The distance between two points $P\left(x_1,y_1\right)$ and $Q\left(x_2,y_2\right)$ is given by

$|PQ|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The area of a triangle, whose vertices are $\left(x_1,y_1\right),\left(x_2,y_2\right)$, and $\left(x_3,y_3\right)$ is

= $\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

The area of polygon whose vertices are $\left(x_1, y_1\right),\left(x_2, y_2\right)$, $\left(x_3, y_3\right), \ldots,\left(x_n, y_n\right)$ is

$$ =\frac{1}{2}\left|\left\{\left(x_1 y_2-x_2 y_1\right)+\left(x_2 y_3-x_3 y_2\right)+\cdots+\left(x_n y_1-x_1 y_n\right)\right\}\right| $$

Coordinates of the point that divides the line segment joining the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ internally in the ratio $m: n$ are given by

$$ \begin{aligned} &x=\frac{m x_2+n x_1}{m+n}, \\\\ &y=\frac{m y_2+n y_1}{m+n} \end{aligned} $$

Coordinates of the point that divides the line segment joining the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ externally in the ratio $m$ : $\mathrm{n}$ are given by

$x=\frac{m{x}_2-n{x}_1}{m-n},$

$y=\frac{m{y}_2-n{y}_1}{m-n}$

The point of concurrency of the medians of a triangle is called the centroid of the triangle. The coordinates of the centroid of the triangle with vertices as $\left(x_1, y_1\right),\left(x_2, y_2\right)$, and $\left(x_3, y_3\right)$ are

$$ \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$

The coordinates of the incentre of a triangle whose vertices are $A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right)$ are

$$ \left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}\right) $$

where, $a, b, c$ are the lengths of sides $B C, C A$ and $A B$ respectively.

If $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right)$ are two points on a line $l$, then the slope $m$ of the line $l$ is given by

$$ m=\frac{y_2-y_1}{x_2-x_1}, x_1 \neq x_2 $$

If $x_1=x_2$, then $m$ is not defined. In that case the line is perpendicular to $X$-axis.

The angle $\theta$ between the lines having slope $m_1$ and $m_2$ is given by

$\tan \theta =\pm \frac{m_2-m_1}{1+m_1{m}_2}$

and the acute angle $\theta$ between the lines having slopes $m_1$ and $m_2$ is given by

$\theta ={\tan }^{-1}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

If two lines of slopes $m_1$ and $m_2$ are parallel, then the angle $\theta$ between is $0^{\circ}$.

$$ \begin{aligned} \therefore \tan \theta=\tan 0^{\circ}=0 \\\\ \Rightarrow \frac{m_2-m_1}{1+m_1 m_2}=0 \\\\ \Rightarrow m_2 = m_1 \end{aligned} $$

Thus, when two lines are parallel, their slopes are equal.

If two lines of slopes $m_1$ and $m_2$ are perpendicular, then

$m_1{m}_2=-1$

Thus, when lines are perpendicular, the product of their slope is $-1$.

If $m$ is the slope of a line, then the slope of a line perpendicular to it is $-\frac{1}{m}$.

The equation of a line with slope $m$ that makes an intercept $c$ on $y$-axis is

$$ y=m x+c $$

The equation of a line which passes through the point $\left(x_1, y_1\right)$ and has the slope '$m$ is

$$ y-y_1=m\left(x-x_1\right) $$

The equation of a line passing through two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is

$$ y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right) $$

The equation of a line which cuts-off intercepts $a$ and $b$, respectively from the $x$ and $y$-axes is

$\frac{x}{a}+\frac{y}{b}=1$

The equation of the straight line upon which the length of the perpendicular from the origin is $p$ and this perpendicular makes an angle $\alpha$ with +ve direction of $x$-axis is

$x\cos \alpha +y\sin \alpha =p$

Let the angle $\theta$ between the lines $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$ is given by

$$ \tan \theta=\left|\frac{a_2 b_1-a_1 b_2}{a_1 a_2+b_1 b_2}\right| $$

If the lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ are parallel, then $m_1=m_2$

$\Rightarrow -\frac{a_1}{b_1}=-\frac{a_2}{b_2}$

$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}$

If the lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ are perpendicular, then

$m_1{m}_2=-1$

$\Rightarrow \left(-\frac{a_1}{b_1}\right)\times \left(-\frac{a_2}{b_2}\right)=-1$

$\Rightarrow a_1{a}_2+b_1{b}_2=0$

The equation of a line parallel to a given line $a x+b y+c=0$ is

$$ a x+b y+\lambda=0 $$

where $\lambda$ is $a$ constant.

