Straight Line Question 93

Question: The length of perpendicular drawn from origin on the line joining $ ({x}’,{y}’) $ and $ ({x}’’,{y}’’) $ , is

Options:

A) $ \frac{x’y’’+x’‘y’}{\sqrt{{{(x’’-x’)}^{2}}+{{(y’’-y’)}^{2}}}} $

B) $ \frac{x’y’’-x’‘y’}{\sqrt{{{(x’’-x’)}^{2}}+{{(y’’-y’)}^{2}}}} $

C) $ \frac{x’x’’+y’y’’}{\sqrt{{{(x’’+x’)}^{2}}+{{(y’’+y’)}^{2}}}} $

D) $ \frac{x’x’’+y’y’’}{\sqrt{{{(x’’-x’)}^{2}}+{{(y’’-y’)}^{2}}}} $

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Answer:

Correct Answer: B

Solution:

  • Straight line $ y-{y}’=\frac{{{y}’}’-{y}’}{{{x}’}’-{x}’}(x-{x}’) $ Length of perpendicular $ =\frac{{x}’({{y}’}’-{y}’)-{y}’({{x}’}’-{x}’)}{\sqrt{{{({{x}’}’-{x}’)}^{2}}+{{({{y}’}’-{y}’)}^{2}}}} $ $ =\frac{{x}’{{y}’}’-{y}’{{x}’}’}{\sqrt{{{({{x}’}’-{x}’)}^{2}}+{{({{y}’}’-{y}’)}^{2}}}} $ Note : Students should remember this question as a formula.