CBSE Class 9 Mathematics Polynomials Worksheet Set 01

Case Based MCQs

Case I : Read the following passage and answer the questions from 1 to 5.
On one day, principal of a particular school visited the classroom. Class teacher was teaching the concept of polynomial to students. He was very much impressed by her way of teaching. To check, whether the students also understand the concept taught by her or not, he asked various questions to students. Some of them are given below. Answer them.

1. Which one of the following is not a polynomial?
(a) 4x² + 2x – 1
(b) y+3/y
(c) x³ – 1
(d) y² + 5y + 1
Answer : B

2. The polynomial of the type ax² + bx + c, a = 0 is called
(a) Linear polynomial
(b) Quadratic polynomial
(c) Cubic polynomial
(d) Biquadratic polynomial
Answer : A

3. The value of k, if (x – 1) is a factor of 4x³ + 3x² – 4x + k, is
(a) 1
(b) –2
(c) –3
(d) 3
Answer : C

4. If x + 2 is the factor of x³ – 2ax² + 16, then value of a is
(a) –7
(b) 1
(c) –1
(d) 7
Answer : B

5. The number of zeroes of the polynomial x² + 4x + 2 is
(a) 1
(b) 2
(c) 3
(d) 4
Answer : B

Assertion & Reasoning Based MCQs

Directions: In these questions, a statement of Assertion is followed by a statement of Reason is given.

Choose the correct answer out of the following choices :
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.

1. Assertion : 3x² + x – 1 = (x + 1) (3x – 2) + 1.
Reason : To factorise ax² + bx + c, write b as sum of two numbers whose product is ac.
Answer : A

2. Assertion : –7 is a constant polynomial.
Reason : Degree of a constant polynomial is zero.
Answer : A

3. Assertion : The degree of the polynomial (x² – 2)(x – 3)(x + 4) is 3.
Reason : A polynomial of degree 3 is called a cubic polynomial.
Answer : D

4. Assertion : If 2x² – 32 is the volume of a cuboid, then length of cuboid can be x – 8.
Reason : Volume of a cuboid = l × b × h.
Answer : D

5. Assertion : The value of 593 × 607 is 359951.
Reason : (a + b) (a – b) = a² – b²
Answer : A

6. Assertion : The value of (–27)³ + (–9)³ is –20412.
Reason : If a + b = 0, then a³ + b³ + c³ = 0
Answer : C

7. Assertion : If x = 3/2 is a zero of polynomial 2x² + kx – 12, then k = 5.
Reason : If x = a is zero of a polynomial f(x), then f(a) = 0.
Answer : A

8. Assertion : The expression 3x⁴ – 4x^3/2 + x² = 2 is not a polynomial because the term – 4x^3/2 contains a rational power of x.
Reason : The highest exponent in various terms of an algebraic expression in one variable is called its degree.
Answer : B

Question. Use suitable identities to find the following products :
(i) (x+4)(x+10)
(ii) (x+8)(x–10)
(iii) (3x+4)(3x–5)
(iv) (y²+3/2)(y²-3/2)
(v) (3–2x)(3+2x)
Answer : 
(i) (x+4)(x+10) = x²+(4+10)x+4×10
                        = x²+14x+40
(ii) (x+8)(x–10) = x²+(8−10)x+8×(−10)
                        = x²−2x−80
(iii) (3x+4)(3x–5) = 3x(3x−5)+4(3x−5)
                           = 3x×3x−3x×5+4×3x−4×5
                           = 9x²−15x+12x−20
                           = 9x²−3x−20
(iv) (y²+3/2)(y²−3/2) = (y²)² −(3/2)²
                                  = y⁴−9/4
(v) (3–2x)(3+2x) = (3)² −(2x)²
                          = 9−4x²

Question. Evaluate the following products without multiplying directly :
(i) 103×107
(ii) 95×96
(iii) 104×96
Answer : (i) 103×107 = (100+3)(100+7)
                         = (100)² +(3+7)(100)+3×7
                         = 100×100+(10)(100)+21
                         = 10000+1000+21=11021
(ii) 95×96 = (100−5)(100−4)
               = (100)² +(−5−4)(100)+(−5)(−4)
               = 100×100+(−9)(100)+20
               = 10000−900+20=9120
(iii) 104×96 = (100+4)(100−4)
                  = (100)² −(4)²
                  = 10000−16=9984

