Question

Line segment AB has endpoints A(-5,-3) and B (9,1). Find the equation for the perpendicular bisector of line segment AB.​

Answer 1

Step-by-step explanation:

Line segment AB has endpoints A(-5,-3) and B (9,1). Find the equation for the perpendicular bisector of line segment AB.

Answer 2

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➨ y = 6x + 12

${ \tt{ \underline{ \underline{ \green{Explanation:}}}}}$

$\tt{→ Slope \: of \: the \: line \: is \: m = \frac{(y_2 - y_1)}{(x_2 - x_1)}}$

→ Point-slope form of linear equation is

$\tt{ = > y - y_1 = m \: (x - x_1)}$

→ Coordinates of midpoint are

$\tt{ \frac{(x_{1}x1 + x_{2}x2 )}{2 (y_{1}y1 + y_{2}y2 ) \frac{2}{...}}}$

→ Slope of perpendicular line is opposite reciprocal

$\tt{ = > m_{AB}</p><p> = \frac{-5+7}{-9-3} </p><p> </p><p> = - \frac{1}{6} </p><p>}$

→ Slope of perpendicular line is 6

→ Coordinates of midpoint are

$\tt{ = > ( \frac{-9+3}{2} </p><p> \frac{-5-7}{2} </p><p> ) = ( - 3, - 6 )</p><p>}$

→ The equation of the perpendicular bisector of AB is

$\tt{ = > y - (-6) = 6(x - (- 3))}$

$\tt{ = > y + 6 = 6(x + 3)}$

$\tt{==> y = 6x + 12}$