1. Define coordinate plane (Cartesian plane).
The plane formed by two straight lines perpendicular to each other is called coordinate plane and the lines $XOX'$ and $YOY'$ are called coordinate axes.
Class 9 Mathematics | Punjab Curriculum and Textbook Board Syllabus 2025
The plane formed by two straight lines perpendicular to each other is called coordinate plane and the lines $XOX'$ and $YOY'$ are called coordinate axes.
An ordered pair of real numbers $x$ and $y$ is a pair $(x,\ y)$ in which elements are written in specific order.
Note: $\left( \mathbf{x,\ y} \right)\mathbf{\neq (y,\ x)}$
The point of intersection of two coordinate axes is called origin.
The $x$-$coordinate$ of the point is called abscissa of the point $P(x,y)$ and the $y$-$coordinate$ is called its ordinate.
If $A\left( x_{1},y_{1} \right)$ and $B\left( x_{2},y_{2} \right)$ are two points and $d$ is the distance between them, then
$$d = |\overline{AB}| = \sqrt{\left( x_{2} - x_{1} \right)^{2} + \left( y_{2} - y_{1} \right)^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d \geq 0$$
Note: $\left| \overline{AB} \right|$ stands for $m\overline{AB}$
If $A\left( x_{1},y_{1} \right)$ and $B\left( x_{2},y_{2} \right)$ are two points in the plane, then the mid-point $M(x,y)$ of line segment $\overline{AB}$ is
$$M(x,y) = \left( \frac{x_{1} + x_{2}}{2}\ ,\frac{y_{1} + y_{2}}{2} \right)$$
The inclination of a line is the angle $\alpha$ ($0^{{^\circ}} < \alpha < \ 180^{{^\circ}}$) measured counterclockwise from the positive $x$-$axis$ to a non-horizontal straight line $l$.
Slope or gradient of an inclined path is a measure of its steepness, denoted by m. It is defined as the ratio of rise to run:
$$m = \frac{rise}{run} = \frac{y}{x} = \tan\alpha$$
In analytical geometry, for a non-vertical line with inclination $\alpha$,
$$m = \tan\alpha$$
Theorem: If a non-vertical line $l$ with inclination $\alpha$ passes through two points $P\left( x_{1},y_{1} \right)$ and $Q\left( x_{2},y_{2} \right)$, then the slope or gradient $m$ of the line is given by:
$$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \tan\alpha$$
Theorem 2: Let two lines $l_{1}$ and $l_{2}$ have slopes $m_{1}$ and $m_{2}$, respectively:
$${Slope\ of\ \overline{AB} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} }{= \frac{2 - 6}{3 - ( - 3)} }{= \frac{- 4}{6} }{= \frac{- 2}{3}}$$
$${Slope\ of\ \overline{BC} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} }{= \frac{0 - 2}{6 - 3} }{= \frac{- 2}{3}}$$
$$\because\ Slope\ of\ \overline{AB} = Slope\ of\ \overline{BC}$$
Hence, the points $\mathbf{A,\ B,}$ and $\mathbf{C}$ are collinear.
$${Slope\ of\ \overline{AB} = m_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} }{= \frac{5 - 1}{4 - 1} }{m_{1} = \frac{4}{3}}$$
$${Slope\ of\ \overline{BC} = m_{2} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} }{= \frac{- 1 - 5}{12 - 4} }{= \frac{- 6}{8} }{m_{2} = \frac{- 3}{4}}$$
$${m_{1} \cdot m_{2} = \left( \frac{4}{3} \right)\left( \frac{- 3}{4} \right) }{m_{1} \cdot m_{2} = - 1}$$
Therefore, $\overline{AB}\bot\overline{BC}$. So $\mathrm{\Delta}ABC$ is a right triangle.
