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NCERT Exemplar · Class 10 Mathematics Coordinate Geometry

58 questions · 58 still being checked

EXERCISE 7.1 1–10 (part 1 of 6)

  1. Choose the correct answer from the given four options:

    Exercise 1

    The distance of the point P(2,3)\displaystyle \mathrm{P}(2,3) from the x\displaystyle x-axis is
    (A)
    2\displaystyle 2 (B) 3\displaystyle 3 (C) 1\displaystyle 1 (D) 5\displaystyle 5

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 3\) NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-1_Q1 \[\text{Distance of } P(x,y) \text{ from the } x\text{-axis} = |y| \] \[P(2,3) \ \Rightarrow\ |y| = 3 \]

  2. Exercise 2

    The distance between the points A(0,6)\displaystyle \mathrm{A}(0,6) and B(0,−2)\displaystyle \mathrm{B}(0,-2) is
    (A)
    6\displaystyle 6 (B) 8\displaystyle 8 (C) 4\displaystyle 4 (D) 2\displaystyle 2

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 8\) NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-1_Q2 \[AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \] \[AB = \sqrt{(0-0)^2+(-2-6)^2} = \sqrt{64} = 8 \]

  3. Exercise 3

    The distance of the point P(−6,8)\displaystyle \mathrm{P}(-6,8) from the origin is
    (A)
    8\displaystyle 8 (B) 27\displaystyle 2 \sqrt{7}
    (C)
    10\displaystyle 10

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    (C)
    (C) \(\displaystyle 10\) NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-1_Q3 \[OP = \sqrt{x^2+y^2} \] \[OP = \sqrt{(-6)^2+8^2} = \sqrt{36+64} = \sqrt{100} = 10 \]

  4. Exercise 4

    The distance between the points (0,5)\displaystyle (0,5) and (−5,0)\displaystyle (-5,0) is
    (A)
    5\displaystyle 5 (B) 52\displaystyle 5 \sqrt{2}
    (C)
    25\displaystyle 2 \sqrt{5}
    (D)
    10\displaystyle 10

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 5\sqrt{2}\) NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-1_Q4 \[d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \] \[d = \sqrt{(-5-0)^2+(0-5)^2} = \sqrt{50} = 5\sqrt{2} \]

  5. Exercise 5

    AOBC is a rectangle whose three vertices are vertices A(0,3),O(0,0)\displaystyle \mathrm{A}(0,3), \mathrm{O}(0,0) and B(5,0)\displaystyle \mathrm{B}(5,0). The length of its diagonal is
    (A)
    5\displaystyle 5 (B) 3\displaystyle 3 (C) 34\displaystyle \sqrt{34}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle \sqrt{34}\) NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-1_Q5 \[C = A + B - O = (5,3) \] \[AB = \sqrt{(5-0)^2+(0-3)^2} = \sqrt{25+9} = \sqrt{34} \]

  6. Exercise 6

    The perimeter of a triangle with vertices (0,4),(0,0)\displaystyle (0,4),(0,0) and (3,0)\displaystyle (3,0) is
    (A)
    5\displaystyle 5 (B) 12\displaystyle 12
    (C)
    11\displaystyle 11
    (D)
    7+5\displaystyle 7+\sqrt{5}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 12\) NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-1_Q6 \[\sqrt{(0-0)^2+(4-0)^2}=4,\quad \sqrt{(3-0)^2+(0-0)^2}=3,\quad \sqrt{(3-0)^2+(0-4)^2}=5 \] \[\text{Perimeter} = 4+3+5 = 12 \]

  7. Exercise 7

    The area of a triangle with vertices A(3,0),B(7,0)\displaystyle \mathrm{A}(3,0), \mathrm{B}(7,0) and C(8,4)\displaystyle \mathrm{C}(8,4) is
    (A)
    14\displaystyle 14
    (B)
    28\displaystyle 28
    (C)
    8\displaystyle 8 (D) 6\displaystyle 6

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 8\) NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-1_Q7 \[\text{Area} = \tfrac12\big|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\big| \] \[= \tfrac12\big|3(0-4)+7(4-0)+8(0-0)\big| = \tfrac12(16) = 8 \]

  8. Exercise 8

    The points (−4,0),(4,0),(0,3)\displaystyle (-4,0),(4,0),(0,3) are the vertices of a
    (A)
    right triangle
    (B)
    isosceles triangle
    (C)
    equilateral triangle
    (D)
    scalene triangle

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    NCERT’s answer
    (B)
    (B) isosceles triangle \[AB=\sqrt{(4-(-4))^2+(0-0)^2}=8 \] \[AC=\sqrt{(0-(-4))^2+(3-0)^2}=5 \] \[BC=\sqrt{(0-4)^2+(3-0)^2}=5 \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-1_Q8 \(\displaystyle AC=BC\ne AB\), so the triangle is isosceles.

  9. Exercise 9

    The point which divides the line segment joining the points (7,−6)\displaystyle (7,-6) and (3,4)\displaystyle (3,4) in ratio 1\displaystyle 1 : 2\displaystyle 2 internally lies in the
    (A)
    I quadrant
    (B)
    II quadrant
    (C)
    III quadrant
    (D)
    IV quadrant

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    NCERT’s answer
    (D)
    (D) IV quadrant \[x=\frac{1\cdot3+2\cdot7}{1+2}=\frac{17}{3} \] \[y=\frac{1\cdot4+2\cdot(-6)}{1+2}=-\frac{8}{3} \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-1_Q9 \(\displaystyle x>0\) and \(\displaystyle y<0\), so the point lies in the fourth quadrant.

  10. Exercise 10

    The point which lies on the perpendicular bisector of the line segment joining the points A(−2,−5)\displaystyle \mathrm{A}(-2,-5) and B(2,5)\displaystyle \mathrm{B}(2,5) is
    (A)
    (0,0)\displaystyle (0,0)
    (B)
    (0,2)\displaystyle (0,2)
    (C)
    (2,0)\displaystyle (2,0)
    (D)
    (−2,0)\displaystyle (-2,0)

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    NCERT’s answer
    (A)
    (A) \(\displaystyle (0,0)\) \[\text{Midpoint of }AB=\left(\frac{-2+2}{2},\frac{-5+5}{2}\right)=(0,0) \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-1_Q10 The perpendicular bisector always passes through the midpoint of the segment.