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

  • A

    $\frac{x}{4} + \frac{y}{3} = \pm \;1$

  • B

    $\frac{x}{4} - \frac{y}{3} = \pm \;3$

  • C

    $\frac{x}{6} + \frac{y}{1} = \pm \;1$

  • D

    $\frac{x}{1} - \frac{y}{6} = \pm \;1$

Similar Questions

Co-ordinates of the orthocentre of the triangle whose vertices are $A(0, 0) , B(3, 4)$ and $C(4, 0)$ is

The line $\frac{x}{a} + \frac{y}{b} = 1$ moves in such a way that $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{2{c^2}}}$, where $a, b, c \in R_0$ and $c$ is constant, then locus of the foot of the perpendicular from the origin on the given line is -

The area of triangle formed by the lines $x + y - 3 = 0 , x - 3y + 9 = 0$ and $3x - 2y + 1= 0$

Two vertices of a triangle are $(5, - 1)$ and $( - 2,3)$. If orthocentre is the origin then coordinates of the third vertex are

  • [IIT 1979]

The co-ordinates of three points $A(-4, 0) ; B(2, 1)$ and $C(3, 1)$ determine the vertices of an equilateral trapezium $ABCD$ . The co-ordinates of the vertex $D$ are :