Q.Find the equation of the lines passing through the point $(1, 1)$

The general (linear) equation of a straight line is $ax+by+c=0$, where $a,b$ are not both zero; the set of solutions of any such equation is a straight line in the plane. Because dividing through by $b$ (or $a$) removes one constant, every line's equation genuinely contains only two independent arbitrary constants — so exactly two independent pieces of information (two points, or a point and a slope, or two intercepts, etc.) are enough to pin a line down uniquely.

Slope. The angle of inclination $\theta$ of a line is the angle it makes with the $x$-axis, measured counter-clockwise; the slope $m=\tan\theta$ (undefined when $\theta=\pi/2$, i.e. for a vertical line). Equivalently, through two points $(x_1,y_1),(x_2,y_2)$ with $x_1\ne x_2$, $m=\dfrac{y_2-y_1}{x_2-x_1}$; from the general form, $m=-a/b$ ($b\ne0$). Three points are collinear exactly when the slope of any one pair equals the slope of another pair sharing a point.

Intercepts. The $x$-intercept is where a line meets the $x$-axis ($y=0$); the $y$-intercept is where it meets the $y$-axis ($x=0$). ($x=0$ is itself the equation of the $y$-axis; $y=0$ is the equation of the $x$-axis.)

The six forms (two conditions each, all interconvertible by algebra):

Special cases of slope-intercept form: $b=0,\ m\ne0$ gives a line through the origin, $y=mx$; $b=0,\ m=0$ gives the $x$-axis itself, $y=0$; $b\ne0,\ m=0$ gives a horizontal line $y=b$. The point-slope form breaks down for a line parallel to the $y$-axis (slope undefined); such a line is simply $x=x_1$. A line through the origin, or a horizontal/vertical line, cannot be written in intercept form (an intercept would be $0$ or undefined).

General form to other forms. For $Ax+By+C=0$ ($A,B$ not both $0$): slope $=-A/B$, $y$-intercept $=-C/B$ (when $B\ne0$); $x$-intercept $=-C/A$, $y$-intercept $=-C/B$ (when $A,B,C$ all nonzero); normal form is obtained by dividing through by $\pm\sqrt{A^2+B^2}$, choosing the sign so the resulting constant (the normal length $p$) comes out positive: $\cos\alpha=\mp A/\sqrt{A^2+B^2}$, $\sin\alpha=\mp B/\sqrt{A^2+B^2}$, $p=|C|/\sqrt{A^2+B^2}$.