A line that goes left to right (horizontal) is called a line parallel to the x-axis. It is also said to be perpendicular to the y-axis. The equation of this line is:
$$y = a$$
A line that goes up and down (vertical) is called a line parallel to the y-axis. It is also said to be perpendicular to the x-axis. The equation of this line is:
$$x = b$$
The x-intercept of a line is the point where it crosses the x-axis. If the line crosses the x-axis at $(a,0)$, then $\mathbf{x}$-$\mathbf{intercept = a}$.
The y-intercept of a line is the point where it crosses the y-axis. If the line crosses the y-axis at $(0,b)$, then $\mathbf{y}$-$\mathbf{intercept = b}$.
The equation of a non-vertical straight line with slope $m$ and $y$-$intercept$ $c$ is:
$$y = mx + c$$
Note: If the line passes through the origin, then $c = 0$. So, the equation becomes: $y = mx$
If a non-vertical line with slope $m$ passes through a point $Q\left( x_{1},y_{1} \right)$, then its equation is:
$$y - y_{1} = m\left( x - x_{1} \right)$$
The equation of the line passing through two given points can be derived using the point-slope form:
Step 1: Calculate the slope (m) between the two points:
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
Step 2: Using point-slope form with point Q:
$$y - y_1 = m(x - x_1)$$
$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$
Alternative Form: Using point R instead:
$$y - y_2 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_2)$$
Key Notes:
Example: Find the equation through points (2,3) and (4,7)
$$m = \frac{7-3}{4-2} = 2$$
Using first point: $$y - 3 = 2(x - 2)$$
Simplifies to: $$y = 2x - 1$$
The equation of the line passing through two given points is:
$${y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\left( x - x_{1} \right) }{or\ \ \ \ \ \ \ \ \ \ \ \ \ \ y - y_{2} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\left( x - x_{2} \right)\ }$$
The equation of a line with non-zero $\mathbf{x}$-$intercept$ $a$ and $y$-$intercept$ $b$ is:
$$\frac{x}{a} + \frac{y}{b} = 1$$
The equation of the line is:
$$x\cos\alpha + y\sin\alpha = p$$
Starting with general form:
$$ax + by + c = 0$$
Solve for y:
$$by = -ax - c$$
$$y = \frac{-ax - c}{b}$$
$$y = -\frac{a}{b}x - \frac{c}{b}$$
Compare with $y = mx + c_1$:
$$m = -\frac{a}{b}$$
$$c_1 = -\frac{c}{b}$$
From slope-intercept form, we have slope:
$$m = -\frac{a}{b}$$
Take a known point on the line (x-intercept):
$$Q\left(-\frac{c}{a}, 0\right)$$
Using point-slope form:
$$y - y_1 = m(x - x_1)$$
$$y - 0 = -\frac{a}{b}\left(x - \left(-\frac{c}{a}\right)\right)$$
$$y = -\frac{a}{b}\left(x + \frac{c}{a}\right)$$
$$y = -\frac{a}{b}x - \frac{c}{b}$$
Starting with general form:
$$ax + by + c = 0$$
Move constant term:
$$ax + by = -c$$
Divide by -c:
$$\frac{ax}{-c} + \frac{by}{-c} = 1$$
Simplify:
$$\frac{x}{-\frac{c}{a}} + \frac{y}{-\frac{c}{b}} = 1$$
Where:
x-intercept $= -\frac{c}{a}$
y-intercept $= -\frac{c}{b}$
Starting with general form:
$$ax + by + c = 0$$
Divide by $\pm\sqrt{a^2 + b^2}$:
$$\frac{ax}{\pm\sqrt{a^2 + b^2}} + \frac{by}{\pm\sqrt{a^2 + b^2}} = \frac{-c}{\pm\sqrt{a^2 + b^2}}$$
Let:
$$\cos\alpha = \frac{a}{\pm\sqrt{a^2 + b^2}}$$
$$\sin\alpha = \frac{b}{\pm\sqrt{a^2 + b^2}}$$
$$p = \frac{-c}{\pm\sqrt{a^2 + b^2}}$$
Resulting in normal form:
$$x\cos\alpha + y\sin\alpha = p$$