Question. Factorise the following using appropriate identifies:
(i) 9x²+6xy+y²
(ii) 4y²−4y+1
(iii) x²−y²/100
Answer : a) 9x²+6xy+y² = (3x)² +2(3x)(y)+(y)²   [(a+b)² =a²+2ab+b²]
                              = (3x+y)²
                              = (3x+y)(3x+y)
(b) 4y²−4y+1 = (2y)² −2(2y)(1)+(1)²           [(a−b)² =a²−2ab+b²]
                     = (2y−1)²
                     = (2y−1)(2y−1)
(c) x²−y²/100 = (x)2−(y/10)2                       [(x+y)(x−y)=x²−y²]
                      = (x−y/10)(x+y/10)

Question. Expand each of the following using suitable identifies:
(i) (x+2y+4z)²
(ii) (2x−y+z)²
(iii) (−2x+3y+2z)²
(iv) (3a−7b−c)²
Answer : 
(i) (x+2y+4z)²
   = x²+(2y)² +(4z)²+2(x)(2y)+2(2y)(4z)+2(4z)(x)
   = x²+4y²+16z²+4xy+16yz+8zx
(ii) (2x−y+z)²
   = [2x+(−y)+z]²
   = (2x)²+(−y)²+z²+2(2x)(−y)+2(−y)(z)+2(z)(2x)
   = 4x²+y²+z²−4xy−2yz+4zx
(iii) (−2x+3y+2z)²
   = [(−2x)+3y+2z]²
   = (−2x)²+(3y)² +(2z)² +2(−2x)(3y)+2(3y)(2z)+2(2z)(−2x)
   = 4x²+9y²+4z²−12xy+12yz−8zx
(iv) (3a−7b−c)²
    = [3a+(−7b)+(−c)]²
    = (3a)² +(−7b)²+(−c)² +2(3a)(−7b)+2(−7b)(−c)+2(−c)(3a)
    = 9a²+49b²+c²−42ab+14bc−6ca

Question. Factorise:
(i) 4x²+9y²+16z²+12xy−24yz−16xz
(ii) 2x²+y²+8z²−2√2xy+4√2yz−8xz
Answer : (i) 4x²+9y²+16z²+12xy−24yz−16xz
       = (2x)2 +(3y)² +(−4z)² +2(2x)(3y)+2(3y)(−4z)+2(2x)(−4z)
       = [2x+3y+(−4z)]²
       = (2x+3y−4z)2
(ii) 2x²+y²+8z²−2√2xy+4√2yz−8xz
    = (√2x)² +(−y)² +(−2√2z)²+2(√2x)(−y)+2(−y)(−2√2z)+2(-2√2z)(√2x)
    = [√2x+(−y)+(−2√2z)]2
   = (2√x−y−2√2z)2

Assignment 2: Polynomials

Question 1. If two polynomials \(ax^3 + 4x^2 + 3x - 4\) & \(x^3 - 4x + a\) leave the same remainder when divided by \((x - 3)\), find the value of a.
Answer: Let \(p(x) = ax^3 + 4x^2 + 3x - 4\) and \(q(x) = x^3 - 4x + a\).
By the Remainder Theorem, when a polynomial is divided by \((x - 3)\), the remainder is obtained by substituting \(x = 3\).
For \(p(x)\), remainder \(R_1 = p(3) = a(3)^3 + 4(3)^2 + 3(3) - 4\)
\(R_1 = 27a + 36 + 9 - 4 = 27a + 41\)
For \(q(x)\), remainder \(R_2 = q(3) = (3)^3 - 4(3) + a\)
\(R_2 = 27 - 12 + a = 15 + a\)
Since both remainders are equal, \(R_1 = R_2\):
\(27a + 41 = 15 + a\)
\(27a - a = 15 - 41\)
\(26a = -26\)
\(a = -1\)
In simple words: Substitute \(x = 3\) into both polynomials, set the resulting expressions equal to each other, and solve for \(a\).

