The equations of two sides $\mathrm{AB}$ and $\mathrm{AC}$ of a triangle $\mathrm{ABC}$ are $4 \mathrm{x}+\mathrm{y}=14$ and $3 \mathrm{x}-2 \mathrm{y}=5$, respectively. The point $\left(2,-\frac{4}{3}\right)$ divides the third side $\mathrm{BC}$ internally in the ratio $2: 1$. The equation of the side $\mathrm{BC}$ is :
$x-6 y-10=0$
$x-3 y-6=0$
$x+3 y+2=0$
$x+6 y+6=0$
Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.
The base of an equilateral triangle is along the line given by $3x + 4y\,= 9$. If a vertex of the triangle is $(1, 2)$, then the length of a side of the triangle is
The number of possible straight lines , passing through $(2, 3)$ and forming a triangle with coordinate axes, whose area is $12 \,sq$. units , is
Three lines $x + 2y + 3 = 0 ; x + 2y - 7 = 0$ and $2x - y - 4 = 0$ form the three sides of two squares. The equation to the fourth side of each square is