Question

Given triangle ABC with vertices A(−3, 0), B(0, 6), and C(4, 6).

Find the equations of the three perpendicular bisectors of the same triangle

Answer 1

Answer:

X=1.5

Step-by-step explanation:

1)x=3/2

2)

line passing through (5,3) and perpendicular to (4,6) and (6,0)

slope of line - 6-0/4-6=-6/2=-3

1/3 slope of line and it passes through (5,3)

so y-3=1/3(x-5)

3y-9=x-5

3y-x=4

3)third is also same

Answer 2

Answer:

Step-by-step explanation:

Coordinates of midpoint ( $\frac{x_{1} +x_{2} }{2}$ , $\frac{y_{1} +y_{2} }{2}$ )

m = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

y - $y_{1}$ = m(x - $x_{1}$ )

Slope of perpendicular lines are opposite reciprocals.

~~~~~~~~~~~~~~

A(−3, 0), B(0, 6), and C(4, 6).

1). $m_{a}$ = $\frac{6-6}{4-0}$ = 0

Coordinates of midpoint (2, 6)  

The equations of the perpendicular bisector to BC is x = 2

2). $m_{b}$ = $\frac{6-0}{4+3}$ = $\frac{6}{7}$

Coordinates of midpoint ( $\frac{-3+4}{2}$ , $\frac{0+6}{2}$ ) = ( $\frac{1}{2}$ , 3)

The equations of the perpendicular bisector to AC is

y - 3 = - $\frac{7}{6}$ ( x - $\frac{1}{2}$) ⇔ y = - $\frac{7}{6}$ x + $\frac{43}{12}$  

3). $m_{c}$ = $\frac{6-0}{0+3}$ = 2

Coordinates of midpoint ( $\frac{-3+0}{2}$ , $\frac{0+6}{2}$ ) = ( $\frac{-3}{2}$ , 3 )

The equations of the perpendicular bisector to AB is

y - 3 = - $\frac{1}{2}$ (x + $\frac{3}{2}$ ) ⇔ y = - $\frac{1}{2}$ x + $\frac{9}{4}$