Exam Tip: Always apply the Remainder Theorem directly by substituting the zero of the divisor divisor root into the polynomial.

Question 2. Evaluate using identities:- (a) \(103 \times 97\) (b) \((0.99)^2\) (c) \(105^3\)
Answer: (a) \(103 \times 97 = (100 + 3)(100 - 3)\)
Using identity \((a + b)(a - b) = a^2 - b^2\):
\(= 100^2 - 3^2 = 10000 - 9 = 9991\)
(b) \((0.99)^2 = (1 - 0.01)^2\)
Using identity \((a - b)^2 = a^2 - 2ab + b^2\):
\(= 1^2 - 2(1)(0.01) + (0.01)^2 = 1 - 0.02 + 0.0001 = 0.9801\)
(c) \(105^3 = (100 + 5)^3\)
Using identity \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\):
\(= 100^3 + 3(100)^2(5) + 3(100)(5^2) + 5^3\)
\(= 1000000 + 150000 + 7500 + 125 = 1157625\)
In simple words: Break numbers into round tens or hundreds and apply standard algebraic expansion formulas.

Exam Tip: Clearly write out the algebraic identity used before substituting values to ensure full credit.

Question 3. Find the remainder when \(4x^3 - 3x^2 + 2x - 4\) is divided by \(x + 2\).
Answer: Let \(p(x) = 4x^3 - 3x^2 + 2x - 4\).
The zero of the divisor \((x + 2)\) is \(x = -2\).
By the Remainder Theorem, the remainder is \(p(-2)\):
\(p(-2) = 4(-2)^3 - 3(-2)^2 + 2(-2) - 4\)
\(= 4(-8) - 3(4) - 4 - 4\)
\(= -32 - 12 - 4 - 4 = -52\)
In simple words: Put \(x = -2\) into the polynomial expression and evaluate the arithmetic.

Exam Tip: Pay special attention to negative signs when raising negative numbers to odd and even powers.

Question 4. Show that \((x - 1)\) is a factor of \(x^{10} - 1\).
Answer: Let \(p(x) = x^{10} - 1\).
By the Factor Theorem, \((x - 1)\) is a factor of \(p(x)\) if \(p(1) = 0\).
Substituting \(x = 1\):
\(p(1) = (1)^{10} - 1 = 1 - 1 = 0\)
Since the remainder \(p(1) = 0\), \((x - 1)\) is indeed a factor of \(x^{10} - 1\).
In simple words: Substitute \(x = 1\) into the polynomial and check if the result equals zero.

Exam Tip: State the Factor Theorem clearly as part of your proof.

Question 5. Find the value of \(a\), if \((x - a)\) is a factor of \(x^3 - a^2x + x + 2\).
Answer: Let \(p(x) = x^3 - a^2x + x + 2\).
Since \((x - a)\) is a factor, by the Factor Theorem, \(p(a) = 0\).
Substituting \(x = a\):
\(a^3 - a^2(a) + a + 2 = 0\)
\(a^3 - a^3 + a + 2 = 0\)
\(a + 2 = 0\)
\(a = -2\)
In simple words: Substitute \(x = a\) into the polynomial, equate it to zero, and solve the linear equation.

Exam Tip: Be careful to replace every instance of \(x\) with \(a\) before simplifying.

Question 6. Determine the value of \(a\) for which the polynomial \(2x^4 - ax^3 + 4x^2 + 2x + 1\) is divisible by \((1 - 2x)\).
Answer: Let \(p(x) = 2x^4 - ax^3 + 4x^2 + 2x + 1\).
The zero of the divisor \((1 - 2x)\) is found by setting \(1 - 2x = 0\), which gives \(x = \frac{1}{2}\).
Since \(p(x)\) is divisible by \((1 - 2x)\), the remainder must be zero, so \(p\left(\frac{1}{2}\right) = 0\).
\(2\left(\frac{1}{2}\right)^4 - a\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) + 1 = 0\)
\(2\left(\frac{1}{16}\right) - a\left(\frac{1}{8}\right) + 4\left(\frac{1}{4}\right) + 1 + 1 = 0\)
\(\frac{1}{8} - \frac{a}{8} + 1 + 1 + 1 = 0\)
\(\frac{1 - a}{8} + 3 = 0\)
\(\frac{1 - a}{8} = -3\)
\(1 - a = -24\)
\(-a = -25 \implies a = 25\)
In simple words: Find the root \(x = \frac{1}{2}\) from the divisor, substitute it into the polynomial, set it to zero, and solve for \(a\).

