Alright class, let's embark on a deep dive into the fascinating world of
Cartesian Coordinates in a Plane. This topic isn't just a fundamental building block for coordinate geometry; it's the very language we use to translate geometric problems into algebraic equations and vice-versa. Think of it as giving every point in a flat world a unique address!
### 1. Introduction: The Need for Coordinates
Imagine you're trying to describe the location of a specific spot on a flat map to someone. How would you do it? You might say, "Go 5 units east, then 3 units north." This is essentially what a coordinate system does! It provides a systematic way to pinpoint any location in a given space.
The system we primarily use, especially in JEE Mathematics, is the
Cartesian Coordinate System, named after the brilliant French mathematician and philosopher RenΓ© Descartes.
#### 1.1 From Number Line to Coordinate Plane
You're already familiar with the
number line. It's a straight line where every point corresponds to a unique real number. We typically mark a central point as 0 (the origin), positive numbers to the right, and negative numbers to the left.
To move from a one-dimensional line to a two-dimensional plane, Descartes' ingenious idea was to use *two* perpendicular number lines.
* We draw a horizontal line, which we call the
X-axis (or abscissa axis).
* Perpendicular to it, passing through its origin, we draw a vertical line, called the
Y-axis (or ordinate axis).
The point where these two axes intersect is called the
Origin, denoted by $mathbf{O(0, 0)}$.
Key Idea: Every point in this plane can now be uniquely identified by an ordered pair of real numbers, $mathbf{(x, y)}$, where 'x' tells you its horizontal position relative to the origin, and 'y' tells you its vertical position.
#### 1.2 Quadrants and Sign Conventions
The two axes divide the plane into four regions, called
quadrants. These are numbered counter-clockwise:
Quadrant |
X-coordinate (Abscissa) |
Y-coordinate (Ordinate) |
Example Point |
|---|
I |
Positive (+) |
Positive (+) |
(3, 5) |
II |
Negative (-) |
Positive (+) |
(-2, 4) |
III |
Negative (-) |
Negative (-) |
(-6, -1) |
IV |
Positive (+) |
Negative (-) |
(7, -2) |
* Points on the X-axis have a y-coordinate of 0 (e.g., (5, 0)).
* Points on the Y-axis have an x-coordinate of 0 (e.g., (0, -3)).
* The origin is (0, 0).
CBSE vs. JEE Focus: While CBSE focuses on plotting points and identifying quadrants, JEE often uses these basic concepts as a starting point for more complex problems involving transformations, loci, and properties of geometric figures. A strong grasp of quadrant signs is crucial for visualizing functions and geometric transformations.
### 2. The Distance Formula
One of the most fundamental applications of coordinate geometry is finding the distance between two points in a plane.
#### 2.1 Derivation of the Distance Formula
Let's say we have two points, $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, and we want to find the distance between them, denoted by $d$.
1. Draw a horizontal line through $P_1$ and a vertical line through $P_2$. These lines intersect at a point, let's call it $Q$.
2. The coordinates of $Q$ will be $(x_2, y_1)$. (Think about it: its x-coordinate is the same as $P_2$, and its y-coordinate is the same as $P_1$).
3. Now, we have a right-angled triangle $P_1QP_2$.
* The length of the horizontal side $P_1Q$ is the absolute difference in their x-coordinates: $|x_2 - x_1|$.
* The length of the vertical side $QP_2$ is the absolute difference in their y-coordinates: $|y_2 - y_1|$.
4. According to the
Pythagoras Theorem, in a right-angled triangle, the square of the hypotenuse (the distance $P_1P_2$) is equal to the sum of the squares of the other two sides.
$d^2 = (P_1Q)^2 + (QP_2)^2$
$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$
5. Taking the square root of both sides, we get the
Distance Formula:
$mathbf{d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$
Important Note: Since $(x_2 - x_1)^2 = (x_1 - x_2)^2$ and $(y_2 - y_1)^2 = (y_1 - y_2)^2$, the order of subtraction doesn't matter for the final distance value.
#### 2.2 Examples of Distance Formula
Example 1: Find the distance between the points $A(3, 4)$ and $B(-1, 7)$.
- Let $(x_1, y_1) = (3, 4)$ and $(x_2, y_2) = (-1, 7)$.
- Apply the distance formula:
$d = sqrt{(-1 - 3)^2 + (7 - 4)^2}$
$d = sqrt{(-4)^2 + (3)^2}$
$d = sqrt{16 + 9}$
$d = sqrt{25}$
$mathbf{d = 5}$ units.
Example 2: Show that the points $P(1, -1)$, $Q(5, 2)$, and $R(9, 5)$ are collinear using the distance formula.
Concept: Three points are collinear if the sum of the distances between two pairs of points equals the distance of the third pair. For example, if $PQ + QR = PR$, then P, Q, R are collinear.
