The equation of the base BC of an equilateral | vedclass

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

Question

(A) Let the vertex be $A(-3, 2)$ and the base be $BC$ with equation $3x + 4y - 1 = 0$.The altitude $h$ from $A$ to $BC$ is the perpendicular distance

Vedclass · Accepted Answer

$\frac{4\sqrt{3}}{75}$

Vedclass · Answer

(A) Let the vertex be $A(-3, 2)$ and the base be $BC$ with equation $3x + 4y - 1 = 0$.The altitude $h$ from $A$ to $BC$ is the perpendicular distance from $A$ to the line $BC$:$h = \frac{|3(-3) + 4(2) - 1|}{\sqrt{3^2 + 4^2}} = \frac{|-9 + 8 - 1|}{\sqrt{25}} = \frac{|-2|}{5} = \frac{2}{5}$.In an equilateral triangle with side length $s$,the altitude $h = \frac{\sqrt{3}}{2}s$,so $s = \frac{2h}{\sqrt{3}} = \frac{2(2/5)}{\sqrt{3}} = \frac{4}{5\sqrt{3}}$.The area of an equilateral triangle is $\frac{\sqrt{3}}{4}s^2$.Area $= \frac{\sqrt{3}}{4} 	imes \left(\frac{4}{5\sqrt{3}}ight)^2 = \frac{\sqrt{3}}{4} 	imes \frac{16}{25 	imes 3} = \frac{4\sqrt{3}}{75}$.

The equation of the base $BC$ of an equilateral triangle is $3x + 4y = 1$ and the vertex $A$ is $(-3, 2)$. Then the area of the triangle is-

Similar Questions

Let the midpoints of the sides of a triangle $ABC$ be $(\frac{5}{2}, 7)$, $(\frac{5}{2}, 3)$, and $(4, 5)$. If its incentre is $(h, k)$, then $3h + k$ is equal to:

The vertex of the right angle of a right-angled triangle lies on the straight line $2x + y - 10 = 0$. If the two other vertices are at points $(2, -3)$ and $(4, 1)$,then the area of the triangle in sq. units is:

$A$ square is formed by the lines $x=0, y=0, x=1, y=1$. Then,the equations of its diagonals will be

The sides of a triangle are $x = 2$,$y + 1 = 0$,and $x + 2y = 4$. Find its circumcenter.

The circumcentre of the triangle formed by the lines $xy+2x+2y+4=0$ and $x+y+2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module