The vertices of a triangle are A(-4,0), B(2,0) and C(0,6). Let Msub1, Msub2 and Msub3 be the midpoints of sides AB, BC and AC, respectively. Let Hsub1, Hsub2 and Hsub3 be the feet of the altitudes on sides AB, BC and AC, respectively.
Find equation of circle passing through Hsub1, Hsub2 and Hsub3.
Find equation of circle passing through Hsub1, Hsub2 and Hsub3.
-
If I didn't make a mistake, the equation is
(x + 1/2)^2 + (y - 11/6)^2 = 65/18
and I found these points: H1 (0,0), H2 (7/5, 9/5), H3 (-14/13, 18/13)
I'll post some details after dinner.
(x + 1/2)^2 + (y - 11/6)^2 = 65/18
and I found these points: H1 (0,0), H2 (7/5, 9/5), H3 (-14/13, 18/13)
I'll post some details after dinner.
-
You awarded "best answer" before I could provide details. Bring up my profile and email me. I'll respond. Basically, I found the base of the three altitudes H1 (0,0), H2, H3, then the midpoints of two chords, then the intersection of perpendicular bisectors. That's the center of the circle.
Report Abuse