Let R be the point (3,7) and let P and Q be t | vedclass

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

Question

(D) The perpendicular distance $h$ from point $R(3,7)$ to the line $x+y-5=0$ is given by:
$h = \frac{|3+7-5|}{\sqrt{1^2+1^2}} = \frac{5}{\sqrt{2}}$

Vedclass · Accepted Answer

$\frac{25 \sqrt{3}}{6}$

Vedclass · Answer

(D) The perpendicular distance $h$ from point $R(3,7)$ to the line $x+y-5=0$ is given by:
$h = \frac{|3+7-5|}{\sqrt{1^2+1^2}} = \frac{5}{\sqrt{2}}$
In an equilateral triangle with side length $a$,the height $h$ is given by $h = \frac{\sqrt{3}}{2} a$.
Therefore,$\frac{\sqrt{3}}{2} a = \frac{5}{\sqrt{2}}$,which implies $a = \frac{10}{\sqrt{6}} = \frac{5 \sqrt{6}}{3}$.
The area of the equilateral triangle is $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \left( \frac{100}{6} \right) = \frac{100 \sqrt{3}}{24} = \frac{25 \sqrt{3}}{6}$.

Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $\triangle PQR$ is an equilateral triangle. Then the area of $\triangle PQR$ is

Similar Questions

If $12a + 5b = 9$,where $a, b \in \mathbb{R}$,then the minimum value of $a^2 + b^2$ is:

The coordinates of the point on the line $x+y+3=0$,whose distance from the line $x+2y+2=0$ is $\sqrt{5}$ units,are

Given the lines $y + 2x = 3$ and $y + 2x = 5$ cut the axes at $A, B$ and $C, D$ respectively.
Statement-$1$ : $ABDC$ forms a quadrilateral and the point $(2, 3)$ lies inside the quadrilateral.
Statement-$2$ : The point $(2, 3)$ lies between the two parallel lines.

The distance of the point $(-2, 3)$ from the line $x - y - 5 = 0$ is

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the lines $x \cos \theta - y \sin \theta = k \cos 2 \theta$ and $x \sec \theta + y \csc \theta = k$ respectively,prove that $p^{2} + 4q^{2} = k^{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module