- Calculate $PQ$:
$PQ = sqrt{(5 - 1)^2 + (2 - (-1))^2} = sqrt{4^2 + 3^2} = sqrt{16 + 9} = sqrt{25} = 5$
- Calculate $QR$:
$QR = sqrt{(9 - 5)^2 + (5 - 2)^2} = sqrt{4^2 + 3^2} = sqrt{16 + 9} = sqrt{25} = 5$
- Calculate $PR$:
$PR = sqrt{(9 - 1)^2 + (5 - (-1))^2} = sqrt{8^2 + 6^2} = sqrt{64 + 36} = sqrt{100} = 10$
- Check for collinearity:
$PQ + QR = 5 + 5 = 10$
Since $PQ + QR = PR$, the points $P, Q, R$ are collinear.
### 3. The Section Formula
The section formula helps us find the coordinates of a point that divides a line segment joining two given points in a specific ratio. This is a very powerful tool in coordinate geometry.
#### 3.1 Internal Division - Derivation
Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be two points. Let $P(x, y)$ be a point that divides the line segment $AB$ internally in the ratio $m : n$. This means $AP/PB = m/n$.
1. Draw perpendiculars from $A, P, B$ to the X-axis, meeting it at $A'$, $P'$, $B'$ respectively.
2. Draw a line through $A$ parallel to the X-axis, intersecting $PP'$ at $C$ and $BB'$ at $D$.
3. We now have two similar triangles: $ riangle APC$ and $ riangle BPD$. (They are similar because $AC parallel PD parallel BB'$, so corresponding angles are equal).
4. From similarity, the ratio of corresponding sides is equal to the ratio of division:
$frac{AP}{PB} = frac{AC}{CD} = frac{PC}{DB} = frac{m}{n}$
5. Let's look at the x-coordinates:
$AC = x - x_1$
$CD = x_2 - x$
So, $frac{x - x_1}{x_2 - x} = frac{m}{n}$
$n(x - x_1) = m(x_2 - x)$
$nx - nx_1 = mx_2 - mx$
$nx + mx = mx_2 + nx_1$
$x(m + n) = mx_2 + nx_1$
$mathbf{x = frac{mx_2 + nx_1}{m + n}}$
6. Similarly, for the y-coordinates:
$PC = y - y_1$
$DB = y_2 - y$
So, $frac{y - y_1}{y_2 - y} = frac{m}{n}$
$n(y - y_1) = m(y_2 - y)$
$ny - ny_1 = my_2 - my$
$ny + my = my_2 + ny_1$
$y(m + n) = my_2 + ny_1$
$mathbf{y = frac{my_2 + ny_1}{m + n}}$
Combining these, the coordinates of $P(x, y)$ are:
$mathbf{P(x, y) = left( frac{mx_2 + nx_1}{m + n}, frac{my_2 + ny_1}{m + n}
ight)}$
#### 3.2 Special Case: Midpoint Formula
If $P$ is the
midpoint of $AB$, then it divides $AB$ in the ratio $1:1$ (i.e., $m=1, n=1$).
Substituting $m=1, n=1$ into the section formula:
$mathbf{x = frac{1 cdot x_2 + 1 cdot x_1}{1 + 1} = frac{x_1 + x_2}{2}}$
$mathbf{y = frac{1 cdot y_2 + 1 cdot y_1}{1 + 1} = frac{y_1 + y_2}{2}}$
So, the midpoint $M(x, y)$ is $mathbf{left( frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2}
ight)}$.
#### 3.3 External Division
If point $P(x, y)$ divides the line segment $AB$ externally in the ratio $m:n$, it means $P$ lies on the line containing $AB$ but outside the segment $AB$, such that $AP/PB = m/n$.
The formula for external division is:
$mathbf{P(x, y) = left( frac{mx_2 - nx_1}{m - n}, frac{my_2 - ny_1}{m - n}
ight)}$
Tip for JEE: Sometimes, external division can be thought of as internal division with a negative ratio. If $P$ divides $AB$ externally in ratio $m:n$, it divides $AB$ internally in ratio $m:(-n)$. Substituting $n
ightarrow -n$ in the internal division formula gives the external division formula.
#### 3.4 Examples of Section Formula
Example 3: Find the coordinates of the point that divides the line segment joining $A(-1, 7)$ and $B(4, -3)$ in the ratio $2:3$ internally.
- Here, $(x_1, y_1) = (-1, 7)$, $(x_2, y_2) = (4, -3)$, $m=2$, $n=3$.
- Using the internal section formula:
$x = frac{mx_2 + nx_1}{m + n} = frac{2(4) + 3(-1)}{2 + 3} = frac{8 - 3}{5} = frac{5}{5} = 1$
$y = frac{my_2 + ny_1}{m + n} = frac{2(-3) + 3(7)}{2 + 3} = frac{-6 + 21}{5} = frac{15}{5} = 3$
- The point is $mathbf{(1, 3)}$.
Example 4: In what ratio does the X-axis divide the line segment joining the points $P(2, -3)$ and $Q(-4, 6)$?
Concept: Any point on the X-axis has its y-coordinate as 0. Let the ratio be $k:1$.
- Let the point of division be $R(x, 0)$. Let it divide $PQ$ in the ratio $k:1$.
