On comparing the ratios a1/a2, b1/b2 and c1/c2, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: 5x – 4y + 8 = 0; 7x + 6y – 9 = 0

For the equations 5x – 4y + 8 = 0 and 7x + 6y – 9 = 0, we have a1/a2 = 5/7 and b1/b2 = -4/6 = -2/3. Since a1/a2 ≠ b1/b2, the lines are intersecting at a point.

NCERT Solutions Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables

Q: On comparing the ratios a1/a2, b1/b2 and c1/c2, find out whether the following pair of linear equations are consistent, or inconsistent: 3x + 2y = 5; 2x – 3y = 7

Here a1/a2 = 3/2 and b1/b2 = 2/-3. Since a1/a2 ≠ b1/b2, the pair of linear equations has a unique solution and is therefore consistent.

Detailed solutions for Exercise 3.1 (Graphical Representation of Linear Equations).

Ex 3.1 Ex 3.2 Ex 3.3

#### **Q1. Form the pair of linear equations in the following problems, and find their solutions graphically.** **(i)** 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. **(ii)** 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen. #### **A1. Solution:** **(i) Solution:** Let the number of girls be $x$ and the number of boys be $y$. **Algebraic Representation:** 1. $x + y = 10$ 2. $x - y = 4$ **Graphical Representation:** * **$x + y = 10$:** Points $(5, 5), (6, 4)$ * **$x - y = 4$:** Points $(5, 1), (4, 0)$ **Result:** The lines intersect at **$(7, 3)$**. **Answer:** Number of girls = **7**, Number of boys = **3**. **(ii) Solution:** Let the cost of 1 pencil be ₹$x$ and 1 pen be ₹$y$. **Algebraic Representation:** 1. $5x + 7y = 50$ 2. $7x + 5y = 46$ **Graphical Representation:** * **$5x + 7y = 50$:** Points $(3, 5), (10, 0)$ * **$7x + 5y = 46$:** Points $(3, 5), (8, -2)$ **Result:** The lines intersect at **$(3, 5)$**. **Answer:** Cost of one pencil = **₹3**, Cost of one pen = **₹5**. #### **Q2. On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:** **(i)** $5x – 4y + 8 = 0; \quad 7x + 6y – 9 = 0$ **(ii)** $9x + 3y + 12 = 0; \quad 18x + 6y + 24 = 0$ **(iii)** $6x – 3y + 10 = 0; \quad 2x – y + 9 = 0$ #### **A2. Solution:** **(i) $5x – 4y + 8 = 0$ and $7x + 6y – 9 = 0$** * $\frac{a_1}{a_2} = \frac{5}{7}$ * $\frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}$ * **Ratio Comparison:** $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ * **Conclusion:** The lines **intersect at a point**. **(ii) $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$** * $\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}$ * $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$ * $\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$ * **Ratio Comparison:** $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ * **Conclusion:** The lines are **coincident** (overlapping). **(iii) $6x – 3y + 10 = 0$ and $2x – y + 9 = 0$** * $\frac{a_1}{a_2} = \frac{6}{2} = 3$ * $\frac{b_1}{b_2} = \frac{-3}{-1} = 3$ * $\frac{c_1}{c_2} = \frac{10}{9}$ * **Ratio Comparison:** $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ * **Conclusion:** The lines are **parallel**. #### **Q3. On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, find out whether the following pair of linear equations are consistent, or inconsistent.