Question

BEING TIMED

Answer 1

Answer:

Therefore orthocentre is ( 5 , 1.5 )

Step-by-step explanation:

From the general equation of a circle,

${x}^{2} + {y}^{2} + 2gx + 2fy + c = 0$

Substitute K(2, -2), L(4,6), M(8,-2) in (x,y)

K(2, -2): 2^2 +(-2)^2 + 2g(2) + 2f(-2) + c = 0

4g - 4f + c = -8

Divide by 4: g - f + c = -2 -----------(1)

L(4,6): 4^2 + 6^2 + 2g(4) + 2f(6) + c = 0

8g + 12f + c = -52

Divide by 4: 2g + 3f + c = -13 ----------(2)

M(8,-2): 8^2 +(-2)^2 + 2g(8) + 2f(-2) + c = 0

16g - 4f + c = -68

Divide by 4: 4g - f + c = -17 ----------(3)

Equation (2) - (1)

g + 4f = -11

g = -11 - 4f -----------(a)

Equation (3) - (2)

2g - 4f = -4

2g = -4 + 4f ----------(b)

Substitute for g in eqn (b)

2(-11 - 4f) = -4 + 4f

-12f = 18

f = -1.5

From eqn (a)

g = -11 - 4(-1.5)

g = -5

Since the orthocentre is given by (-g,-f)

Therefore orthocentre is (5,1.5)

Answer 2

Answer:

(5,2)

Step-by-step explanation:

you need to go two half way from the farthest point from the left to the right so that would be from 2 to 8 which is a distance of 6 so you go halfway over and that puts you at 5 then you find the distance from the lowest point to the highest point which goes from -2 to 6 with a distance of 8 so that means you go up for which puts you at 2 which means the center of the triangle is at the point (5, 2)