the equation of the perpendicular bisectors of sides AB and AC of a triangle ABC are x-y+5=0 and x+2y=0
respectively. if the point is A (1,-2), find the equation of line BC.
Hint:
Find the intersection of the two perpendicular bisectors, which gives the circumcentre O.
The distance AO=r is the radius of the circumscribed circle.
The circle with radius r, and centred at O will intersect AB and AC at B and C respectively.
To find the equation of the line BC, we can use the fact that the perpendicular bisectors of two sides of a triangle intersect at the circumcenter of the triangle. First, let's find the coordinates of the circumcenter.
Given that the equation of the perpendicular bisector of side AB is x - y + 5 = 0, we can find its slope. The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept. Rearranging the equation, we have y = x + 5. Comparing this with the slope-intercept form, we see that the slope of this line is 1.
The slope of the perpendicular bisector of side AC is the negative reciprocal of the slope of the line AC. The equation of line AC is given as x + 2y = 0, which can be rearranged as y = -0.5x. So the slope of the perpendicular bisector of side AC is -1/(-0.5) = 2.
Now, let's find the equations of the perpendicular bisectors of sides AB and AC in general form.
For the perpendicular bisector of side AB, we have the slope (m = 1) and a point (1, -2) that lies on this line. Using the point-slope form of a line, we have:
y - y1 = m(x - x1)
y - (-2) = 1(x - 1)
y + 2 = x - 1
x - y - 3 = 0
For the perpendicular bisector of side AC, we have the slope (m = 2) and a point (1, -2) that lies on this line. Using the point-slope form, we have:
y - y1 = m(x - x1)
y - (-2) = 2(x - 1)
y + 2 = 2x - 2
2x - y - 4 = 0
Now, we have two equations representing the perpendicular bisectors of sides AB and AC: x - y - 3 = 0 and 2x - y - 4 = 0.
To find the coordinates of the circumcenter, we solve these two equations simultaneously. Subtracting the first equation from the second equation, we get:
(2x - y - 4) - (x - y - 3) = 0
2x - y - 4 - x + y + 3 = 0
x - 1 = 0
x = 1
Substituting x = 1 into the first equation, we find:
1 - y - 3 = 0
y - 3 = 0
y = 3
So the circumcenter of triangle ABC is (1, 3).
Finally, to find the equation of line BC passing through points B(1, -2) and C(x, y), we can use the slope-intercept form. The slope, m, is given by (y2 - y1)/(x2 - x1):
m = (y - (-2))/(x - 1)
m = (y + 2)/(x - 1)
Substituting the coordinates of the circumcenter (1, 3), we get:
m = (3 + 2)/(x - 1)
m = 5/(x - 1)
Now we have the slope. To find the equation, we use the point-slope form of a line with the point B(1, -2):
y - y1 = m(x - x1)
y - (-2) = (5/(x - 1))(x - 1)
y + 2 = 5
y = 3
Therefore, the equation of line BC is y = 3.
side AB is a line ┴ to x-y+5=0, so it has slope -1. It is thus x+y+1=0
side AC is ┴ to x+2y=0, so it has slope 2. It is thus 2x-y-4=0
x-y+5=0 and x+y+1=0 intersect at P=(-3,2).
Since (-3,2)-A = (-4,4), B=P+(-4,4) = (-7,6)
x+2y=0 and 2x-y-4=0 intersect at Q=(8/5,-4/5).
Since Q-A=(3/5,6/5), C=Q+(3/5,6/5) = (11/5,2/5)
So, now you have B and C, and the line through those points is
y-6 = (-14/23)(x+7)