NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1

Jasmine Grover logo

Jasmine Grover

Education Journalist | Study Abroad Lead

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1 is given in this article. Class 10 Maths Chapter 7 is an important chapter and its Exercise 7.1 has about 10 questions related to the distance formula.

Read More: Coordinate Geometry MCQs


Download PDF: NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1

Check out the solutions of Class 10 Maths NCERT solutions chapter 7 Coordinate Geometry Exercise 7.1

Read More: NCERT Solutions For Class 10 Maths Coordinate Geometry

Check out other exercise solutions of Class 10 Maths Chapter 7


Class 10 Chapter 7 Topics:

CBSE Class 10 Maths Study Guides:

CBSE X Related Questions

  • 1.
    There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

      • 144
      • 2
      • 420
      • 272

    • 2.
      For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

        • \(0\)
        • \(\frac{80}{81}\)
        • \(1\)
        • \(\frac{82}{80}\)

      • 3.
        If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

          • \( -\frac{2}{3} \)
          • \( \frac{2}{3} \)
          • \( \frac{1}{3} \)
          • \( -\frac{1}{3} \)

        • 4.
          The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are

            • \((x + 2), (x - 5)\)
            • \((x + 5), (x - 2)\)
            • \((x + 1), (x - 5)\)
            • \((x - 1), (x + 2)\)

          • 5.
            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

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

            • 6.
              Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

                • 34
                • 43
                • 38.5
                • 41.5

              Comments


              No Comments To Show