Exam Tip: Always solve for the exact zero of the divisor when it has a coefficient of \(x\) other than \(1\).

Question 7. Factorize the polynomials:-
(a) \(x^3 - 6x^2 + 11x - 6\)
(b) \((a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3\)
(c) \(x^3 + 13x^2 + 31x - 45\) given that \(x + 9\) is a factor
(d) \(8x^3 + 27y^3 + z^3 - 18xyz\)
(e) \((a + b)^3 + (b + c)^3 + (c + a)^3 - 3(a + b)(b + c)(c + a)\)
Answer:
(a) Let \(p(x) = x^3 - 6x^2 + 11x - 6\). Factors of the constant term \(6\) are \(\pm 1, \pm 2, \pm 3, \pm 6\).
Testing \(x = 1\): \(p(1) = 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0\). Thus \((x - 1)\) is a factor.
Dividing \(p(x)\) by \((x - 1)\) using synthetic division or long division gives the quotient \(x^2 - 5x + 6\).
Splitting the middle term of \(x^2 - 5x + 6\): \((x - 2)(x - 3)\).
Complete factorization: \((x - 1)(x - 2)(x - 3)\).

(b) Let \(x = a^2 - b^2\), \(y = b^2 - c^2\), and \(z = c^2 - a^2\).
Notice that \(x + y + z = (a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) = 0\).
Using the algebraic identity: if \(x + y + z = 0\), then \(x^3 + y^3 + z^3 = 3xyz\).
Therefore, \((a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3 = 3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2)\)
\(= 3(a - b)(a + b)(b - c)(b + c)(c - a)(c + a)\).

(c) Given \(x + 9\) is a factor of \(x^3 + 13x^2 + 31x - 45\).
Dividing the polynomial by \((x + 9)\) using long division:
\(x^3 + 13x^2 + 31x - 45 = (x + 9)(x^2 + 4x - 5)\).
Now factorize the quadratic part \(x^2 + 4x - 5 = (x + 5)(x - 1)\).
Complete factorization: \((x + 9)(x + 5)(x - 1)\).

(d) \(8x^3 + 27y^3 + z^3 - 18xyz = (2x)^3 + (3y)^3 + z^3 - 3(2x)(3y)(z)\)
Using identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\):
\(= (2x + 3y + z)((2x)^2 + (3y)^2 + z^2 - (2x)(3y) - (3y)(z) - z(2x))\)
\(= (2x + 3y + z)(4x^2 + 9y^2 + z^2 - 6xy - 3yz - 2zx)\).

(e) Let \(X = a + b\), \(Y = b + c\), and \(Z = c + a\).
The expression becomes \(X^3 + Y^3 + Z^3 - 3XYZ\).
Using identity: \(= (X + Y + Z)(X^2 + Y^2 + Z^2 - XY - YZ - ZX)\).
Substitute back \(X, Y, Z\):
\(X + Y + Z = (a + b) + (b + c) + (c + a) = 2a + 2b + 2c = 2(a + b + c)\).
Complete factorization: \((2)(a + b + c)((a+b)^2 + (b+c)^2 + (c+a)^2 - (a+b)(b+c) - (b+c)(c+a) - (c+a)(a+b))\).
In simple words: Apply trial-and-error root finding for cubic polynomials, and use standard sum-of-cubes identities for algebraic expressions.

Exam Tip: Memorize standard cubic identities thoroughly as they appear frequently in factorization problems.

Question 8. Factorize:-
(a) \(a^3 - 0.216\)
(b) \(2x^2 - \frac{5}{6}x + \frac{1}{12}\)
(c) \((x + 1)^3 + (x - 1)^3\)
Answer:
(a) \(a^3 - 0.216 = a^3 - (0.6)^3\)
Using identity \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\):
\(= (a - 0.6)(a^2 + 0.6a + 0.36)\).

