Question

determining the perpendicular bisector
which points are on the perpendicular bisector of the
given segment? check all that apply.
(-8, 19)
(1, -8)
(0, 19)
(-5, 10)
(2, -7)

Explanation:

Step1: Find mid - point of the segment

Let the two endpoints of the segment be \(A(-15,5)\) and \(B(15,15)\). The mid - point formula is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).

$$(\frac{-15 + 15}{2},\frac{5+15}{2})=(0,10)$$

The perpendicular bisector of a segment passes through its mid - point. Also, the slope of the line joining \(A(-15,5)\) and \(B(15,15)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{15 - 5}{15+15}=\frac{10}{30}=\frac{1}{3}\). The slope of the perpendicular bisector is \(m'=- 3\) (since \(m\times m'=-1\) for perpendicular lines). But we can also use the property that a point \((x,y)\) on the perpendicular bisector of the segment with endpoints \((x_1,y_1)\) and \((x_2,y_2)\) satisfies the equation \((x - x_1)^2+(y - y_1)^2=(x - x_2)^2+(y - y_2)^2\)

Step2: Check each point

• For point \((-8,19)\):

Let \((x_1,y_1)=(-15,5)\) and \((x_2,y_2)=(15,15)\)
\((-8 + 15)^2+(19 - 5)^2=7^2+14^2=49 + 196=245\)
\((-8 - 15)^2+(19 - 15)^2=(-23)^2+4^2=529+16 = 545
eq245\)

• For point \((1,-8)\):

\((1 + 15)^2+(-8 - 5)^2=16^2+(-13)^2=256 + 169=425\)
\((1 - 15)^2+(-8 - 15)^2=(-14)^2+(-23)^2=196+529 = 725
eq425\)

• For point \((0,19)\):

\((0 + 15)^2+(19 - 5)^2=15^2+14^2=225+196 = 421\)
\((0 - 15)^2+(19 - 15)^2=(-15)^2+4^2=225 + 16=241
eq421\)

• For point \((-5,10)\):

\((-5+15)^2+(10 - 5)^2=10^2+5^2=100 + 25=125\)
\((-5 - 15)^2+(10 - 15)^2=(-20)^2+(-5)^2=400+25 = 425
eq125\)

• For point \((2,-7)\):

\((2 + 15)^2+(-7 - 5)^2=17^2+(-12)^2=289+144 = 433\)
\((2 - 15)^2+(-7 - 15)^2=(-13)^2+(-22)^2=169+484=653
eq433\)

Answer:

None of the given points \((-8,19)\), \((1,-8)\), \((0,19)\), \((-5,10)\), \((2,-7)\) are on the perpendicular bisector of the segment.