Co-ordinates of the orthocentre of the triangle whose vertices are $A(0, 0) , B(3, 4)$ and $C(4, 0)$ is
$(3, 1)$
$(3, 4)$
$(3, 3)$
$\left( {3,\frac{3}{4}\,} \right)$
A triangle is formed by $X -$ axis, $Y$ - axis and the line $3 x+4 y=60$. Then the number of points $P ( a, b)$ which lie strictly inside the triangle, where $a$ is an integer and $b$ is a multiple of $a$, is $...........$
The vertex of an equilateral triangle is $(2,-1)$ and the equation of its base in $x + 2y = 1$. The length of its sides is
Locus of the image of point $ (2,3)$ in the line $\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,k \in R$ is a:
Without using distance formula, show that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are vertices of a parallelogram.
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 :