Three lines $x + 2y + 3 = 0$,$x + 2y - 7 = 0$,and $2x - y - 4 = 0$ form three sides of two squares. Find the equation of the fourth side of each square.

The line $x + y = p$ meets the $x$ and $y$ axes at $A$ and $B$ respectively. $A$ triangle $APQ$ is inscribed in the triangle $OAB$,where $O$ is the origin,with a right angle at $Q$. $P$ and $Q$ lie on $OB$ and $AB$ respectively. If the area of triangle $APQ$ is $3/8$ of the area of triangle $OAB$,then $\frac{AQ}{BQ}$ is equal to:

Suppose that the points $(h, k)$,$(1, 2)$,and $(-3, 4)$ lie on the line $l_1$. If a line $l_2$ passing through the points $(h, k)$ and $(4, 3)$ is perpendicular to $l_1$,then $\left(\frac{k}{h}\right)$ equals

$A$ is the point of intersection of the lines $3x + y - 4 = 0$ and $x - y = 0$. If a line having a negative slope makes an angle of $45^{\circ}$ with the line $x - 3y + 5 = 0$ and passes through $A$,then its equation is:

The equation of the line passing through the point of intersection of the lines $2x + y - 4 = 0$ and $x - 3y + 5 = 0$ and lying at a distance of $\sqrt{5}$ units from the origin is:

$A$ straight line $x/a - y/b = 1$ passes through the point $(8, 6)$ and cuts a triangle of area $12 \text{ sq units}$ from the axes of coordinates. The equations of the straight lines are