 6 + 4x -3y - 21 = 0

=> 5x - 15 = 0

=> x = 15/5`

=> x = 3

putting the value of x in eq (2) and 

we get

`=> 4 x 3 - 3y - 21 = 0` 

`=> 12 - 3y - 21 = 0`

`=> -3y - 9 = 0`

`=> y = (-9)/3 = -3`

So the coordinates of the  circumcentre P are (3, -3)

  Is there an error in this question or solution?

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Solution Find the Circumcenter of the Triangle Whose Vertices Are (-2, -3), (-1, 0), (7, -6). Concept: Distance Formula.
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