The equation of the line,which bisects the li | vedclass

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Question

(A) Step $1$: Find the midpoint of the line segment joining $(2, -19)$ and $(6, 1)$.Midpoint $= (\frac{2+6}{2}, \frac{-19+1}{2}) = (4, -9)$.Step $2$:

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$3x - 2y = 30$

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(A) Step $1$: Find the midpoint of the line segment joining $(2, -19)$ and $(6, 1)$.Midpoint $= (\frac{2+6}{2}, \frac{-19+1}{2}) = (4, -9)$.Step $2$: Find the slope of the line joining $(-1, 3)$ and $(5, -1)$.Slope $(m_1) = \frac{-1-3}{5-(-1)} = \frac{-4}{6} = -\frac{2}{3}$.Step $3$: The required line is perpendicular to this line,so its slope $(m_2)$ is $-\frac{1}{m_1} = \frac{3}{2}$.Step $4$: Use the point-slope form $(y - y_1) = m(x - x_1)$ with point $(4, -9)$ and slope $\frac{3}{2}$.$y - (-9) = \frac{3}{2}(x - 4)$$2(y + 9) = 3(x - 4)$$2y + 18 = 3x - 12$$3x - 2y = 30$.

The equation of the line,which bisects the line segment joining the points $(2, -19)$ and $(6, 1)$ and is perpendicular to the line joining the points $(-1, 3)$ and $(5, -1)$,is

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