ABC is a triangle with coordinates A =(2, 6), B =(0, 0), and C =(14, 0). (a) Let P be the midpoint of AB. Determine the equation of the line perpendicular to AB passing through P. (b) Let Q be the point on line BC for which PQ is perpendicular to AB. Determine the length of AQ. (c) There is a (unique) circle passing through the points A, B, and C. Determine the radius of this circle.
Problem
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Tags: analytic geometry, graphing lines, slope, geometry, circumcircle
10000th User
04.11.2012 18:19
Find slope $m_{AB}$ of $AB$ and recall $m_{AB}\cdot m_\perp =?$. The rest is easy analytic geo.
Point $Q$ is of the form $(q,0)$ and use that on the result from part (a), then the rest is again easy.
Geometric construction: perpendicular bisectors. 'nuff said.
ptes77
05.11.2012 15:12
For a, the midpoint of the line is (1,3). So the point P is (1,3)
For the line to be perpendicular to AB, it has to have a slope of -1/3 because line AB has a slope of 3. So the equation of the line will be y=-x/3+10/3.
For b, the line BC is always on the x-axis so all you need to find is when y=-x/3+10/3=0.
10/3=x/3
10=x
So the point Q is at (10,0).
Point A is at (2,6). Use the distance formula to get the distance of AQ is 10.
For c, you can use abc/4R=K. Make the equation equal abc/4K=R. Therefore (2sqrt10*14*3sqrt20)/4*42 = (60sqrt2*14)/4*42=5sqrt2. You now that the radius is 5sqrt2.
10000th User
05.11.2012 16:02
ptes77 wrote:
For c, you can use abc/4R=K. Make the equation equal abc/4K=R. Therefore (2sqrt10*14*3sqrt20)/4*42 = (60sqrt2*14)/4*42=5sqrt2. You now that the radius is 5sqrt2.
You've got the same part (a) and part (b) as the hints I gave (the way I would've done). However, here is a part (c) comparison.
Instead of finding each individual sides and the area of the triangle, we can work with part (a)'s answer. Since the perpendicular bisectors are concurrent at the center of the circle, the line $x=7$ is one such perpendicular bisector. The intersection between this line and the line found in part (a) gives the center $O$. Thus, for $x=7$, $y=1$ using part (a). Therefore, the circumradius $BO=\sqrt{7^2+1^2}=5\sqrt2$, which avoided much of the unnecessary calculations.
ptes77
05.11.2012 17:31
10000th User wrote: ptes77 wrote:
For c, you can use abc/4R=K. Make the equation equal abc/4K=R. Therefore (2sqrt10*14*3sqrt20)/4*42 = (60sqrt2*14)/4*42=5sqrt2. You now that the radius is 5sqrt2.
You've got the same part (a) and part (b) as the hints I gave (the way I would've done). However, here is a part (c) comparison.
Instead of finding each individual sides and the area of the triangle, we can work with part (a)'s answer. Since the perpendicular bisectors are concurrent at the center of the circle, the line $x=7$ is one such perpendicular bisector. The intersection between this line and the line found in part (a) gives the center $O$. Thus, for $x=7$, $y=1$ using part (a). Therefore, the circumradius $BO=\sqrt{7^2+1^2}=5\sqrt2$, which avoided much of the unnecessary calculations.
Thanks for the much easier way