Coordinate Geometry Exercise 7.2 Class 10 NCERT Solutions

Understanding the Section Formula

The primary focus of NCERT Class 10 Exercise 7.2 is the Section Formula. Suppose we have two points $A(x_1, y_1)$ and $B(x_2, y_2)$ joined by a line segment. If a third point $P(x, y)$ lies on the line segment $AB$ such that it divides $AB$ internally in the ratio $m_1 : m_2$ (i.e., $AP : PB = m_1 : m_2$), then the coordinates of $P$ are given by:

$P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right)$

If $P$ is the midpoint of $AB$, it divides the line in the ratio $1:1$. Substituting $m_1 = 1$ and $m_2 = 1$ simplifies the formula to:

$P(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

To find an unknown ratio, we often assume the ratio to be $k : 1$. This reduces our variables and makes solving equations involving coordinates much faster.

Key Concepts and Terminology

Internal Division

When a point lies on the line segment joining two points and divides the segment into two parts internally.

Section Formula

A mathematical formula used to find the coordinates of a point that divides a line segment in a given ratio.

Trisection Points

Two points that divide a line segment into three equal parts, dividing it in ratios of 1:2 and 2:1 respectively.

Midpoint

The exact middle point of a line segment that divides it into two equal parts (ratio 1:1).

How to Find the Ratio When Point Coordinates are Given

1. Assume the ratio as k : 1 — Instead of using $m_1:m_2$, let the ratio be $k:1$. This simplifies calculation because we only need to solve for one variable, $k$.
2. Apply Section Formula — Write down the Section Formula for either the x-coordinate or the y-coordinate: $x = \frac{kx_2 + x_1}{k+1}$.
3. Equate and Solve for k — Substitute the known value of the coordinate ($x$ or $y$) and solve the linear equation to find the value of $k$.
4. Write the final ratio — Once $k$ is found (e.g., $k = 2/3$), write the final ratio as $2:3$. Verify by checking the calculation with the other coordinate.

Common Board Exam Mistakes & Tips

1. Cross-Multiplication Error: A very common mistake is multiplying $m_1$ with $x_1$ and $m_2$ with $x_2$. Always remember to cross-multiply! $m_1$ multiplies with the coordinates of point $B$ (the second point), and $m_2$ multiplies with the coordinates of point $A$ (the first point).
2. Trisection Points Ratio: If a question asks for points of trisection, remember there are two points. The first point divides the segment in the ratio $1:2$, and the second point divides it in the ratio $2:1$.
3. Parallelogram Property: For questions involving the vertices of a parallelogram, use the property that the diagonals of a parallelogram bisect each other. This means the midpoint of diagonal $AC$ is equal to the midpoint of diagonal $BD$.

Practice Questions with Solutions

• Q: Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$. A: Step 1: Identify the given points and ratio. Here, $A(x_1, y_1) = (-1, 7)$, $B(x_2, y_2) = (4, -3)$, and ratio $m_1:m_2 = 2:3$. Step 2: Apply the Section Formula for x-coordinate. $x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 - 3}{5} = \frac{5}{5} = 1$. Step 3: Apply the Section Formula for y-coordinate. $y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$. Final answer: The coordinates of the required point are $(1, 3)$.
• Q: Find the coordinates of the points of trisection of the line segment joining $(4, -1)$ and $(-2, -3)$. A: Step 1: Understand trisection. Let points $P$ and $Q$ trisect $AB$ where $A = (4, -1)$ and $B = (-2, -3)$. $P$ divides $AB$ in the ratio $1:2$ and $Q$ divides $AB$ in the ratio $2:1$. Step 2: Find coordinates of $P$ using ratio $1:2$. $x_P = \frac{1(-2) + 2(4)}{1 + 2} = \frac{-2 + 8}{3} = \frac{6}{3} = 2$. $y_P = \frac{1(-3) + 2(-1)}{1 + 2} = \frac{-3 - 2}{3} = \frac{-5}{3}$. So, $P$ is $(2, \frac{-5}{3})$. Step 3: Find coordinates of $Q$ using ratio $2:1$. $x_Q = \frac{2(-2) + 1(4)}{2 + 1} = \frac{-4 + 4}{3} = 0$. $y_Q = \frac{2(-3) + 1(-1)}{2 + 1} = \frac{-6 - 1}{3} = \frac{-7}{3}$. Final answer: The points of trisection are $(2, \frac{-5}{3})$ and $(0, \frac{-7}{3})$.
• Q: Find the ratio in which the line segment joining the points $(-3, 10)$ and $(6, -8)$ is divided by $(-1, 6)$. A: Step 1: Let the required ratio be $k:1$. Given points: $A(-3, 10)$, $B(6, -8)$ and the point of division $P(-1, 6)$. Step 2: Use the section formula for x-coordinate of $P$. $x = \frac{kx_2 + x_1}{k + 1} \implies -1 = \frac{k(6) + (-3)}{k + 1}$. Step 3: Solve for $k$. $-1(k + 1) = 6k - 3 \implies -k - 1 = 6k - 3 \implies 7k = 2 \implies k = \frac{2}{7}$. Step 4: Verify with y-coordinate. $y = \frac{k(-8) + 10}{k+1} = \frac{\frac{2}{7}(-8) + 10}{\frac{2}{7} + 1} = \frac{-16 + 70}{9} = \frac{54}{9} = 6$. This is correct. Final answer: The required ratio is $2:7$.
• Q: If $(1, 2)$, $(4, y)$, $(x, 6)$ and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$. A: Step 1: Use diagonal midpoint property. Let vertices be $A(1, 2)$, $B(4, y)$, $C(x, 6)$, and $D(3, 5)$. The diagonals $AC$ and $BD$ of a parallelogram bisect each other. Step 2: Find midpoint of $AC$. Midpoint of $AC = \left( \frac{1 + x}{2}, \frac{2 + 6}{2} \right) = \left( \frac{1 + x}{2}, 4 \right)$. Step 3: Find midpoint of $BD$. Midpoint of $BD = \left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right) = \left( \frac{7}{2}, \frac{y + 5}{2} \right)$. Step 4: Equate midpoints. $\frac{1+x}{2} = \frac{7}{2} \implies 1 + x = 7 \implies x = 6$. $\frac{y+5}{2} = 4 \implies y + 5 = 8 \implies y = 3$. Final answer: $x = 6$ and $y = 3$.

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