Question

The equation of the base of an equilateral triangle is X +Y =2 and vertex is (2, -1) length of its side is :

• A. $\sqrt{\left(\frac{1}{2}\right)}$
• B. $\sqrt{\left(\frac{3}{2}\right)}$
• C. $\sqrt{\left(\frac{2}{3}\right)}$
• D. $\sqrt{2}$

Answer 1

The correct option is C $\sqrt{\left(\frac{2}{3}\right)}$

$Given-P(x_1,y_1)=(2,-1)isthevetexofanequilateraltriangle.Theequationofitsbaseisx+y=2⟹x+y-2=0.Tofindout-thesideofthegivenequilateraltriangle=?Solution-Thepependiculardistaceofthevertexfromthebaseofatriangleisitsheight.Foranequilateraltriangleofsidea,theheighth=\frac{\sqrt{3}}{2}a⟹a=\frac{2h}{\sqrt{3}}.Nowweapplytheformulap=\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}whenP(x_1,y_1)isthepointandax+by+c=0istheequationoftheline.HereP(x_1,y_1)=(2,-1),a=1,b=1\&c=-2.\therefore p=h=\left|\frac{1\times 2+1\times (-1)+(-2)}{\sqrt{1^2+1^2}}\right|units=\frac{1}{\sqrt{2}}units.\therefore Thesidea=\frac{2h}{\sqrt{3}}=\frac{2\times \frac{1}{\sqrt{2}}}{\sqrt{3}}units=\sqrt{\frac{2}{3}}units.$