Cartesian Plane & Coordinate Geometryकार्तीय तल एवं निर्देशांक ज्यामिति
Mastery of Cartesian coordinates, distance formula, section formula, midpoint rule, and area of triangles in the Cartesian plane as required for the Super TET Junior Mathematics exam. This module carries substantial weight in Part 2C, demanding high speed and absolute accuracy under negative marking penalties.
Mark complete once covered. Progress syncs automatically with the main syllabus hub.
Syllabus Focus & High-Yield Strategy: Mastery of Cartesian coordinates, distance formula, section formula, midpoint rule, and area of triangles in the Cartesian plane as required for the Super TET Junior Mathematics exam. This module carries substantial weight in Part 2C, demanding high speed and absolute accuracy under negative marking penalties.
1. Fundamental Axioms, Definitions & Foundational Properties
ASTC — All Silver Tea Cups (Quadrants I to IV for trigonometric ratios, but visually map quadrant signs as Right-Up, Left-Up, Left-Down, Right-Down).2. Formulas, Standard Theorems & Algorithmic Procedures
| Category / Case | Formula / Mathematical Rule | Topper Calculation Shortcut |
|---|---|---|
| Distance Formula | $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ | For points with equal ordinates or abscissas, simply take the absolute difference of the varying coordinates. |
| Section Formula | $(x, y) = \left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)$ | For midpoint ratio $1:1$, average the coordinates: $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. |
3. Word Problem Modeling & Practical Applications
4. Mathematical Edge Cases, Sign Conventions & Common Traps
MID — Minus In Distance, Plus In Midpoint $(x_2 - x_1)$ vs $(x_1 + x_2)$.Comprehensive Mathematical Matrix & Method Comparison
| Parameter / Dimension | Method / Concept A | Method / Concept B | Exam Speed & Accuracy Takeaway |
|---|---|---|---|
| Distance Formula | Cartesian Distance $d = \sqrt{\Delta x^2 + \Delta y^2}$ | Origin Distance $d = \sqrt{x^2 + y^2}$ | Use origin shortcut when one point is $(0,0)$. |
| Ratio Division | Internal Division $(+) | External Division $(-) | Super TET primarily tests internal division ($m:n$). |
| Collinearity | Distance Sum ($AB+BC=AC$) | Area of Triangle ($\Delta = 0$) | Area method is algebraically faster to compute. |
| Triangle Center | Centroid ($\sum x / 3$) | Midpoint ($\sum x / 2$) | Do not confuse divisor 3 (centroid) with divisor 2 (midpoint). |
1. What is the distance between the points $P(-6, 7)$ and $Q(-1, -5)$?
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Substituting values: $d = \sqrt{(-1 - (-6))^2 + (-5 - 7)^2} = \sqrt{(5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ units.
2. Find the coordinates of the point which divides the line segment joining $(4, -3)$ and $(8, 5)$ in the ratio $3:1$ internally.
Using section formula $(x, y) = \left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)$. Here $m_1=3, m_2=1, x_1=4, y_1=-3, x_2=8, y_2=5$. $x = \frac{3(8) + 1(4)}{3 + 1} = \frac{24 + 4}{4} = \frac{28}{4} = 7$. $y = \frac{3(5) + 1(-3)}{3 + 1} = \frac{15 - 3}{4} = \frac{12}{4} = 3$. Point is $(7, 3)$.
3. If the points $A(1, 2), B(0, 0)$ and $C(a, b)$ are collinear, which of the following relations holds true?
For collinearity with one point as origin $(0,0)$, the condition is $x_1 y_2 - x_2 y_1 = 0$ or slope equality. Slope of $AB = \frac{0-2}{0-1} = 2$. Slope of $BC = \frac{b-0}{a-0} = \frac{b}{a}$. Equating slopes: $\frac{b}{a} = 2 \implies b = 2a$ or $2a = b$.
4. Find the point on the X-axis which is equidistant from $(2, -5)$ and $(-2, 9)$.
Let the point on X-axis be $(x, 0)$. Equating distances from $(2, -5)$ and $(-2, 9)$: $(x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2$. Expanding: $x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$. Simplifying: $-4x + 29 = 4x + 85 \implies -8x = 56 \implies x = -7$. Point is $(-7, 0)$.