- Using the section formula for the y-coordinate:
$y = frac{ky_2 + 1 cdot y_1}{k + 1}$
$0 = frac{k(6) + 1(-3)}{k + 1}$
$0 = 6k - 3$
$6k = 3$
$k = frac{3}{6} = frac{1}{2}$
- The ratio is $k:1 = 1/2 : 1$, which is $mathbf{1:2}$. Since $k$ is positive, the division is internal.
JEE Advanced Callout: The section formula is crucial for deriving coordinates of important geometric centers like the Centroid (intersection of medians, divides median in 2:1 ratio), Incenter (intersection of angle bisectors), and Excenters. For these, understanding the ratio in which these points divide the respective segments is key. This will be explored in more detail in chapters on triangles.
### 4. Area of a Triangle in Coordinate Geometry
When the vertices of a triangle are given by their coordinates, we can calculate its area using a simple formula. This is a powerful application, as it eliminates the need to calculate base and height explicitly.
#### 4.1 Derivation Intuition (using Trapezoids)
Imagine a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.
1. Draw perpendiculars from $A, B, C$ to the X-axis, meeting it at $A'$, $B'$, $C'$.
2. The area of $ riangle ABC$ can be found by summing and subtracting areas of trapezoids:
Area($ riangle ABC$) = Area(Trapezoid $AA'C'C$) + Area(Trapezoid $CC'B'B$) - Area(Trapezoid $AA'B'B$)
This approach is a bit cumbersome. A more direct and commonly used formula is often introduced without full derivation at this stage, but it stems from vector cross products or determinant properties, which you'll learn later.
#### 4.2 Area Formula
The area of a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ is given by:
$mathbf{ ext{Area} = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|}$
The absolute value sign ensures that the area is always positive. The formula can be remembered using a determinant-like structure or the "shoelace" method.
Shoelace Formula (for polygons with N vertices):
For a polygon with vertices $(x_1, y_1), (x_2, y_2), dots, (x_n, y_n)$ listed in counter-clockwise (or clockwise) order:
Area $= frac{1}{2} |(x_1y_2 + x_2y_3 + dots + x_ny_1) - (y_1x_2 + y_2x_3 + dots + y_nx_1)|$
This is extremely useful for quadrilaterals, pentagons, etc.
#### 4.3 Condition for Collinearity
If three points $A, B, C$ are collinear, they cannot form a triangle. Therefore, the area of the "triangle" formed by them must be zero.
$mathbf{ ext{Condition for collinearity}: x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0}$
This is a much more efficient way to check for collinearity compared to the distance formula method.
#### 4.4 Examples of Area Formula
Example 5: Find the area of the triangle whose vertices are $A(1, -1)$, $B(-4, 6)$, and $C(-3, -5)$.
- Let $(x_1, y_1) = (1, -1)$, $(x_2, y_2) = (-4, 6)$, $(x_3, y_3) = (-3, -5)$.
- Apply the area formula:
Area $= frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area $= frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|$
Area $= frac{1}{2} |1(6 + 5) - 4(-5 + 1) - 3(-7)|$
Area $= frac{1}{2} |1(11) - 4(-4) - 3(-7)|$
Area $= frac{1}{2} |11 + 16 + 21|$
Area $= frac{1}{2} |48|$
$mathbf{ ext{Area} = 24}$ square units.
Example 6: Find the value of $k$ if the points $A(2, 3)$, $B(4, k)$, and $C(6, -3)$ are collinear.
- For the points to be collinear, the area of the triangle formed by them must be zero.
- Set the area formula to 0:
$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$
$2(k - (-3)) + 4(-3 - 3) + 6(3 - k) = 0$
$2(k + 3) + 4(-6) + 6(3 - k) = 0$
$2k + 6 - 24 + 18 - 6k = 0$
$-4k + 0 = 0$
$-4k = 0$
$mathbf{k = 0}$
### 5.
JEE Focus and Advanced Insights
*
Locus: While not directly a "basics" topic, understanding Cartesian coordinates is foundational for the concept of
locus. A locus is the path traced by a point that moves under certain geometric conditions. For example, the locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining them. You'll translate these geometric conditions into algebraic equations using distance, section, or area formulas.
*
Transformations: A basic understanding of how coordinates change under simple transformations like translation (shifting the origin) or reflection (across an axis or a line) can simplify problems. For instance, translating the origin to $(h,k)$ changes coordinates from $(x,y)$ to $(X,Y)$ where $x = X+h$ and $y = Y+k$.
*
Geometric Properties via Algebra: In JEE, you'll constantly be asked to prove geometric properties (e.g., a quadrilateral is a parallelogram, a triangle is isosceles, lines are perpendicular) using coordinate geometry. This means converting geometric statements into algebraic equations (using distance, slope, section formula) and solving them.
*
Visualization is Key: Always try to sketch the points and figures on a coordinate plane. This helps in understanding the problem, identifying potential symmetries, and verifying your results. A rough sketch can prevent silly errors.
This deep dive into Cartesian coordinates provides a robust foundation. Master these basic formulas and their derivations, practice with varied examples, and you'll be well-prepared for the more advanced topics in coordinate geometry that are integral to JEE. Keep translating geometric intuition into algebraic expressions, and you'll excel!