$A$ line $4x+y=1$ passes through the point $A(2,-7)$ and meets the line $BC$,whose equation is $3x-4y+1=0$,at the point $B$. The equation of the line $AC$ such that $AB=AC$ is

$A$ triangle $ABC$ lying in the first quadrant has two vertices as $A(1, 2)$ and $B(3, 1)$. If $\angle BAC = 90^{\circ}$ and $\text{ar}(\Delta ABC) = 5\sqrt{5}$ sq. units,then the abscissa of the vertex $C$ is

Let $ABC$ be a triangle with $A(-3, 1)$ and $\angle ACB = \theta$,where $0 < \theta < \frac{\pi}{2}$. If the equation of the median through $B$ is $2x + y - 3 = 0$ and the equation of the angle bisector of $C$ is $7x - 4y - 1 = 0$,then $\tan \theta$ is equal to:

If $P(1, 2), Q(4, 6), R(5, 7)$ and $S(a, b)$ are the vertices of a parallelogram $PQRS$,then:

The equations of the sides of a triangle are $x - 2y = 0$,$4x + 3y = 5$,and $2x + y = 0$. Through which point does the line $3y - 4x = 0$ pass?

If a diagonal of a square is along the line $8x - 15y = 0$ and one of its vertices is $(1, 2)$,then the equations of the sides of the square passing through this vertex are