5. What is the area of the triangle formed by the vertices $(0, 0), (3, 0)$ and $(0, 4)$?
Using right-angled triangle formula or coordinate area: Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6$ square units.
6. If the midpoint of the line segment joining $(3, 4)$ and $(k, 6)$ is $(5, k)$, find the value of $k$.
Midpoint y-coordinate: $\frac{4 + 6}{2} = k \implies \frac{10}{2} = k \implies k = 5$. Checking x-coordinate: $\frac{3 + 5}{2} = 4$, which matches.
7. Determine the centroid of a triangle whose vertices are $(1, 4), (-2, 6)$ and $(3, 5)$.
Centroid formula: $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{1 - 2 + 3}{3}, \frac{4 + 6 + 5}{3}\right) = \left(\frac{2}{3}, \frac{15}{3}\right) = \left(\frac{2}{3}, 5\right)$.
8. The perimeter of the triangle with vertices $(0, 4), (0, 0)$ and $(3, 0)$ is:
Side lengths: Distance from $(0,0)$ to $(0,4)$ is $4$. Distance from $(0,0)$ to $(3,0)$ is $3$. Distance from $(0,4)$ to $(3,0)$ is $\sqrt{3^2 + 4^2} = 5$. Perimeter = $4 + 3 + 5 = 12$ units.
9. If the distance between the points $(x, -1)$ and $(3, 2)$ is $5$ units, what are the possible values of $x$?
Distance $d = \sqrt{(3 - x)^2 + (2 - (-1))^2} = 5$. Squaring both sides: $(3 - x)^2 + 3^2 = 25 \implies (3 - x)^2 + 9 = 25 \implies (3 - x)^2 = 16$. Taking square root: $3 - x = \pm 4$. If $3 - x = 4 \implies x = -1$. If $3 - x = -4 \implies x = 7$. Thus $x = 7$ or $x = -1$.
10. What type of quadrilateral is formed by joining the points $A(1, 7), B(4, 2), C(-1, -1)$ and $D(-4, 4)$ in order?
Calculating side lengths: $AB = \sqrt{(4-1)^2 + (2-7)^2} = \sqrt{9 + 25} = \sqrt{34}$. $BC = \sqrt{(-1-4)^2 + (-1-2)^2} = \sqrt{25 + 9} = \sqrt{34}$. $CD = \sqrt{(-4-(-1))^2 + (4-(-1))^2} = \sqrt{9 + 25} = \sqrt{34}$. $DA = \sqrt{(1-(-4))^2 + (7-4)^2} = \sqrt{25 + 9} = \sqrt{34}$. All four sides are equal. Diagonals $AC = \sqrt{(-1-1)^2 + (-1-7)^2} = \sqrt{4 + 64} = \sqrt{68}$ and $BD = \sqrt{(-4-4)^2 + (4-2)^2} = \sqrt{64 + 4} = \sqrt{68}$. Equal diagonals with equal sides prove it is a square.
5-Minute Exam Simulation Mode
Experience the real pressure of Super TET Junior (Part 2C Mathematics). You will face 10 randomized speed-math questions with negative marking (-1 for incorrect answers) and a strict 5-minute countdown.
High-Yield Rapid Recall Formulas & Laws (Must-Remember Pointers)
Speed Math & Memory Mnemonics Vault (Exam Retention Anchors)
Coordinate Geometry Distance Mnemonic
DASH
D - Difference in Abscissa squared, S - Sum with, H - Hypotenuse (Square root of sum of squares).
Midpoint Rule Mnemonic
HALF
H - Halve, A - Addition, L - Line, F - Fractions (of coordinates).
Negative Marking Traps & Common Blunders (-1 Penalty Avoidance)
Negative Marking Trap 1: Forgetting to square the negative signs inside binomial expansions like $(x_2 - x_1)^2$.
Negative Marking Trap 2: Mixing up $x$ and $y$ coordinates when substituting into the section formula.
Negative Marking Trap 3: Omitting the absolute value bars in the triangle area formula, leading to negative areas.
Negative Marking Trap 4: Confusing internal division with external division formula signs (+ vs -).