The perpendicular bisector of the line segmen | vedclass

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(A) $1$. Find the midpoint $P$ of the line segment $AB$:$P = \left( \frac{1+4}{2}, \frac{5+6}{2} \right) = \left( \frac{5}{2}, \frac{11}{2} \right)$$2

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$(0, 13)$

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(A) $1$. Find the midpoint $P$ of the line segment $AB$:$P = \left( \frac{1+4}{2}, \frac{5+6}{2} \right) = \left( \frac{5}{2}, \frac{11}{2} \right)$$2$. Find the slope of the line segment $AB$ $(m_{AB})$:$m_{AB} = \frac{6-5}{4-1} = \frac{1}{3}$$3$. The slope of the perpendicular bisector $(m_{\perp})$ is the negative reciprocal of $m_{AB}$:$m_{\perp} = -\frac{1}{m_{AB}} = -3$$4$. Use the point-slope form to find the equation of the perpendicular bisector:$y - y_1 = m_{\perp}(x - x_1)$$y - \frac{11}{2} = -3(x - \frac{5}{2})$$y - \frac{11}{2} = -3x + \frac{15}{2}$$y = -3x + \frac{15}{2} + \frac{11}{2}$$y = -3x + \frac{26}{2}$$y = -3x + 13$$5$. To find where it cuts the $y$-axis,set $x = 0$:$y = -3(0) + 13 = 13$Thus,the perpendicular bisector cuts the $y$-axis at $(0, 13)$.

The perpendicular bisector of the line segment joining the points $A(1, 5)$ and $B(4, 6)$ cuts the $y$-axis at

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