Question

Circumcentre of a triangle, which is formed by the lines y=x, y=2x and y= 3x+4 is

• A. (6,8)
• B. (6,-8)
• C. (3,4)
• D. (-3,-4)

Answer 1

The correct option is B (6,-8)

Let's find the intersection points of these lines

y = x and y = 2x will intersect at A(0,0) (By subsituting the value of y = x)
y=x and y=3x+4

Hence we have x = 3x + 4 ; -2x = 4 ; x = -2

Since y = x ; hence y = -2

Hence the points will intersect at B(-2,-2)

y = 2x and y = 3x + 4
2x=3x+4
x = 4
y = -8
(-4,-8)
Vertices of triangle are A(0,0), B(-2,-2), C(-4,-8)

Circumcentre P(x,y) should be equidistant from the vertices of the triangle
Let(x,y) be the co-ordinates of the circum centre.
AP=CP and AP=BP
$x^2+y^2=(x+4{)}^2+(y+8{)}^2$

$\Rightarrow$ 8x+16y+80=0
Or 4x+8y+40=0 .....(1)

AP=BP
$x^2+y^2=(x+2{)}^2+(y+2{)}^2$

$\Rightarrow$ 4x+4y+8=0 .....(2)
By solving (1) and (2), we get x=6 and y=-8
Coordinates of circumcentre P=(6,-8)