The equation of a line perpendicular to a given line $ax+by$ $+c=0$ is

$bx-ay+\lambda =0$

where $\lambda$ is a constant.

The equation of the straight line passing through $\left(x_1,y_1\right)$ and making an angle $\theta$ with the positive direction of $x$-axis is

$\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }=r$

where $r$ is the distance of the point $(x,y)$ on the line from point $\left(x_1,y_1\right)$.

The equation of the straight line upon which the length of perpendicular from the origin is $p$ and this normal makes an angle $\alpha$ with the positive direction of $X$-axis is

$$ x \cos \alpha+y \sin \alpha=p . $$

The length of the perpendicular from a point $\left(x_1, y_1\right)$ to a line $a x+b y+c=0$ is

$$ \left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right| $$

Let the two parallel lines be

$$ a x+b y+c=0 \text { and } a x+b y+c_1=0 $$

The distance between the parallel lines is the perpendicular distance of any point on one line from the other line.

Let $\left(x_1, y_1\right)$ be any point on $a x+b y+c=0$

$\therefore a x_1+b y_1+c=0$ .......... (i)

Now, perpendicular distance of the point $\left(x_1, y_1\right)$ from the line $a x+b y+c_1=0$ is

$$ \frac{\left|a x_1+b y_1+c_1\right|}{\sqrt{\left(a^2+b^2\right)}}=\frac{\left|c_1-c\right|}{\sqrt{\left(a^2+b^2\right)}} $$ [from Eq. (i)]

This is required distance between the given parallel lines.

Area of parallelogram $A B C D$ whose sides $A B, B C, C D$ and $D A$ are represented by

$a_1 x+b_1 y+c_1=0$, $a_2 x+b_2 y+c_2=0, a_1 x+b_1 y+d_1=0$ and $a_2 x+b_2 y+d_2=0$ is

$$ \frac{p_1 p_2}{\sin \theta} \text { or } \frac{\left|c_1-d_1\right|\left|c_2-d_2\right|}{|\left|\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right| \mid} $$

where, $p_1$ and $p_2$ are the distances between parallel sides and $\theta$ is the angle between two adjacent sides.

The three given lines are concurrent, if they meet in a point. Hence to prove that three given lines are concurrent, we proceed as follows :

Method 1 : Find the point of intersection of any two lines by solving them simultaneously. If this point satisfies the third equation also, then the given lines are concurrent.

Method 2 : The three lines $a_i x+b_i y+c_i=0, i=1,2,3$ are concurrent if

$$ \left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=0 $$

Method 3 : The condition for the lines $P=0, Q=0$ and $R=0$ to be concurrent is that three constants $l, m, n$ (not all zeros at the same time) can be obtained such that

$$ l P+m Q+n R=0 $$

Any line through the point of intersection of the lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ can be represented by the equation

$\left(a_{1}x+b_{1}y+c_1\right)+\lambda \left(a_{2}x+b_{2}y+c_2\right)=0$

where $\lambda$ is a parameter which depends on the other property of line.

The equations of the straight lines which pass through a given point $\left(x_1, y_1\right)$ and make a given angle $\alpha$ with the given straight line $y=m x+c$ are

$ (y-y_1) =\tan (\theta \pm \alpha)\left(x-x_1\right) \\ \text { where, } m =\tan \theta$

If two lines with slopes $m_1$ and $m_2$ be equally inclined to a line with slope $m$, then

$\left(\frac{m_1-m}{1+m{m}_1}\right)=-\left(\frac{m_2-m}{1+m{m}_2}\right)$

Prove that the equation of the bisectors of the angles between the lines

$a_{1}x+b_{1}y+c_1=0\text{ and }{a}_{2}x+b_{2}y+c_2=0$ are given by

$\frac{\left(a_{1}x+b_{1}y+c_1\right)}{\sqrt{\left(a_1^2+b_1^2\right)}}=\pm \frac{\left(a_{2}x+b_{2}y+c_2\right)}{\sqrt{\left(a_2^2+b_2^2\right)}}$

Let the image of $A\left(x_1,y_1\right)$ with respect to the line mirror $ax+by+c=0$ be $B\left(x_2,y_2\right)$ then it is given by

$\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{-2\left(a{x}_1+b{y}_1+c\right)}{\left(a^2+b^2\right)}$

The point of intersection of the two lines represented by

$\begin{array}{rl}& a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0\text{ is }\\ & \\ & \left(\frac{bg-hf}{h^2-ab},\frac{af-gh}{h^2-ab}\right)\end{array}$