Triangle PQR has vertices P(0,4)Q (8, -2) and R (7, -5).

Question

Write equations for the perpendicular bisectors of the 3 sides of the triangle? 

Answer 1

The vertices of the triangle PQR are P = (0 , 4), Q = (8 ,-2) and R = (7 ,-5).

Let X, Y and Z are mid points of lines PQ, QR and RP respectively.

Mid point formula  = [(x1 +x2)/2 , (y1 + y2)/2].

X = [(0+8)/2, (4+(-2))/2] = (4, 1)

Y = [(8+7)/2, (-2+(-5))/2] = (15/2, -7/2)

Z = [(0+7)/2, (4+(-5))/2] = (7/2, -1/2).

Two points form of line equation : y - y1 = [(y2 - y1) / (x2 - x1)] (x - x1).

The equation of line PY :

Substitute the values of P = (x1, y1) = (0 , 4) and Y = (x2, y2) = (15/2, -7/2) in the two points form of line equation.

y - 4 = [(-7/2 - 4) / (15/2 - 0)] (x - 0)

y - 4 = - x

y = - x + 4.

The equation of line QZ :

Substitute the values of Q = (x1, y1) = (8 ,-2) and Z = (x2, y2) = (7/2, -1/2) in the two points form of line equation.

y - (-2) = [(-1/2-(-2)) / (7/2 - 8)] (x - 8)

y +2 = [(-1/2+2) / (7/2 - 8)] (x - 8)

y +2 = [(3/2) / (-9/2)] (x - 8)

y + 2 =  (-1/3) (x - 8)

y = (-1/3)x + 1/3(8) - 2

y = - (1/3)x + 8/3 - 2

y = - x/3 + 8/3 - 2

y =(-1/3)x+(2/3)

The equation of line RX :

Substitute the values of R = (x1, y1) = (7 ,-5) and X = (x2, y2) = (4, 1) in the two points form of line equation.

y - (-5) = [(1-(-5)) / (4 - 7)] (x - 7)

y + 5 =  (-2) (x - 7)

y + 5 =  -2x + 14

y =  -2x + 14 - 5

y =  -2x + 9