The equation of the line bisecting the line s | vedclass

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Question

(A) The slope of the line segment joining $(a, b)$ and $(a', b')$ is $m_1 = \frac{b' - b}{a' - a}$.Since the required line is the perpendicular bisect

Vedclass · Accepted Answer

$2(a - a')x + 2(b - b')y = a^2 + b^2 - a'^2 - b'^2$

Vedclass · Answer

(A) The slope of the line segment joining $(a, b)$ and $(a', b')$ is $m_1 = \frac{b' - b}{a' - a}$.Since the required line is the perpendicular bisector,its slope $m$ is $-\frac{1}{m_1} = \frac{a - a'}{b' - b} = \frac{a' - a}{b - b'}$.The midpoint of the segment is $\left( \frac{a + a'}{2}, \frac{b + b'}{2} \right)$.The equation of the line is $y - \frac{b + b'}{2} = \frac{a' - a}{b - b'} \left( x - \frac{a + a'}{2} \right)$.Multiplying by $2(b - b')$,we get $2(b - b')y - (b^2 - b'^2) = 2(a' - a)x - (a'^2 - a^2)$.Rearranging terms: $2(a - a')x + 2(b - b')y = a^2 + b^2 - a'^2 - b'^2$.

The equation of the line bisecting the line segment joining the points $(a, b)$ and $(a', b')$ at a right angle is

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