The equation of the line which makes right angled triangle with axes whose area is $6$ sq. units and whose hypotenuse is of $5$ units, is
$\frac{x}{4} + \frac{y}{3} = \pm \;1$
$\frac{x}{4} - \frac{y}{3} = \pm \;3$
$\frac{x}{6} + \frac{y}{1} = \pm \;1$
$\frac{x}{1} - \frac{y}{6} = \pm \;1$
A point moves so that square of its distance from the point $(3, -2)$ is numerically equal to its distance from the line $5x - 12y = 13$. The equation of the locus of the point is
Without using distance formula, show that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are vertices of a parallelogram.
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
Consider a triangle $\mathrm{ABC}$ having the vertices $\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$ and $\mathrm{C}(\gamma, \delta)$ and angles $\angle \mathrm{ABC}=\frac{\pi}{6}$ and $\angle \mathrm{BAC}=\frac{2 \pi}{3}$. If the points $\mathrm{B}$ and $\mathrm{C}$ lie on the line $\mathrm{y}=\mathrm{x}+4$, then $\alpha^2+\gamma^2$ is equal to....................
The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta ,2\sin \theta )$ internally in the ratio $2 : 3$ for all $\theta $, is a