The equation of a line which makes an angle of $45^{\circ}$ with each of the pair of lines $xy-x-y+1=0$ is

If a pair of lines drawn through the origin forms an isosceles right-angled triangle with the line $2x + 3y = 6$,then those lines are

Let $P$ be the pair of lines represented by $2x^2 - 5xy + 2y^2 + 6x - 3y = 0$. Consider the following independent statements:
$(i)$ $\alpha$ is the $x$-coordinate of the point of intersection of the pair of lines $P$.
(ii) $\beta$ is the slope of one of the lines of $P$ passing through the origin.
(iii) $\gamma$ is the constant term in the equation of the pair of angular bisectors of $P$.
Then,

If $(a, b)$ is the centroid of the triangle formed by the lines $4x^2 - 17xy + 4y^2 = 0$ and $x + y - 5 = 0$,and $c$ is the numerical value of the area of the triangle,then $a + b + c =$

The normal to the curve $x^2 + 2xy - 3y^2 = 0$ at the point $(1,1)$ intersects the curve again in which quadrant?

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a