(b) \(2x^2 - \frac{5}{6}x + \frac{1}{12} = \frac{1}{12}(24x^2 - 10x + 1)\)
Splitting the middle term of \(24x^2 - 10x + 1\) (factors of \(24\) that add to \(-10\) are \(-6\) and \(-4\)):
\(= \frac{1}{12}(24x^2 - 6x - 4x + 1)\)
\(= \frac{1}{12}[6x(4x - 1) - 1(4x - 1)]\)
\(= \frac{1}{12}(6x - 1)(4x - 1)\).

(c) \((x + 1)^3 + (x - 1)^3\)
Using identity \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\) where \(a = (x + 1)\) and \(b = (x - 1)\):
\(= [(x + 1) + (x - 1)][(x + 1)^2 - (x + 1)(x - 1) + (x - 1)^2]\)
\(= (2x)[(x^2 + 2x + 1) - (x^2 - 1) + (x^2 - 2x + 1)]\)
\(= (2x)[x^2 + 2x + 1 - x^2 + 1 + x^2 - 2x + 1]\)
\(= (2x)(x^2 + 3)\).
In simple words: Factor fractions by taking out the common denominator first, and apply sum/difference of cubes formulas directly.

Exam Tip: Factoring out fractions at the beginning simplifies middle term splitting significantly.

Question 9. Give possible expressions for the length and breadth of a rectangle having \(A = 35y^2 + 13y - 12\) (Area).
Answer: Area of rectangle = \(\text{Length} \times \text{Breadth}\).
We need to factorize the quadratic expression \(35y^2 + 13y - 12\).
Find two numbers whose product is \(35 \times (-12) = -420\) and whose sum is \(13\). These numbers are \(28\) and \(-15\).
\(35y^2 + 28y - 15y - 12\)
\(= 7y(5y + 4) - 3(5y + 4)\)
\(= (7y - 3)(5y + 4)\)
Therefore, possible expressions for length and breadth are \((7y - 3)\) and \((5y + 4)\).
In simple words: Factorize the area polynomial into two linear factors, which represent the length and breadth.

Exam Tip: Since dimensions must be positive for valid geometric lengths, you can mention either factor as length or breadth.

Question 10. Evaluate using a suitable identity:- \((1.93)^3 + (0.07)^3 - (2)^3\)
Answer: Let \(a = 1.93\), \(b = 0.07\), and \(c = -2\).
Notice that \(a + b + c = 1.93 + 0.07 - 2 = 2.00 - 2 = 0\).
Using the algebraic identity: if \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\).
\((1.93)^3 + (0.07)^3 + (-2)^3 = 3(1.93)(0.07)(-2)\)
\(= 3 \times 1.93 \times 0.07 \times (-2)\)
\(= -42 \times 1.93 \times 0.01 = -0.8106\)
In simple words: Check if the sum of the three base numbers is zero; if so, use the identity \(a^3 + b^3 + c^3 = 3abc\).

Exam Tip: Always verify if \(a + b + c = 0\) before applying the conditional cube identity.

Question 11. Find the product: \((2x - y + 3z)(4x^2 + y^2 + 9z^2 + 2xy + 3yz - 6xz)\)
Answer: Rewrite the second factor to match the standard identity structure:
\((2x - y + 3z)[(2x)^2 + (-y)^2 + (3z)^2 - (2x)(-y) - (-y)(3z) - (3z)(2x)]\)
This matches the identity: \((a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc\)
where \(a = 2x\), \(b = -y\), and \(c = 3z\).
Product \(= (2x)^3 + (-y)^3 + (3z)^3 - 3(2x)(-y)(3z)\)
\(= 8x^3 - y^3 + 27z^3 + 18xyz\).
In simple words: Recognize this as the expanded form of the sum/difference of three cubes identity.

Exam Tip: Pay careful attention to the signs of \(ab\), \(bc\), and \(ca\) to correctly identify negative terms.

Question 12. Factorize by splitting the middle term :
(a) \(9x^2 - 3x - 2\) (Note: OCR has a typo "-9", corrected to standard quadratic \(-2\) based on standard curriculum)
(b) \(x^2 + 14x + 40\)
(c) \(5x^2 + 16x + 3\)
Answer:
(a) \(9x^2 - 3x - 2\): Product = \(9 \times (-2) = -18\), Sum = \(-3\). Numbers are \(-6\) and \(3\).
\(= 9x^2 - 6x + 3x - 2\)
\(= 3x(3x - 2) + 1(3x - 2)\)
\(= (3x - 2)(3x + 1)\).

