KnowledgeBoat Logo
|

Mathematics

The vertices of a triangle are (5, 1), (11, 1) and (11, 9). Find the co-ordinates of the circumcentre of the triangle.

Distance Formula

10 Likes

Answer

Let the circumcentre of the triangle be P(x, y).

The circumcentre is equidistant from all three vertices of the triangle.

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x2 - x1)^2 + (y2 - y1)^2}

The distance from P to A is equal to the distance from P to B, so:

⇒(x−5)2+(y−1)2=(x−11)2+(y−1)2⇒(x−5)2+(y−1)2=(x−11)2+(y−1)2⇒x2+25−10x+y2+1−2y=x2+121−22x+y2+1−2y⇒x2+26−10x+y2−2y=x2+122−22x+y2−2y⇒26−10x=122−22x⇒22x−10x=122−26⇒12x=96⇒x=9612⇒x=8⇒ \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{(x - 11)^2 + (y - 1)^2}\\[1em] ⇒ (x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 1)^2\\[1em] ⇒ x^2 + 25 - 10x + y^2 + 1 - 2y = x^2 + 121 - 22x + y^2 + 1 - 2y\\[1em] ⇒ x^2 + 26 - 10x + y^2 - 2y = x^2 + 122 - 22x + y^2 - 2y\\[1em] ⇒ 26 - 10x = 122 - 22x\\[1em] ⇒ 22x - 10x = 122 - 26\\[1em] ⇒ 12x = 96\\[1em] ⇒ x = \dfrac{96}{12}\\[1em] ⇒ x = 8

The distance from P to A is equal to the distance from P to C, so:

⇒(x−5)2+(y−1)2=(x−11)2+(y−9)2⇒(x−5)2+(y−1)2=(x−11)2+(y−9)2⇒x2+25−10x+y2+1−2y=x2+121−22x+y2+81−18y⇒x2+26−10x+y2−2y=x2+202−22x+y2−18y⇒26−10x−2y=202−22x−18y⇒26−10x−2y−202+22x+18y=0⇒12x+16y=176⇒3x+4y=44⇒ \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{(x - 11)^2 + (y - 9)^2}\\[1em] ⇒ (x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 9)^2\\[1em] ⇒ x^2 + 25 - 10x + y^2 + 1 - 2y = x^2 + 121 - 22x + y^2 + 81 - 18y\\[1em] ⇒ x^2 + 26 - 10x + y^2 - 2y = x^2 + 202 - 22x + y^2 - 18y\\[1em] ⇒ 26 - 10x - 2y = 202 - 22x - 18y\\[1em] ⇒ 26 - 10x - 2y - 202 + 22x + 18y = 0\\[1em] ⇒ 12x + 16y = 176\\[1em] ⇒ 3x + 4y = 44\\[1em]

Putting the value of x = 8 in the above equation,

⇒3×8+4y=44⇒24+4y=44⇒4y=44−24⇒4y=20⇒y=204⇒y=5⇒ 3 \times 8 + 4y = 44\\[1em] ⇒ 24 + 4y = 44\\[1em] ⇒ 4y = 44 - 24\\[1em] ⇒ 4y = 20\\[1em] ⇒ y = \dfrac{20}{4}\\[1em] ⇒ y = 5

Hence, the co-ordinates of the circumcentre of the triangle is (8, 5).

Answered By

1 Like

Related Questions