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) \[\frac{x'x''+y'y''}{\sqrt{{{(x''-x')}^{2}}+{{(y''-y')}^{2}}}}\]
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.You need to login to perform this action.
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