Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • I have a doubt, kindly clarify. If an equilateral triangle, having centroid at the origin, has a side along the line, x + y = 2, then the area (in sq. units) of thistriangle is :

If an equilateral triangle, having centroid at the origin, has a side along the line, x + y = 2, then the area (in sq. units) of this

triangle is :

  • Option 1)

    3\sqrt{6}

  • Option 2)

    6

  • Option 3)

    6\sqrt{3}

  • Option 4)

    \frac{9}{2}\sqrt{3}

 

Answers (1)

As learnt in

Perpendicular distance of a point from a line -

\rho =\frac{\left | ax_{1}+by_{1}+c\right |}{\sqrt{a^{2}+b^{2}}}

 

 

- wherein

P  is the distance from the line ax+by+c=0 .

 

 AD\:=\:P\:=\frac{(2-1-2)}{\sqrt{1^{2}+1^{2}}}\\=\frac{1}{\sqrt{2}}

\frac{P}{a}=sin60^{\circ}=> \:a=\frac{P}{sin60^{\circ}}\:=\frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{6}}=\frac{\sqrt{2}}{3}

Area = \frac{\sqrt{3}}{4}\:\left (\frac{\sqrt{2}}{3} \right )^{2}

= 6\sqrt{3}


Option 1)

3\sqrt{6}

This option is incorrect.

Option 2)

6

This option is incorrect.

Option 3)

6\sqrt{3}

This option is correct.

Option 4)

\frac{9}{2}\sqrt{3}

This option is incorrect.

Posted by

Sabhrant Ambastha

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 session 1 exam in the last week of January; registration details
anam-khan

Over 656 UPSC, JEE, NEET, other coaching aspirants refunded more than Rs 1 crore: Govt
anam-khan

NTA Exam Calendar 2025: JEE Main, NEET, CUET, UGC NET dates soon; websites to check

Latest Question