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Question

(B) The equation of the line passing through $(x', y')$ and $(x'', y'')$ is given by: $(y - y') = \frac{y'' - y'}{x'' - x'}(x - x')$ $(y - y')(x'' - x

Vedclass · Accepted Answer

$\frac{|x'y'' - y'x''|}{\sqrt{(x'' - x')^2 + (y'' - y')^2}}$

Vedclass · Answer

(B) The equation of the line passing through $(x', y')$ and $(x'', y'')$ is given by: $(y - y') = \frac{y'' - y'}{x'' - x'}(x - x')$ $(y - y')(x'' - x') = (y'' - y')(x - x')$ $x(y'' - y') - y(x'' - x') + y'x'' - y'x' - x'y'' + x'y' = 0$ $x(y'' - y') - y(x'' - x') + (x'y'' - y'x'') = 0$ The length of the perpendicular from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is $\frac{|C|}{\sqrt{A^2 + B^2}}$. Here,$A = (y'' - y')$,$B = -(x'' - x')$,and $C = (x'y'' - y'x'')$. Thus,the length is $\frac{|x'y'' - y'x''|}{\sqrt{(y'' - y')^2 + (-(x'' - x'))^2}} = \frac{|x'y'' - y'x''|}{\sqrt{(x'' - x')^2 + (y'' - y')^2}}$.

The length of the perpendicular drawn from the origin $(0, 0)$ to the line joining the points $(x', y')$ and $(x'', y'')$ is:

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