The base of an equilateral triangle is along | vedclass

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

Question

Shortes distance of a point $\left( {{x_1},{y_1}} \right)$ from line $ax + by = c$ is

$d = \frac{{a{x_1} + b{y_1} - c}}{{\sqrt {{a^2} + {

Vedclass · Accepted Answer

$\frac{{4\sqrt 3 }}{{15}}$

Vedclass · Answer

Shortes distance of a point $\left( {{x_1},{y_1}} ight)$ from line $ax + by = c$ is

$d = \frac{{a{x_1} + b{y_1} - c}}{{\sqrt {{a^2} + {b^2}} }}$

Now shortest distance of $P(1,2)$ from $3x+4y=9$ is

$PC = d = \frac{{3\left( 1 ight) + 4\left( 2 ight) - 9}}{{\sqrt {{3^2} + {4^2}} }} = \frac{2}{5}$

Given that $\Delta APB$ is equilateral triangle

Let $'a'$ be its side

then $PB = a,Cb = \frac{a}{2}$

Now,In $\Delta PCB,{\left( {PB} ight)^2} = {\left( {PC} ight)^2} + {\left( {CB} ight)^2}$

(By Pythagoras theorem )

${a^2} = {\left( {\frac{2}{5}} ight)^2} + \frac{{{a^2}}}{4}$

${a^2} - \frac{{{a^2}}}{4} = \frac{4}{{25}} \Rightarrow \frac{{3{a^2}}}{4} = \frac{4}{{25}}$

${a^2} = \frac{{16}}{{75}} \Rightarrow a = \sqrt {\frac{{16}}{{75}}}  = \frac{4}{{5\sqrt 3 }} 	imes \frac{{\sqrt 3 }}{{\sqrt 3 }} = \frac{{4\sqrt 3 }}{{15}}$

$	herefore $ Lengh of Equilateral triangle $(a) = \frac{{4\sqrt 3 }}{5}$

The base of an equilateral triangle is along the line given by $3x + 4y\,= 9$. If a vertex of the triangle is $(1, 2)$, then the length of a side of the triangle is

Similar Questions

Two mutually perpendicular straight lines through the origin from an isosceles triangle with the line $2x + y = 5$ . Then the area of the triangle is :

lf a line $L$ is perpendicular to the line $5x - y\,= 1$ , and the area of the triangle formed by the line $L$ and the coordinate axes is $5$, then the distance of line $L$ from the line $x + 5y\, = 0$ is

If two vertices of a triangle are $(5, -1)$ and $( - 2, 3)$ and its orthocentre is at $(0, 0)$, then the third vertex

Without using distance formula, show that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are vertices of a parallelogram.

Let $\mathrm{A}(-2,-1), \mathrm{B}(1,0), \mathrm{C}(\alpha, \beta)$ and $\mathrm{D}(\gamma, \delta)$ be the vertices of a parallelogram $A B C D$. If the point $C$ lies on $2 x-y=5$ and the point $D$ lies on $3 x-2 y=6$, then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to_____.