** **(i)** $3x + 2y = 5; \quad 2x – 3y = 7$ **(ii)** $2x – 3y = 8; \quad 4x – 6y = 9$ **(iii)** $\frac{3}{2}x + \frac{5}{3}y = 7; \quad 9x – 10y = 14$ **(iv)** $5x – 3y = 11; \quad -10x + 6y = -22$ **(v)** $\frac{4}{3}x + 2y = 8; \quad 2x + 3y = 12$ #### **A3. Solution:** **(i) $3x + 2y = 5$ and $2x – 3y = 7$** * $\frac{3}{2} \neq \frac{2}{-3}$. * Intersecting lines $\rightarrow$ **Consistent**. **(ii) $2x – 3y = 8$ and $4x – 6y = 9$** * $\frac{2}{4} = \frac{-3}{-6} \neq \frac{8}{9}$ (i.e., $\frac{1}{2} = \frac{1}{2} \neq \frac{8}{9}$). * Parallel lines $\rightarrow$ **Inconsistent**. **(iii) $\frac{3}{2}x + \frac{5}{3}y = 7$ and $9x – 10y = 14$** * $\frac{3/2}{9} = \frac{1}{6}$ and $\frac{5/3}{-10} = -\frac{1}{6}$. * Since $\frac{1}{6} \neq -\frac{1}{6}$, Intersecting lines $\rightarrow$ **Consistent**. **(iv) $5x – 3y = 11$ and $-10x + 6y = -22$** * $\frac{5}{-10} = \frac{-3}{6} = \frac{11}{-22}$ (All equal $-\frac{1}{2}$). * Coincident lines $\rightarrow$ **Consistent (Dependent)**. **(v) $\frac{4}{3}x + 2y = 8$ and $2x + 3y = 12$** * $\frac{4/3}{2} = \frac{2}{3}$ and $\frac{2}{3} = \frac{2}{3}$ and $\frac{8}{12} = \frac{2}{3}$. * Coincident lines $\rightarrow$ **Consistent**. #### **Q4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.** **(i)** $x + y = 5, \quad 2x + 2y = 10$ **(ii)** $x – y = 8, \quad 3x – 3y = 16$ **(iii)** $2x + y – 6 = 0, \quad 4x – 2y – 4 = 0$ **(iv)** $2x – 2y – 2 = 0, \quad 4x – 4y – 5 = 0$ #### **A4. Solution:** **(i) $x + y = 5, \quad 2x + 2y = 10$** * Ratios: $\frac{1}{2} = \frac{1}{2} = \frac{5}{10}$. * **Consistent** (Coincident lines). * **Graph:** Both equations represent the same line. There are **infinite solutions** (any point on $x+y=5$). **(ii) $x – y = 8, \quad 3x – 3y = 16$** * Ratios: $\frac{1}{3} = \frac{-1}{-3} \neq \frac{8}{16}$. * **Inconsistent** (Parallel lines). No solution. **(iii) $2x + y – 6 = 0, \quad 4x – 2y – 4 = 0$** * Ratios: $\frac{2}{4} \neq \frac{1}{-2}$. * **Consistent** (Intersecting lines). * **Graph:** * $2x+y=6$: Points $(0,6), (3,0)$. * $4x-2y=4$: Points $(1,0), (0,-2)$. * **Result:** Intersection at **$(2, 2)$**. **(iv) $2x – 2y – 2 = 0, \quad 4x – 4y – 5 = 0$** * Ratios: $\frac{2}{4} = \frac{-2}{-4} \neq \frac{-2}{-5}$. * **Inconsistent** (Parallel lines). No solution. #### **Q5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.** #### **A5. Solution:** Let Length = $l$ and Width = $w$. 1. Half Perimeter: $l + w = 36$ 2. Relationship: $l = w + 4$ Substitute (2) into (1): $(w + 4) + w = 36$ $2w = 32 \Rightarrow w = 16$ m. $l = 16 + 4 = 20$ m. **Answer:** Length = **20 m**, Width = **16 m**. #### **Q6. Given the linear equation $2x + 3y – 8 = 0$, write another linear equation in two variables such that the geometrical representation of the pair so formed is:** **(i)** intersecting lines **(ii)** parallel lines **(iii)** coincident lines #### **A6. Solution:** *(Note: Multiple answers are possible; these are standard examples)* **(i) Intersecting lines:** Condition: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. **Example:** $3x + 2y - 7 = 0$ **(ii) Parallel lines:** Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. **Example:** $2x + 3y - 12 = 0$ **(iii) Coincident lines:** Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. **Example:** $4x + 6y - 16 = 0$ (Original equation multiplied by 2). #### **Q7. Draw the graphs of the equations $x – y + 1 = 0$ and $3x + 2y – 12 = 0$. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.** #### **A7. Solution:** **Table for $x - y = -1$:** | x | 0 | -1 | 2 | |---|---|----|---| | y | 1 | 0 | 3 | Points: $(0,1), (-1,0), (2,3)$ **Table for $3x + 2y = 12$:** | x | 0 | 4 | 2 | |---|---|---|---| | y | 6 | 0 | 3 | Points: $(0,6), (4,0), (2,3)$ **Vertices of the Triangle:** 1. **Intersection of two lines:** $(2, 3)$ 2. **Line 1 intersection with x-axis ($y=0$):** $(-1, 0)$ 3. **Line 2 intersection with x-axis ($y=0$):** $(4, 0)$ **Answer:** The vertices are **$(2, 3)$**, **$(-1, 0)$**, and **$(4, 0)$**.