Question

Find the equations of perpendicular
bisectors of sides of the triangle whose
vertices are P(-1,8), Q
(4,-2) and R(-5,-3).​

Answer 1

Answer : 42x+2y+26 = 0

Step-by-step explanation:

P (-1,8) , Q(4 , -2) , R(-5,-3)

let us consider the perpendicular bisector is PM

first of all find the coordinates of the perpendicular bisector where they meet to triangles side QR

• midpoint formula, =(x , y ) {(x1+x2)/2 , (y1+y2)/2}

so, by midpoint formula, M(x,y)= {(x1+x2)/2 , (y1+y2)/2}

= {(4+(-5))/2 , (-2+(-3))/2}

= (-1/2 , -5/2 )

• two point formula :

two point formula :{(x-x1)/(x1-x2) = (y-y1)/(y1-y2)}

Now, by two point formula ,

here P(x1, y1) = (-1,8)

M(x2,y2) = (-1/2,-5/2)

(x-(-1))/(-1-(-1/2))=(y-8)/(8-(-5/2))

(x+1)/(-1+(1/2))=(y-8)/(8+(5/2))

(x+1)(8+(5/2))=(y-8)(-1+(1/2))

(x+1){(16+5)/2}=(y-8){(-2+1)/2}

2(x+1)(21)=2(y-8)(-1)

(2x+2)(21)=(2y-16)(-1)

42x+42 = -2y+ 16

42x+2y = 16 -42 = -26

42x+2y+26 = 0

Hence ,the required equation of perpendicular bisector is 42x+2y+26 = 0

similarly find the equations of perpendicular bisectors RS and TQ

Answer 2

upper answer is correct