If the equation of base of an equilateral tri | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(a) $AD = \left| {\,\frac{{ - 2 - 2 - 1}}{{\sqrt {{{(2)}^2} + {{( - 1)}^2}} }}\,} \right| = \left| {\,\frac{{ - 5}}{{\sqrt 5 }}\,} \right| = \sqrt 5 $

Vedclass · Accepted Answer

$\sqrt {\frac{{20}}{3}} $

Vedclass · Answer

(a) $AD = \left| {\,\frac{{ - 2 - 2 - 1}}{{\sqrt {{{(2)}^2} + {{( - 1)}^2}} }}\,} ight| = \left| {\,\frac{{ - 5}}{{\sqrt 5 }}\,} ight| = \sqrt 5 $

$	an {60^o}$ $ = \frac{{AD}}{{BD}} \Rightarrow \sqrt 3 = \frac{{\sqrt 5 }}{{BD}}$==> $BD = \sqrt {\frac{5}{3}} $

$	herefore $ $BC = 2BD = 2\sqrt {\frac{5}{3}} = \sqrt {\frac{{20}}{3}} $.

If the equation of base of an equilateral triangle is $2x - y = 1$ and the vertex is $(-1, 2)$, then the length of the side of the triangle is

Similar Questions

If the middle points of the sides $BC,\, CA$ and $AB$ of the triangle $ABC$ be $(1, 3), \,(5, 7)$ and $(-5, 7)$, then the equation of the side $AB$ is

Given three points $P, Q, R$ with $P(5, 3)$ and $R$ lies on the $x-$ axis. If equation of $RQ$ is $x - 2y = 2$ and $PQ$ is parallel to the $x-$ axis, then the centroid of $\Delta PQR$ lies on the line

A point starts moving from $(1, 2)$ and its projections on $x$ and $y$ - axes are moving with velocities of $3m/s$ and $2m/s$ respectively. Its locus is

The point moves such that the area of the triangle formed by it with the points $(1, 5)$ and $(3, -7)$ is $21$ sq. unit. The locus of the point is

A variable straight line passes through the points of intersection of the lines, $x + 2y = 1$ and $2x - y = 1$ and meets the co-ordinate axes in $A\,\, \&\,\, B$ . The locus of the middle point of $AB$ is :