Circumcentre of the triangle formed by the line y=x,y=2x and y=3x+4 is
The correct option is A (6,−8)
Vertex 1: This is where the lines y=x and y=2x intersect. This happens when x=2x⇒x=0⇒y=0x=2x⇒x=0⇒y=0. Thus the point is (0,0).
Vertex 2: This is where the lines y=x and y=3x+4 intersect. This happens when x=3x+4⇒x=−2⇒y=−2x=3x+4⇒x=−2⇒y=−2. Thus the point is (-2,-2).
Vertex 3: This is where the lines y=2x and y=3x+4 intersect. This happens when 2x=3x+4⇒x=−4⇒y=−82x=3x+4⇒x=−4⇒y=−8. Thus the point is (-4,-8).
The circumcentre of a triangle is the point (a,b)(a,b) such that a circle of radius rrcentred on (a,b)(a,b) passes through the three vertices.
Every point on this circle satisfies the generic equation: (x−a)2+(y−b)2=r2(x−a)2+(y−b)2=r2. Expanding this equation, we have x2−2ax+a2+y2−2by+b2=r2x2−2ax+a2+y2−2by+b2=r2.
At vertex 1, x=0x=0 and y=0y=0, thus we have Equation A: a2+b2=r2a2+b2=r2
At vertex 2, x=−2x=−2 and y=−2y=−2, thus we have Equation B: 4+4a+a2+4+4b+b2=r24+4a+a2+4+4b+b2=r2
At vertex 3, x=−4x=−4 and y=−8y=−8, thus we have Equation C: 16+8a+a2+64+16b+b2=r216+8a+a2+64+16b+b2=r2
Subtracting Equation A from Equation B, we have Equation D: 8+4a+4b=08+4a+4b=0
Subtracting Equation B from Equation C, we have Equation E: 72+4a+12b=072+4a+12b=0
Subtracting Equation D from Equation E, we have Equation F: 64+8b=0⇒b=−864+8b=0⇒b=−8
Substituting this into Equation D: 8+4a−32=0⇒a=68+4a−32=0⇒a=6
So, the circumcentre is (6,−8)(6,−8).
As a bonus, using Equation A, the radius of the circle is 10.
Vertex 1: This is where the lines y=x and y=2x intersect. This happens when x=2x⇒x=0⇒y=0x=2x⇒x=0⇒y=0. Thus the point is (0,0).
Vertex 2: This is where the lines y=x and y=3x+4 intersect. This happens when x=3x+4⇒x=−2⇒y=−2x=3x+4⇒x=−2⇒y=−2. Thus the point is (-2,-2).
Vertex 3: This is where the lines y=2x and y=3x+4 intersect. This happens when 2x=3x+4⇒x=−4⇒y=−82x=3x+4⇒x=−4⇒y=−8. Thus the point is (-4,-8).
The circumcentre of a triangle is the point (a,b)(a,b) such that a circle of radius rrcentred on (a,b)(a,b) passes through the three vertices.
Every point on this circle satisfies the generic equation: (x−a)2+(y−b)2=r2(x−a)2+(y−b)2=r2. Expanding this equation, we have x2−2ax+a2+y2−2by+b2=r2x2−2ax+a2+y2−2by+b2=r2.
At vertex 1, x=0x=0 and y=0y=0, thus we have Equation A: a2+b2=r2a2+b2=r2
At vertex 2, x=−2x=−2 and y=−2y=−2, thus we have Equation B: 4+4a+a2+4+4b+b2=r24+4a+a2+4+4b+b2=r2
At vertex 3, x=−4x=−4 and y=−8y=−8, thus we have Equation C: 16+8a+a2+64+16b+b2=r216+8a+a2+64+16b+b2=r2
Subtracting Equation A from Equation B, we have Equation D: 8+4a+4b=08+4a+4b=0
Subtracting Equation B from Equation C, we have Equation E: 72+4a+12b=072+4a+12b=0
Subtracting Equation D from Equation E, we have Equation F: 64+8b=0⇒b=−864+8b=0⇒b=−8
Substituting this into Equation D: 8+4a−32=0⇒a=68+4a−32=0⇒a=6
So, the circumcentre is (6,−8)(6,−8).
As a bonus, using Equation A, the radius of the circle is 10.
