The equation of the base of an equilateral tr | vedclass

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

Question

(C) Let $	riangle ABC$ be the equilateral triangle with base $BC$ given by $x+y-2=0$ and vertex $A$ as $(2,-1)$.The length of the altitude $h$ from v

Vedclass · Accepted Answer

$\sqrt{2/3}$

Vedclass · Answer

(C) Let $	riangle ABC$ be the equilateral triangle with base $BC$ given by $x+y-2=0$ and vertex $A$ as $(2,-1)$.The length of the altitude $h$ from vertex $A$ to the base $BC$ is the perpendicular distance from $(2,-1)$ to $x+y-2=0$.$h = \frac{|(1)(2) + (1)(-1) - 2|}{\sqrt{1^2 + 1^2}} = \frac{|2 - 1 - 2|}{\sqrt{2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.In an equilateral triangle with side length $s$,the altitude $h$ is given by $h = \frac{\sqrt{3}}{2}s$.Therefore,$\frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2}s$.$s = \frac{2}{\sqrt{2} 	imes \sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}$.

The equation of the base of an equilateral triangle is $x+y=2$ and one vertex is $(2,-1)$. The length of the side of the triangle is:

Similar Questions

The positions of the points $(3, 4)$ and $(2, -6)$ with respect to the line $3x - 4y = 8$ are:

If a line $l$ passes through $(k, 2k), (3k, 3k)$ and $(3, 1)$,where $k \neq 0$,then the distance from the origin to the line $l$ is

An equilateral triangle is constructed between the lines $\sqrt{3} x+y-6=0$ and $\sqrt{3} x+y+9=0$ with the base on one line and the vertex on the other. The area (in sq. units) of the triangle so formed is

The coordinates of the points on the line $2x - y = 5$ which are at a distance of $1$ unit from the line $3x + 4y = 5$ are:

Let two points be $A(1, -1)$ and $B(0, 2)$. If a point $P(x', y')$ is such that the area of $\Delta PAB = 5 \; \text{sq units}$ and it lies on the line $3x + y - 4\lambda = 0$,then a value of $\lambda$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module