Coordinate Geometry Class 10 NCERT Guide

Understanding the Coordinate Plane and Distance Formula

Coordinate geometry is the study of geometry using a coordinate system. A point in a two-dimensional Cartesian plane is represented as an ordered pair $(x, y)$, where $x$ is the abscissa (distance from the y-axis) and $y$ is the ordinate (distance from the x-axis). The distance between any two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by the Distance Formula:

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

This formula is derived directly from the Pythagoras Theorem. If we construct a right-angled triangle with $AB$ as the hypotenuse, the horizontal leg has length $|x_2 - x_1|$ and the vertical leg has length $|y_2 - y_1|$. Square these lengths, add them, and take the square root to find $AB$. This simple formula helps determine collinearity, calculate perimeters, and identify types of triangles or quadrilaterals.

Step-by-Step Guide to Applying the Section Formula

1. Identify the Given Coordinates — Locate the coordinates of the endpoints, say $A(x_1, y_1)$ and $B(x_2, y_2)$ on your grid or question paper.
2. Identify the Ratio — Determine the ratio $m_1 : m_2$ in which the internal division occurs. If the ratio is unknown, assume it is $k : 1$.
3. Apply the X-Coordinate Formula — Calculate the x-coordinate of the dividing point: $x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$. Ensure cross-multiplication of ratios and coordinates is done correctly.
4. Apply the Y-Coordinate Formula — Calculate the y-coordinate of the dividing point: $y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$.
5. Write the Final Coordinates — State the coordinates of the dividing point as $P(x, y)$. Note: For a midpoint, the ratio is $1:1$, simplifying the formula to $P\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} ight)$.

Avoiding Common Board Exam Mistakes

1. Sign Mistakes: Be extremely careful with negative coordinates. Subtracting a negative coordinate like $(x_2 - x_1)$ where $x_1 = -3$ becomes $(x_2 + 3)$.
2. Incorrect Ratio Multiplication: In the section formula, $m_1$ multiplies with $x_2$ and $m_2$ multiplies with $x_1$. Do not cross-multiply them with the wrong endpoints.
3. Collinearity Test: To prove three points $A$, $B$, and $C$ are collinear, show that the sum of two distances equals the third (e.g., $AB + BC = AC$). Note that the area of triangle method is omitted from the current Class 10 rationalized syllabus.
4. Writing Units: Always write 'units' after distance answers (e.g., $5\text{ units}$) as marks can be deducted for missing units.

Practice Questions with Solutions

• Q: Find the distance between the points $A(-1, 5)$ and $B(7, 3)$. A: Step 1: Identify coordinates: $x_1 = -1$, $y_1 = 5$, $x_2 = 7$, $y_2 = 3$. Step 2: Apply the Distance Formula: $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Step 3: Substitute values: $AB = \sqrt{(7 - (-1))^2 + (3 - 5)^2} = \sqrt{(8)^2 + (-2)^2}$. Step 4: Simplify: $AB = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17}$ units. Final answer: The distance between the points is $2\sqrt{17}$ units.
• Q: Find the coordinates of the point which divides the line segment joining the points $(4, -3)$ and $(8, 5)$ in the ratio $3 : 1$ internally. A: Step 1: Identify endpoints and ratio: $A(4, -3)$, $B(8, 5)$, $m_1 = 3$, $m_2 = 1$. Step 2: Apply Section Formula for x-coordinate: $x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{3(8) + 1(4)}{3 + 1} = \frac{24 + 4}{4} = \frac{28}{4} = 7$. Step 3: Apply Section Formula for y-coordinate: $y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{3(5) + 1(-3)}{3 + 1} = \frac{15 - 3}{4} = \frac{12}{4} = 3$. Step 4: Combine coordinates: $P(7, 3)$. Final answer: The coordinates of the point are $(7, 3)$.
• Q: Find the ratio in which the y-axis divides the line segment joining the points $(5, -6)$ and $(-1, -4)$. Also find the point of intersection. A: Step 1: Let the ratio be $k : 1$. Since the point of division lies on the y-axis, its x-coordinate is $0$. Let the point be $P(0, y)$. Step 2: Apply section formula for the x-coordinate: $x = \frac{k(-1) + 1(5)}{k + 1} = 0$. Step 3: Solve for $k$: $-k + 5 = 0 \Rightarrow k = 5$. Thus, the ratio is $5 : 1$. Step 4: Find the y-coordinate using $k = 5$: $y = \frac{5(-4) + 1(-6)}{5 + 1} = \frac{-20 - 6}{6} = \frac{-26}{6} = -\frac{13}{3}$. Final answer: The ratio is $5 : 1$, and the point of intersection is $(0, -13/3)$.
• Q: If the points $A(6, 1)$, $B(8, 2)$, $C(9, 4)$, and $D(p, 3)$ are the vertices of a parallelogram, taken in order, find the value of $p$. A: Step 1: Use the property that diagonals of a parallelogram bisect each other. Therefore, the midpoint of diagonal $AC$ is equal to the midpoint of diagonal $BD$. Step 2: Find the midpoint of $AC$: $M_{AC} = \left(\frac{6 + 9}{2}, \frac{1 + 4}{2}\right) = \left(\frac{15}{2}, \frac{5}{2}\right)$. Step 3: Find the midpoint of $BD$: $M_{BD} = \left(\frac{8 + p}{2}, \frac{2 + 3}{2}\right) = \left(\frac{8 + p}{2}, \frac{5}{2}\right)$. Step 4: Equate the x-coordinates: $\frac{15}{2} = \frac{8 + p}{2} \Rightarrow 15 = 8 + p \Rightarrow p = 7$. Final answer: The value of $p$ is $7$.

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