Question

Determine an equation for the right bisector of the line segment with endpoints J(-5,3) and K(3,-2)

Answer 1

`y=1.6x+2.1`

Explanation:

we know that

The right bisector of the line segment JK is a perpendicular line to the segment JK that pass through the midpoint of segment JK

step 1

Find the midpoint JK

The formula to calculate the midpoint between two points is equal to

`M((x1+x2)/(2),(y1+y2)/(2))`

we have

`J(-5,3),K(3,-2)`

substitute the values

`M((-5+3)/(2),(3-2)/(2))`

`M((-2)/(2),(1)/(2))`

`M(-1,(1)/(2))`

step 2

Find the slope JK

The formula to calculate the slope between two points is equal to

`m=(y2-y1)/(x2-x1)`

we have

`J(-5,3),K(3,-2)`

substitute

`m=(-2-3)/(3+5)`

`m=(-5)/(8)`

`m=-(5)/(8)`

step 3

Find the slope of the line perpendicular to the segment JK

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of the slopes is equal to -1)

`m_1*m_2=-1`

we have

`m_1=-(5)/(8)` ----> slope of segment JK

Find m_2

substitute

`(-(5)/(8))*m_2=-1`

`m_2=(8)/(5)`

step 4

Find the equation for the right bisector of the line segment JK

The equation in point slope form is equal to

`y-y1=m(x-x1)`

we have

`m=(8)/(5)`

`point\ M(-1,(1)/(2))`

substitute

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

Convert to slope intercept form

isolate the variable y

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

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

`y=(8)/(5)x+(21)/(10)`

`y=1.6x+2.1`