(b) \(x^2 + 14x + 40\): Product = \(40\), Sum = \(14\). Numbers are \(10\) and \(4\).
\(= x^2 + 10x + 4x + 40\)
\(= x(x + 10) + 4(x + 10)\)
\(= (x + 10)(x + 4)\).

(c) \(5x^2 + 16x + 3\): Product = \(5 \times 3 = 15\), Sum = \(16\). Numbers are \(15\) and \(1\).
\(= 5x^2 + 15x + x + 3\)
\(= 5x(x + 3) + 1(x + 3)\)
\(= (5x + 1)(x + 3)\).
In simple words: Find two numbers that multiply to give the product of outer coefficients and add to give the middle coefficient.

Exam Tip: Always double-check your factored terms by multiplying them back out.

Assignment 3: Coordinate Geometry

Question 1. Write the coordinates of a point which:-
(a) Lies on the x-axis and is at a distance of 4 units to the right of the origin.
(b) Lies on the y-axis and is at a distance of y units below the x-axis.
(c) Is at a distance of 3 units from the x-axis and 7 units from the y-axis. [there would be four such points]
Answer:
(a) \((4, 0)\)
(b) \((0, -y)\)
(c) Depending on the quadrant, the four points are \((7, 3)\), \((-7, 3)\), \((-7, -3)\), and \((7, -3)\).
In simple words: X-axis points have zero y-coordinate; y-axis points have zero x-coordinate. Distances can be positive or negative depending on the quadrant direction.

Exam Tip: Remember that distance from the x-axis gives the y-coordinate (ordinate), and distance from the y-axis gives the x-coordinate (abscissa).

Question 2. Draw the graphs of the eqs: (a) \(3x - 2y = 7\) (b) \(y = -2\) on the same pair of axes. Read the coordinates of their point of intersection.
Answer: For equation (a) \(3x - 2y = 7\):
When \(x = 3\), \(3(3) - 2y = 7 \implies 9 - 7 = 2y \implies y = 1\). Point: \((3, 1)\).
When \(x = 1\), \(3(1) - 2y = 7 \implies 3 - 7 = 2y \implies y = -2\). Point: \((1, -2)\).
For equation (b) \(y = -2\): it is a horizontal line parallel to the x-axis passing through \(y = -2\).
Solving simultaneously: substitute \(y = -2\) into \(3x - 2(-2) = 7 \implies 3x + 4 = 7 \implies 3x = 3 \implies x = 1\).
Point of intersection: \((1, -2)\).
In simple words: Plot both lines on graph paper and locate where they cross each other.

Exam Tip: Always plot at least three points for linear equations to ensure accuracy.

Question 3. Find the point where the line represented by the equation \(5y - 3x - 10 = 0\) cuts the y-axis.
Answer: Any point on the y-axis has an x-coordinate of \(0\).
Substitute \(x = 0\) into the equation:
\(5y - 3(0) - 10 = 0\)
\(5y = 10 \implies y = 2\)
The point is \((0, 2)\).
In simple words: To find where a line cuts the y-axis, substitute \(x = 0\) and solve for \(y\).

Exam Tip: Cutting the y-axis always means setting \(x = 0\).

Question 4. Draw the graph of the line \(3x + 4y = 18\). With the help of graph find value of y when \(x = 2\). (show this point on the graph)
Answer: For \(3x + 4y = 18\):
When \(x = 2\): \(3(2) + 4y = 18 \implies 6 + 4y = 18 \implies 4y = 12 \implies y = 3\).
When \(x = 6\): \(3(6) + 4y = 18 \implies 18 + 4y = 18 \implies 4y = 0 \implies y = 0\).
Points: \((2, 3)\) and \((6, 0)\).
When \(x = 2\), the value of \(y\) is \(3\).
In simple words: Plot the line using its intercept points and read off the y-value corresponding to \(x = 2\).

