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This is one of our many quizzes on perpendicular lines. Click on this to access more such competitive quizzes on perpendicular lines.

Perpendicular Lines Quiz 2

mathematics

 Let $P\left( 5,3 \right)$ and $Q\left( 1,13 \right)$ be two given points. What will be the slope of a line perpendicular to PQ?  

 Slope $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ and perpendicular lines have negative reciprocal slopes. 

  $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 5,3 \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)=\left( 1,13 \right) \\ \text{Let “m” be the slope of PQ. Then,}\\ m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{13-3}{1-5}=\dfrac{10}{-4}=-\dfrac{5}{2} \\ \text{So, the slope of a line perpendicular to PQ is }-\dfrac{1}{m}=\dfrac{2}{5}$

 Find the equation of a line which intersects x-axis at $\dfrac{-11}{4}$ and perpendicular to line joining the points $\left( 1,\dfrac{3}{4} \right)$ and $\left( 5,\dfrac{7}{2} \right)$.

 First find slope ${{m}_{1}}$ using ${{m}_{1}}{{m}_{2}}=-1$ and then find the equation using slope intercept form i.e., $y={{m}_{1}}(x-d)$

  $\text{Let }{{m}_{1}}\text{ be the slope of required line }y={{m}_{1}}(x-d), \\ \text{ where x-intercept }=d=\dfrac{-11}{4}. \\ \Rightarrow {{m}_{1}}\times {{m}_{2}}=-1,\text{ where }{{m}_{2}}\text{ is the slope of the line joining the points }\left( 1,\dfrac{3}{4} \right)\text{ and }\left( 5,\dfrac{7}{2} \right) \\ \Rightarrow {{m}_{1}}\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)=-1 \\ \Rightarrow {{m}_{1}}\left( \dfrac{\dfrac{7}{2}-\dfrac{3}{4}}{5-1} \right)=-1\Rightarrow {{m}_{1}}=-\dfrac{16}{11} \\ \text{Then the equation of the line is,} \\ \Rightarrow y=\dfrac{-16}{11}\left( x+\dfrac{11}{4} \right) \\ \Rightarrow 16x+11y+44=0$ 

 The perpendicular from the origin to a line meets it at a point $\left( -5,11 \right)$ . What will be the equation of the line? 

 Slope $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ and perpendicular line have negative reciprocal slopes. Equation of a line is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/08e391a4-fb89-4108-a967-fde38676dc176218135259037329443.png"/>$\\ \text{We have, } {{m}_{1}}=\text{ Slope of OM}=\dfrac{11-0}{-5-0}=-\dfrac{11}{5} \\ \text{Let }m_2\text{ be the slope of line PQ,} \\ \text{Then, Slope of line PQ }\times \text{ Slope of line OM }= -1 \\ \Rightarrow {{m}_{2}}\times -\dfrac{11}{5}=-1 \Rightarrow {{m}_{2}}=\dfrac{5}{11} \\ \text{Thus the equation of the line passing through point }\left( -5,11 \right)\text{ and having a slope }\dfrac{5}{11}\text{ is:} \\ y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ \Rightarrow y-11=\dfrac{5}{11}\left( x+5 \right) \\ \Rightarrow 5x-11y+146=0 $ 

 A line perpendicular to the line segment joining the points $\left( 5,3 \right)$ and $\left( 2,4 \right)$ divides it in the ratio $2 : p$. What will be the equation of the line?  

 If a point divides any line joining  and $({{x}_{2}},{{y}_{2}})$ in the ratio ${{n}_{1}}:{{n}_{2}}$ , then coordinate of that point is $\left( \dfrac{{{n}_{1}}{{x}_{2}}+{{n}_{2}}{{x}_{1}}}{{{n}_{1}}+{{n}_{2}}},\dfrac{{{n}_{1}}{{y}_{2}}+{{n}_{2}}{{y}_{1}}}{{{n}_{1}}+{{n}_{2}}} \right)$.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3b2f82db-1359-4bab-9ac7-ad458dc7a8733996908887483898724.png"/>$\text{Let line ST }\bot \text{ UV join two points U }\left( 5,3 \right)\text{ and V }\left( 2,4 \right). \\ \text{The slope of the line joining U }\left( 5,3 \right)\text{ and V }\left( 2,4 \right)\text{ is } \\ \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} = \dfrac{4-3}{2-5}=-\dfrac{1}{3}\\ \text{If two lines are perpendicular then the product of their slope is }-1. \\ \text{So, slope of line ST }= 3 \\ \text{O divides UV joining }\left( 5,3 \right)\text{ and }\left( 2,4 \right)\text{ in the ratio of }2 : p. \\ \text{Equation of line ST passing through point }O \\ = \left( \dfrac{2\times 2+p\times 5}{2+p},\dfrac{2\times 4+p\times 3}{2+p} \right) \\ \Rightarrow \left( \dfrac{4+5p}{2+p},\dfrac{8+3p}{2+p} \right) \\ \text{And having slope 3 is:} \\ y-\left( \dfrac{8+3p}{2+p} \right)=3\left( x-\left( \dfrac{4+5p}{2+p} \right) \right) \\ \Rightarrow \left( \dfrac{2y+py-8-3p}{2+p} \right)=\left( \dfrac{6x+3px-12-15p}{2+p} \right) \\ \Rightarrow 3x\left( 2+p \right)-y\left( 2+p \right)=4\left( 1+3p \right) $

 Line A is passing through $\left( -3g,\dfrac{1}{9} \right)$ and $\left( g,\dfrac{7}{9} \right)$ while line B is passing through $\left( h,\dfrac{-5}{6} \right)$ and $\left( -2h,\dfrac{1}{2} \right)$. If lines A and B are perpendicular to each other, then which of the following conditions is true?

 Find the value of slopes ${{m}_{1}}\text{ and }{{m}_{2}}$ and then use ${{m}_{1}}{{m}_{2}}=-1$ to find the condition.

  $\text{Slope of line A }({{m}_{1}})=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\ \Rightarrow {{m}_{1}}=\dfrac{\dfrac{7}{9}-\dfrac{1}{9}}{g+3g}\Rightarrow {{m}_{1}}=\dfrac{1}{6g} \\ \text{Slope of line B }({{m}_{2}})=\dfrac{\dfrac{1}{2}+\dfrac{5}{6}}{-2h-h} \\ \Rightarrow {{m}_{2}}=\dfrac{-4}{9h} \\ \text{Now, we have }{{m}_{1}}.{{m}_{2}}=-1 \\ \Rightarrow \dfrac{1}{6g}\times \dfrac{-4}{9h}=-1 \\ \Rightarrow \dfrac{2}{27gh}=1\\ \Rightarrow 27gh-2=0 $