Class 11 Maths Chapter 9 Straight Lines – Exercise 9.3 NCERT Solutions

Introduction

Exercise 9.3 focuses on the distance of a point from a line and related applications. You will learn how to calculate perpendicular distance, check collinearity, and solve problems involving shortest distance between points and lines. This exercise strengthens your understanding of coordinate geometry.

Key Concepts

1. Equation of Line (General Form):

$$ax+by+c=0$$

2. Distance of Point $(x_1,y_1)$ from Line $ax+by+c=0$:

$$d=\frac{\mathrm{\mid }a{x}_1+b{y}_1+c\mathrm{\mid }}{\sqrt{a^2+b^2}}$$

3. Condition for Collinearity: If distance = 0, point lies on line.

4. Shortest Distance Between Parallel Lines: For lines $ax+by+c_1=0$ and $ax+by+c_2=0$:

$$d=\frac{\mathrm{\mid }{c}_1-c_2\mathrm{\mid }}{\sqrt{a^2+b^2}}$$

NCERT Questions with Solutions (10)

Q1. Find distance of point (3,4) from line $3x+4y-5=0$.

$$d=\frac{\mathrm{\mid }3\cdot 3+4\cdot 4-5\mathrm{\mid }}{\sqrt{3^2+4^2}}=\frac{\mathrm{\mid }9+16-5\mathrm{\mid }}{5}=\frac{20}{5}=4$$

Q2. Find distance of point (2,-3) from line $x-y+1=0$.

$$d=\frac{\mathrm{\mid }2-(-3)+1\mathrm{\mid }}{\sqrt{1^2+(-1{)}^2}}=\frac{\mathrm{\mid }6\mathrm{\mid }}{\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2}$$

Q3. Find distance of point (1,2) from line $2x+3y-6=0$.

$$d=\frac{\mathrm{\mid }2\cdot 1+3\cdot 2-6\mathrm{\mid }}{\sqrt{2^2+3^2}}=\frac{\mathrm{\mid }2+6-6\mathrm{\mid }}{\sqrt{13}}=\frac{2}{\sqrt{13}}$$

Q4. Find distance of point (-2,5) from line $x+2y-3=0$.

$$d=\frac{\mathrm{\mid }-2+10-3\mathrm{\mid }}{\sqrt{1^2+2^2}}=\frac{5}{\sqrt{5}}=\sqrt{5}$$

Q5. Find distance of point (0,0) from line $4x+3y-12=0$.

$$d=\frac{\mathrm{\mid }0+0-12\mathrm{\mid }}{\sqrt{16+9}}=\frac{12}{5}=\frac{12}{5}$$

Q6. Find distance of point (1,-1) from line $5x+12y+6=0$.

$$d=\frac{\mathrm{\mid }5\cdot 1+12(-1)+6\mathrm{\mid }}{\sqrt{25+144}}=\frac{\mathrm{\mid }5-12+6\mathrm{\mid }}{13}=\frac{-1}{13}=\frac{1}{13}$$

Q7. Find distance between parallel lines $2x+3y-7=0$ and $2x+3y+5=0$.

$$d=\frac{\mathrm{\mid }-7-5\mathrm{\mid }}{\sqrt{4+9}}=\frac{12}{\sqrt{13}}$$

Q8. Find distance between parallel lines $x+2y-3=0$ and $x+2y+7=0$.

$$d=\frac{\mathrm{\mid }-3-7\mathrm{\mid }}{\sqrt{1+4}}=\frac{10}{\sqrt{5}}=2\sqrt{5}$$

Q9. Find distance of point (4,0) from line $x=3$. Equation: $x-3=0$.

$$d=\mathrm{\mid }4-3\mathrm{\mid }=1$$

Q10. Find distance of point (0,5) from line $y=2$. Equation: $y-2=0$.

$$d=\mathrm{\mid }5-2\mathrm{\mid }=3$$

FAQs (10)

1. What is formula for distance of point from line? $\frac{\mathrm{\mid }a{x}_1+b{y}_1+c\mathrm{\mid }}{\sqrt{a^2+b^2}}$.

2. What is shortest distance between parallel lines? $\frac{\mathrm{\mid }{c}_1-c_2\mathrm{\mid }}{\sqrt{a^2+b^2}}$.

3. When is point on line? If distance = 0.

4. What is slope of line $ax+by+c=0$? $-\frac{a}{b}$.

5. What is intercept form of line? $\frac{x}{a}+\frac{y}{b}=1$.

6. Can distance be negative? No, always non‑negative.

7. What is distance of origin from line $ax+by+c=0$? $\frac{\mathrm{\mid }c\mathrm{\mid }}{\sqrt{a^2+b^2}}$.

8. What is perpendicular distance? Shortest distance from point to line.

9. Can two parallel lines have zero distance? Only if they coincide.

10. Why is this exercise important? It builds skills in calculating distances in coordinate geometry.

Conclusion

Exercise 9.3 has 10 solved questions and 10 FAQs that strengthen your understanding of distance of a point from a line and distance between parallel lines. This builds the foundation for advanced coordinate geometry in Class 11 Maths.

visit:www.fuzymathacademy.com

New Syllabus-Class 11 Maths Chapter 9 Straight Lines – Exercise 9.3 NCERT Solutions