Find the coordinates of the circumcentre of a triangle whose vertices are (-3,1), (0,-2) and (1,3)
7. Find the coordinates of the circumcentre of a triangle whose vertices are (-3,1), (0,-2) and (1,3)
Let A(−3, 1), B(0, −2), C(1, 3) and the circumference O(a, b).
The distance of point O will be equal from A, B and C.
OA = OC ...(Radii of the same circle)
∴ `sqrt([a - (-3)]^2 + (b - 1)^2) = sqrt((a - 1)^2 + (b - 3)^2)` ...(Distance formula)
Squaring both the sides,
`(a + 3)^2 + (b - 1)^2 = (a - 1)^2 + (b - 3)^2`
`∴ a^2 + 6a +9 + b^2 − 2b + 1 = a^2 − 2a + 1 + b^2 − 6b + 9`
∴ 6a − 2b = −2a − 6b
∴ 6a + 2a = − 6b + 2b
∴ 8a = −4b
∴ 8a + 4b = 0
∴ 4(2a + b) = 0
∴ 2a + b = 0 ....(I)
OB = OC ...(Radii of the same circle)
∴ `sqrt((a - 0)^2 + [b - (- 2)]^2) = sqrt((a - 1)^2 + (b - 3)^2)` ...(Distance formula)
Squaring both the sides,
`(a - 0)^2 + (b + 2)^2 = (a - 1)^2 + (b - 3)^2`
`∴ cancela^2 + cancelb^2 + 4b + 4 = cancela^2 − 2a + 1 + cancelb^2 − 6b + 9`
∴ 4b + 4 = − 2a − 6b + 10
∴ 4b + 2a + 6b = 10 - 4
∴ 2a + 10b = 6 ...(II)
Subtracting I from II, we get,
\[\begin{array}{1}
\phantom{\texttt{}}\texttt{2a + b = 0}\\ \phantom{\texttt{}}\texttt{- 2a + 10b = 6}\\ \hline\phantom{\texttt{}}\texttt{(-) (-) (-)}\\ \hline \end{array}\]
∴ - 9b = - 6
∴ b = `(-6)/(-9)`
∴ b = `2/3`
Substituting b = `2/3` in equation I, we get,
∴ 2a + b = 0
∴ 2a + `2/3` = 0
∴ 2a = `-2/3`
∴ a = `-2/3 xx 1/2`
∴ a = `(-1)/3`
Coordinates of the circumcentre are `((-1)/3, 2/3)`.
Explanation:-
The problem involves finding the circumcentre of a triangle with vertices A(-3, 1), B(0, -2), and C(1, 3).
The circumcentre is the centre of the circle passing through the three vertices of the triangle. Since the circumcentre is equidistant from the three vertices of the triangle, we can use the distance formula to set up equations and solve for the coordinates of the circumcentre.
Let the coordinates of the circumcentre be O(a, b). Then, we have:
OA = OC = radius (since they are radii of the same circle)
Using the distance formula, we can set up the following equation:
sqrt[(a – (-3))^2 + (b – 1)^2] = sqrt[(a – 1)^2 + (b – 3)^2]
Squaring both sides, we get:
(a + 3)^2 + (b – 1)^2 = (a – 1)^2 + (b – 3)^2
Expanding and simplifying, we get:
6a – 2b = -2a – 6b
8a = -4b
2a + b = 0 …(I)
Similarly, using the distance formula, we can set up another equation:
sqrt[(a – 0)^2 + (b – (-2))^2] = sqrt[(a – 1)^2 + (b – 3)^2]
Squaring both sides, we get:
a^2 + (b + 2)^2 = (a – 1)^2 + (b – 3)^2
Expanding and simplifying, we get:
2a + 10b = 6 …(II)
Now, we can solve the two equations (I) and (II) simultaneously to find the values of a and b.
Multiplying equation (I) by 5, we get:
10a + 5b = 0
Subtracting equation (II) from this, we get:
8a = -6
a = -3/4
Substituting this value of a in equation (I), we get:
2(-3/4) + b = 0
b = 3/2
Therefore, the coordinates of the circumcentre are O(-3/4, 3/2).
Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)