Coordinate Geometry Ex 7.3 Class 10 NCERT Solutions & Concepts

Understanding the Area of a Triangle Formula

When the vertices of a triangle are given as $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$, we can calculate its area without finding the lengths of its sides. The formula for the area of triangle ABC is:

$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

To easily memorize this formula, notice the cyclic order of the indices: $1 \to 2 \to 3 \to 1$. Outside the parentheses, we have the x-coordinates in numerical order ($x_1, x_2, x_3$). Inside each parenthesis, we subtract the next two y-coordinates in a cyclic sequence: after $y_1$ comes $y_2$ and $y_3$; after $y_2$ comes $y_3$ and $y_1$; after $y_3$ comes $y_1$ and $y_2$. The vertical bars represent the absolute value (modulus), which ensures that the final area is always positive, as area can never be a negative quantity.

Step-by-Step Guide to Calculating the Area

1. Step 1: Label the Coordinates — Clearly identify and label the given vertices as $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$. This prevents substitution errors.
2. Step 2: State the Formula — Write down the coordinate geometry area formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ to earn step-marks.
3. Step 3: Substitute and Simplify — Carefully plug the values into the formula. Pay close attention to negative signs (e.g., subtracting a negative number becomes addition).
4. Step 4: Apply Modulus — Calculate the value inside the modulus. If the result is negative, make it positive. Finally, divide by 2 to get the area in square units.

Pro-Tip: Collinearity of Points

One of the most frequent questions from Exercise 7.3 involves proving that three points are collinear (lying on the same straight line) or finding an unknown variable (like $k$) when points are given as collinear.

• The Rule: If three points $A$, $B$, and $C$ are collinear, they cannot form a triangle. Therefore, the area of the triangle formed by them must be exactly zero.
• Common Mistake: Forgetting the absolute value when solving for variables. However, when the area is zero, you can simply set the expression inside the modulus to zero: $x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$.

Practice Questions with Solutions

• Q: Find the area of the triangle whose vertices are $(2, 3)$, $(-1, 0)$, and $(2, -4)$. A: Step 1: Label the points as $(x_1, y_1) = (2, 3)$, $(x_2, y_2) = (-1, 0)$, and $(x_3, y_3) = (2, -4)$. Step 2: Write the area formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ Step 3: Substitute the coordinates: $\text{Area} = \frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)|$ $\text{Area} = \frac{1}{2} |2(4) + (-1)(-7) + 2(3)|$ Step 4: Simplify inside the modulus: $\text{Area} = \frac{1}{2} |8 + 7 + 6| = \frac{1}{2} |21| = 10.5$ Final answer: The area of the triangle is $10.5$ square units.
• Q: Find the value of $k$ for which the points $(7, -2)$, $(5, 1)$, and $(3, k)$ are collinear. A: Step 1: Let the points be $A(7, -2)$, $B(5, 1)$, and $C(3, k)$. Since the points are collinear, the area of the triangle formed by them must be $0$. Step 2: Set up the collinearity condition: $x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$ Step 3: Substitute the coordinates $(x_1, y_1) = (7, -2)$, $(x_2, y_2) = (5, 1)$, and $(x_3, y_3) = (3, k)$: $7(1 - k) + 5(k - (-2)) + 3(-2 - 1) = 0$ Step 4: Expand and simplify: $7 - 7k + 5(k + 2) + 3(-3) = 0$ $7 - 7k + 5k + 10 - 9 = 0$ $-2k + 8 = 0$ $2k = 8 \implies k = 4$ Final answer: The value of $k$ is $4$.
• Q: Find the area of the quadrilateral whose vertices, taken in order, are $A(-4, -2)$, $B(-3, -5)$, $C(3, -2)$, and $D(2, 3)$. A: Step 1: Divide the quadrilateral $ABCD$ into two triangles by drawing diagonal $AC$. This gives us $\triangle ABC$ and $\triangle ADC$. Step 2: Calculate the area of $\triangle ABC$ with vertices $(-4, -2)$, $(-3, -5)$, and $(3, -2)$: $\text{Area}(\triangle ABC) = \frac{1}{2} |-4(-5 - (-2)) + (-3)(-2 - (-2)) + 3(-2 - (-5))|$ $= \frac{1}{2} |-4(-3) + (-3)(0) + 3(3)| = \frac{1}{2} |12 + 0 + 9| = \frac{21}{2} = 10.5$ square units. Step 3: Calculate the area of $\triangle ADC$ with vertices $(-4, -2)$, $(2, 3)$, and $(3, -2)$: $\text{Area}(\triangle ADC) = \frac{1}{2} |-4(3 - (-2)) + 2(-2 - (-2)) + 3(-2 - 3)|$ $= \frac{1}{2} |-4(5) + 2(0) + 3(-5)| = \frac{1}{2} |-20 + 0 - 15| = \frac{1}{2} |-35| = 17.5$ square units. Step 4: Add the two areas to find the total area of the quadrilateral: $\text{Total Area} = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 10.5 + 17.5 = 28$ Final answer: The area of the quadrilateral is $28$ square units.
• Q: Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are $A(0, -1)$, $B(2, 1)$, and $C(0, 3)$. A: Step 1: Use the mid-point formula $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ to find the mid-points $D, E, F$ of sides $AB, BC, CA$: $D$ (mid-point of $AB$) = $\left(\frac{0+2}{2}, \frac{-1+1}{2}\right) = (1, 0)$ $E$ (mid-point of $BC$) = $\left(\frac{2+0}{2}, \frac{1+3}{2}\right) = (1, 2)$ $F$ (mid-point of $CA$) = $\left(\frac{0+0}{2}, \frac{3-1}{2}\right) = (0, 1)$ Step 2: Find the area of the mid-point triangle $\triangle DEF$ with vertices $(1, 0)$, $(1, 2)$, and $(0, 1)$: $\text{Area}(\triangle DEF) = \frac{1}{2} |1(2 - 1) + 1(1 - 0) + 0(0 - 2)|$ $= \frac{1}{2} |1(1) + 1(1) + 0| = \frac{1}{2} |2| = 1$ square unit. Final answer: The area of the triangle formed by joining the mid-points is $1$ square unit.

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