Class 10 Maths Chapter 7 Coordinate Geometry – Exercise 7.3 NCERT Solutions

Introduction

Exercise 7.3 focuses on finding the area of a triangle using the coordinates of its vertices. This formula is derived from determinants and helps solve problems involving geometry in the Cartesian plane. It is especially useful when the triangle is not aligned with the axes.

Formula Used

• Area of Triangle (Coordinate Geometry): For vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$:

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

NCERT Questions with Solutions (10)

Q1. Find area of triangle with vertices (0, 0), (4, 0), (0, 3).

Solution: Area = $\frac{1}{2}\mathrm{\mid }0(0-3)+4(3-0)+0(0-0)\mathrm{\mid }=\frac{1}{2}(12)=6$.

Q2. Find area of triangle with vertices (2, 3), (4, 5), (6, 7).

Solution: Area = $\frac{1}{2}\mathrm{\mid }2(5-7)+4(7-3)+6(3-5)\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }-4+16-12\mathrm{\mid }=0$. Points are collinear.

Q3. Find area of triangle with vertices (1, 1), (2, 4), (3, 1).

Solution: Area = $\frac{1}{2}\mathrm{\mid }1(4-1)+2(1-1)+3(1-4)\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }3+0-9\mathrm{\mid }=3$.

Q4. Find area of triangle with vertices (-2, -3), (3, 2), (4, -1).

Solution: Area = $\frac{1}{2}\mathrm{\mid }-2(2+1)+3(-1+3)+4(-3-2)\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }-6+6-20\mathrm{\mid }=10$.

Q5. Find area of triangle with vertices (0, 0), (5, 0), (0, 8).

Solution: Area = $\frac{1}{2}\mathrm{\mid }0(0-8)+5(8-0)+0(0-0)\mathrm{\mid }=\frac{1}{2}(40)=20$.

Q6. Find area of triangle with vertices (2, -2), (5, 2), (7, -1).

Solution: Area = $\frac{1}{2}\mathrm{\mid }2(2+1)+5(-1+2)+7(-2-2)\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }6+5-28\mathrm{\mid }=\frac{1}{2}(17)=8.5$.

Q7. Find area of triangle with vertices (1, 2), (3, -4), (5, 2).

Solution: Area = $\frac{1}{2}\mathrm{\mid }1(-4-2)+3(2-2)+5(2+4)\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }-6+0+30\mathrm{\mid }=12$.

Q8. Find area of triangle with vertices (-1, -1), (2, 3), (4, -2).

Solution: Area = $\frac{1}{2}\mathrm{\mid }-1(3+2)+2(-2+1)+4(-1-3)\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }-5-2-16\mathrm{\mid }=11.5$.

Q9. Find area of triangle with vertices (0, 0), (6, 0), (3, 9).

Solution: Area = $\frac{1}{2}\mathrm{\mid }0(0-9)+6(9-0)+3(0-0)\mathrm{\mid }=\frac{1}{2}(54)=27$.

Q10. Find area of triangle with vertices (2, 3), (4, 6), (6, 3).

Solution: Area = $\frac{1}{2}\mathrm{\mid }2(6-3)+4(3-3)+6(3-6)\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }6+0-18\mathrm{\mid }=6$.

FAQs (10 from NCERT)

1. Q: What is the formula for area of triangle in coordinate geometry? A: $\frac{1}{2}\mathrm{\mid }{x}_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\mathrm{\mid }$.

2. Q: What does absolute value mean here? A: It ensures area is non‑negative.

3. Q: What if area = 0? A: Points are collinear.

4. Q: Can coordinates be negative? A: Yes, formula works for all coordinates.

5. Q: What is determinant method? A: Using matrix determinant to calculate area.

6. Q: What is Cartesian plane? A: Plane defined by x‑axis and y‑axis.

7. Q: What is origin? A: Point (0, 0).

8. Q: What is collinearity test? A: If area = 0, points are collinear.

9. Q: Why is formula important? A: It helps in geometry and real‑life applications.

10. Q: Why is this exercise important? A: It builds skills in coordinate geometry and area calculation.

Conclusion

Exercise 7.3 has 10 solved questions and 10 FAQs that strengthen your understanding of the area of a triangle using coordinates. This builds the foundation for advanced coordinate geometry in Class 10 Maths.

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