Question 1208636: “Find the equation of a circle inscribed in a triangle with the sides on the lines x-3y=-5, 3x+y=1 and 3x-y=-11.”
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The equations of the three lines, and those equations in slope-intercept form:
(1) x-3y=-5 --> y=(1/3)x+5/3
(2) 3x+y=1 --> y=-3x+1
(3) 3x-y=-11 --> y=3x+11
Solve the pairs of equations simultaneously to find the coordinates of the vertices of the triangle (I leave the details to the student)
(1) and (2): (-0.2,1.6)
(1) and (3): (-3.5,0.5)
(2) and (3): (-5/3,6)
The center of the inscribed circle is the intersection of the angle bisectors of the three angles of the triangle.
The slopes of equations (2) and (3) are -3 and 3. That means the angle bisector of that angle is a vertical line. Since the intersection of those to lines is at x=-5/3, the equation of that angle bisector is x=-5/3.
The slopes of equations (1) and (3) are 1/3 and 3. That means the angle bisector of that angle has slope 1. Knowing that those two lines intersect at (-3.5,0.5), we find the equation of that angle bisector is y=x+4 (again I leave the details to the student.)
The center of the inscribed circle is the intersection of those two angle bisectors: (-5/3,7/3). And once again I leave the details to the student.
The radius of the inscribed circle is the distance from the center of the circle to any of the three lines that form the triangle. To find that distance, use the point-to-line distance formula:
The distance from the point (p,q) to the line with equation Ax+By+C=0 is
Using (p,q) = (-5/3,7/3) and equation (2), the radius is
The square of the radius is then 121/90, so finally the equation of the inscribed circle is
You can use the graphing utility at geogebra.com to graph the equations of the three given lines and this circle to see that the solution is correct.
Answer by math_tutor2020(3817)  (Show Source):
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