Exam Tip: Draw dotted perpendicular lines from \(x = 2\) to the graph line to clearly indicate how you read the y-value.

Question 5. On a graph draw a quadrilateral whose vertices are \((1, 1)\), \((2, 4)\), \((8, 4)\) and \((10, 1)\). Justify the quadrilateral.
Answer: Let the vertices be \(A(1, 1)\), \(B(2, 4)\), \(C(8, 4)\), and \(D(10, 1)\).
Notice that the y-coordinates of \(A\) and \(D\) are both equal to \(1\), so line segment \(AD\) lies along the horizontal line \(y = 1\).
The y-coordinates of \(B\) and \(C\) are both equal to \(4\), so line segment \(BC\) lies along the horizontal line \(y = 4\).
Since both \(AD\) and \(BC\) are horizontal lines, they are parallel to each other (\(AD \parallel BC\)).
Thus, the quadrilateral \(ABCD\) is a trapezium.
In simple words: Plot the four points and connect them; since one pair of opposite sides is horizontal and parallel, it forms a trapezium.

Exam Tip: Always check the y-coordinates of opposite vertices to easily identify parallel horizontal lines.

Question 6. How will you describe the position of the table lamp on your study table to another person?
Answer: Consider the table as a Cartesian plane where one corner of the table is the origin \((0, 0)\).
Measure the perpendicular distance of the table lamp from the shorter edge (let it be \(x\) cm) and from the longer adjacent edge (let it be \(y\) cm).
The position of the lamp is then given by the ordered pair \((x, y)\).
In simple words: Treat the table edge as axes and give the perpendicular distances from two adjacent edges as coordinates.

Exam Tip: Mentioning the reference corner (origin) and adjacent edges is crucial for a complete answer.

Question 7. Draw the graph of \(y = 2x + 4\). Use the graph to find the area between the line and the axes.
Answer: For \(y = 2x + 4\):
When \(x = 0\), \(y = 4\). (y-intercept: \((0, 4)\))
When \(y = 0\), \(0 = 2x + 4 \implies x = -2\). (x-intercept: \((-2, 0)\))
The line forms a right-angled triangle with the coordinate axes, having base \(b = 2\) units (from \(x = -2\) to origin) and height \(h = 4\) units (from origin to \(y = 4\)).
Area \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4\) square units.
In simple words: Find where the line crosses both axes, treat those intercepts as triangle legs, and calculate the triangle area.

Exam Tip: The distance from origin to intercept is always taken as positive length when calculating area.

Question 8. In which quadrant will the point lie, if:-
(a) ordinate is \(3\) and abscissa is \(-7\)
(b) abscissa is \(-10\) and ordinate is \(-4\)
(c) Ordinate is \(4\) and abscissa is \(-6\).
Answer:
(a) Abscissa (\(x\)) is \(-7\), Ordinate (\(\text{y}\)) is \(3 \implies (-7, 3)\) lies in Quadrant II.
(b) Abscissa (\(x\)) is \(-10\), Ordinate (\(y\)) is \(-4 \implies (-10, -4)\) lies in Quadrant III.
(c) Abscissa (\(x\)) is \(-6\), Ordinate (\(y\)) is \(4 \implies (-6, 4)\) lies in Quadrant II.
In simple words: Check the signs of \((x, y)\): \((+,+)\) is Q1, \((-,+)\) is Q2, \((-,-)\) is Q3, \((+,-)\) is Q4.

Exam Tip: Remember that abscissa is always the x-coordinate and ordinate is always the y-coordinate.

Question 9. Fill in the blanks:-
(a) The coordinates of the origin \(0\) are ..............................
(b) The y coordinate of every point on the x-axis is ...............
(c) Distance along the x-axis is called ..............................
(d) Distance along the y-axis is called ..............................
(e) The point \((x, y) = (y, x)\) only if ..............................
Answer:
(a) \((0, 0)\)
(b) \(0\)
(c) abscissa (or x-coordinate)
(d) ordinate (or y-coordinate)
(e) \(x = y\)
In simple words: Recall the basic definitions and terminology of Cartesian coordinates.

Exam Tip: Memorize basic definitions of coordinate axes terms to quickly answer fill-in-the-blank questions.