# Coordinates of a point on the plane

Let $XOX’$ and $YOY’$ be the two mutually perpendicular straight lines representing $\text{X-axis}$ and $\text{Y-axis}$ respectively meeting at a point $O$ which is known as the origin. Let $P$ be a point on the plane.

From $P$, draw $PM$ perpendicular to $XOX’$ meeting at the point $M$. If $OM=x$ and $PM=y$, then the point $P$ is associated with the numbers $x$ and $y$. We say that $P$ has the coordinates $(x,y)$, $x$ being the x-coordinate (abscissa) and $y$ being the y-coordinate (ordinate).

# Distance Formula

If $P(x_1,y_1)$ and $Q(x_2,y_2)$ be two points on the plane then the distance between them denoted by $d$ is given by $d=PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

# Slope Formula

If $P(x_1,y_1)$ and $Q(x_2,y_2)$ be two points then $\text{rise}=y_2-y_1$ and $\text{run}=x_2-x_1$. The slope of the line $PQ$ is defined as the ratio of the rise to the run. $\therefore\text{Slope of the line }(m)=\frac{y_2-y_1}{x_2-x_1}$ The slope of a line is also defined as the tangent of the angle made by the line $PQ$ with the positive x-axis. Thus, if $\theta$ is the angle made by the line $PQ$ with positive x-axis, then $\text{Slope of the line }(m)=\tan\theta=\frac{y_2-y_1}{x_2-x_1}$

# Section Formula

## Internal Division

If $R(x,y)$ be a point dividing the line joining the points $P(x_1,y_1)$ and $Q(x_2,y_2)$ internally in the ratio $m_1:m_2$, then $x=\frac{m_1x_2+m_2x_1}{m_1+m_2}\text{ and }y=\frac{m_1y_2+m_2y_1}{m_1+m_2}$ If $R(x,y)$ is the middle point of the line $PQ$, then $x=\frac{x_1+x_2}{2}\text{ and }y=\frac{y_1+y_2}{2}$

## External Division

If $R(x,y)$ be a point dividing the line joining the points $P(x_1,y_1)$ and $Q(x_2,y_2)$ externally in the ratio of $m_1:m_2$, then $x=\frac{m_1x_2-m_2x_1}{m_1-m_2}\text{ and }y=\frac{m_1y_2-m_2y_1}{m_1-m_2}$

# Centroid Formula

Let $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$ be the vertices of a triangle. If $G(x,y)$ be the centroid of the triangle, then $x=\frac{x_1+x_2+x_3}{3}\text{ and }y=\frac{y_1+y_2+y_3}{3}$

# Area of a triangle

If $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$ be the vertices of a triangle $ABC$, then the area of the triangle $ABC$ $=\frac{1}{2}(x_1y_2-x_2y_1+x_2y_3-x_3y_2+x_3y_1-x_1y_3)$ $=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$ The expression for the area of the triangle may also be expressed in the following determinant form $\frac{1}{2}\left|\begin{array}{c} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right|$ If the three points $A$, $B$ and $C$ are collinear, then $\text{Area of }\Delta ABC=0$ $\text{i.e. }x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$

# Relation between cartesian coordinates and polar coordinates

If $(x,y)$ be the cartesian coordinates of a point on the plane and $(r,\theta)$ be the polar coordinates of the same point, then $x=r\cos\theta\text{ and }y=r\sin\theta$ $\text{where, }r^2=x^2+y^2\text{ and }\theta=\tan^{-1}\frac{y}{x}$

# Equations of straight lines parallel to the axes

We know that the equation of x-axis is $y=0$, since any point on the x-axis will have its ordinate equal to zero. Similarly, the equation of y-axis can similarly be written as $x=0$.

Now consider a straight line parallel to x-axis at a distance $a$ units from it. Then, any point on the straight line has its ordinate equal to $a$. Hence, the equation of the straight line parallel to x-axis is $y=a$.

Similarly, the equation of a straight line parallel to y-axis at a distance $b$ from it is $x=b$.

## Three Standard Forms of Equation of Straight Lines

### Special Cases

More on Equation of Straight Lines: Linear Equation $Ax+By+C=0$

Next: Slope Intercept Form