Find the value of $k$ for which the line $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$ passes through the origin.

The vertex $A$ of a triangle lies on the lines $x+y=1$ and $2x+3y=6$. If the orthocentre of the triangle is $O\left(\frac{3}{7}, \frac{22}{7}\right)$,then the equation of $OA$ in the normal form is

The polar equation $\cos \theta + 7 \sin \theta = \frac{1}{r}$ represents a

In the equation $y - y_1 = m(x - x_1)$,if $m$ and $x_1$ are fixed and different lines are drawn for different values of $y_1$,then

If a line intercepted between the coordinate axes is trisected at a point $A(4, 3)$,which is nearer to the $x$-axis,then its equation is

If the midpoints of the sides $BC, CA$ and $AB$ of the triangle $ABC$ are $(1, 3), (5, 7)$ and $(-5, 7)$ respectively,then the equation of the side $AB$ is