The equation of the line passing through (1, | vedclass

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Question

(B) Let the equation of the line passing through $(1, 2)$ be $y - 2 = m(x - 1)$,which simplifies to $mx - y + (2 - m) = 0$.The perpendicular distance

Vedclass · Accepted Answer

$y = 2$

Vedclass · Answer

(B) Let the equation of the line passing through $(1, 2)$ be $y - 2 = m(x - 1)$,which simplifies to $mx - y + (2 - m) = 0$.The perpendicular distance from the point $(8, 9)$ to this line is given by the formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Substituting the values,we get $7 = \frac{|m(8) - 9 + (2 - m)|}{\sqrt{m^2 + (-1)^2}}$.$7 = \frac{|7m - 7|}{\sqrt{m^2 + 1}}$.Squaring both sides: $49(m^2 + 1) = (7m - 7)^2$.$49(m^2 + 1) = 49(m - 1)^2$.$m^2 + 1 = m^2 - 2m + 1$.$-2m = 0$,which gives $m = 0$.Substituting $m = 0$ into the line equation: $y - 2 = 0(x - 1) \Rightarrow y = 2$.

The equation of the line passing through $(1, 2)$ and having a distance equal to $7$ units from the point $(8, 9)$ is:

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