A straight line cuts off the intercepts OA = | vedclass

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

Question

Equation of straight line : $\frac{ x }{ a }+\frac{ y }{ b }=1$

Or $x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=p$

$\frac{x}{2}+\frac{y \sqrt{3}}{

Vedclass · Accepted Answer

$\frac{392}{3}$

Vedclass · Answer

Equation of straight line : $\frac{ x }{ a }+\frac{ y }{ b }=1$

Or $x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=p$

$\frac{x}{2}+\frac{y \sqrt{3}}{2}=p$

$\frac{x}{3 p}+\frac{y}{2 p}=1$

Comparing both $: a =2 p , b =\frac{2 p }{\sqrt{3}}$

Now area of $	riangle OAB =\frac{1}{2} \cdot ab =\frac{98}{3} \cdot \sqrt{3}$

$\frac{1}{2} \cdot 2 p \cdot \frac{2 p }{\sqrt{3}}=\frac{98}{3} \cdot \sqrt{3}$

$p^2=49$

$a^2-b^2=4 p^2-\frac{4 p^2}{3}=\frac{2}{3} 4 p^2$

$=\frac{8}{3} \cdot 49=\frac{392}{3}$

Similar Questions

Area of the rhombus bounded by the four lines, $ax \pm by \pm c = 0$ is :

The equation of perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x - y + 5 = 0$ and $x + 2y = 0$ respectively. If the point $A$ is $(1,\; - \;2)$, then the equation of line $BC$ is

Triangle formed by the lines $3x + y + 4 = 0$ , $3x + 4y -15 = 0$ and $24x -7y = 3$ is a/an

$A(-1, 1)$, $B(5, 3)$ are opposite vertices of a square in $xy$-plane. The equation of the other diagonal (not passing through $(A, B)$ of the square is given by

If $A$ and $B$ are two points on the line $3x + 4y + 15 = 0$ such that $OA = OB = 9$ units, then the area of the triangle $OAB$ is