NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.4 Solutions

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NCERT Solutions for Class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.4 Solutions are based on the Equations of Lines Parallel to the x-axis and y-axis respectively .

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Exercise Solutions of Class 9 Maths Chapter 4 Linear Equation in Two Variables

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CBSE X Related Questions

  • 1.
    In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

      • $6$
      • $4$
      • $\pm 4$
      • $3$

    • 2.
      The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

        • 2
        • \(-6\)
        • \(-2\)
        • 6

      • 3.
        If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

          • \(\frac{7}{2}\)
          • \(\frac{6}{7}\)
          • \(\frac{-7}{6}\)
          • \(\frac{7}{6}\)

        • 4.
          If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

            • \(4 \text{ cm}\)
            • \(4\sqrt{2} \text{ cm}\)
            • \(8 \text{ cm}\)
            • \(2\sqrt{2} \text{ cm}\)

          • 5.
            Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
            Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

              • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
              • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
              • Assertion (A) is true, but Reason (R) is false.
              • Assertion (A) is false, but Reason (R) is true.

            • 6.
              It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

                • \(\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle EDF} = \frac{AB}{ED}\)
                • \(\frac{AB}{ED} = \frac{AC}{EF}\)
                • \(\angle A = \angle D, \angle C = \angle F\)
                • \(\frac{AB + BC}{AC} = \frac{ED + DF}{EF}\)

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