Question

NO LINKS!! Find a formula that expresses the fact that an arbitrary point P(x, y). is on the perpendicular bisector l of segment AB

A(-4,3), B(12,-11)​

Answer 1

`\textsf{Slope-intercept form}: \quad y=(8)/(7)x-(60)/(7)`

`\textsf{Standard form}: \quad 8x-7y=60`

Explanation:

A perpendicular bisector is a line that intersects another line segment at 90°, dividing it into two equal parts.

If two lines are perpendicular to each other, their slopes are negative reciprocals.

To find the perpendicular bisector of segment AB, first find its midpoint and its slope.

Define the endpoints of AB:

• (x₁, y₁) = (-4, 3)
• (x₂, y₂) = (12, -11)

`\boxed{\begin{minipage}{7.4 cm}\underline{Midpoint between two points}\\\\Midpoint $=\left((x_2+x_1)/(2),(y_2+y_1)/(2)\right)$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the endpoints.\\\end{minipage}}`

`\implies \textsf{Midpoint} =\left((12+(-4))/(2),(-11+3)/(2)\right)=\left((8)/(2),(-8)/(2)\right)=\left(4,-4}\right)`

`\boxed{\begin{minipage}{4.4cm}\underline{Slope Formula}\\\\Slope $(m)=(y_2-y_1)/(x_2-x_1)$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ \\are two points on the line.\\\end{minipage}}`

`\implies \textsf{slope}\:(m)=(-11-3)/(12-(-4))=(-14)/(16)-(7)/(8)`

Therefore, the slope of the perpendicular bisector is ⁸/₇.

`\boxed{\begin{minipage}{5.8 cm}\underline{Point-slope form of a linear equation}\\\\$y-y_1=m(x-x_1)$\\\\where:\\ \phantom{ww}$\bullet$ $m$ is the slope. \\ \phantom{ww}$\bullet$ $(x_1,y_1)$ is a point on the line.\\\end{minipage}}`

Substitute the slope ⁸/₇ and midpoint (4, -4) into the formula to create an equation for the line that is the perpendicular bisector of segment AB:

`\implies y-(-4)=(8)/(7)(x-4)`

`\implies y+4=(8)/(7)x-(32)/(7)`

`\implies y=(8)/(7)x-(60)/(7)`

`\implies 7y=8x-60`

`\implies 8x-7y=60`