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NCERT Exemplar · Class 9 Mathematics Coordinate Geometry

42 questions · 42 still being checked

EXERCISE 3.1 11–20 (part 2 of 6)

  1. Write the correct answer in each of the following :

    Exercise 11

    The points (−5,2)\displaystyle (-5,2) and (2,−5)\displaystyle (2,-5) lie in the (A) same quadrant (B) II and III quadrants, respectively (C) II and IV quadrants, respectively (D) IV and II quadrants, respectively

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    NCERT’s answer
    (C)
    (C) II and IV quadrants, respectively.\[(-5,2): \quad x<0,\ y>0 \ \Rightarrow\ \text{II} \]\[(2,-5): \quad x>0,\ y<0 \ \Rightarrow\ \text{IV} \]NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-1_Q11

  2. Exercise 12

    If the perpendicular distance of a point P from the x\displaystyle x-axis is 5\displaystyle 5 units and the foot of the perpendicular lies on the negative direction of x\displaystyle x-axis, then the point P has (A) x\displaystyle x coordinate =−5\displaystyle =-5 (B) y\displaystyle y coordinate =5\displaystyle =5 only (C) y\displaystyle y coordinate =−5\displaystyle =-5 only (D) y\displaystyle y coordinate =5\displaystyle =5 or -5\displaystyle 5

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    NCERT’s answer
    (D)
    (D) \(\displaystyle y\) coordinate \(\displaystyle =5\) or \(\displaystyle -5\).\[|y| = 5 \ \Rightarrow\ y = 5 \ \text{or}\ -5 \]The foot on the negative \(\displaystyle x\)-axis fixes only the sign of \(\displaystyle x\); \(\displaystyle P\) may sit above or below it.NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-1_Q12

  3. Exercise 13

    On plotting the points O(0,0),A(3,0),B(3,4),C(0,4)\displaystyle \mathrm{O}(0,0), \mathrm{A}(3,0), \mathrm{B}(3,4), \mathrm{C}(0,4) and joining OA,AB,BC\displaystyle \mathrm{OA}, \mathrm{AB}, \mathrm{BC} and CO which of the following figure is obtained? (A) Square (B) Rectangle (C) Trapezium (D) Rhombus

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    NCERT’s answer
    (B)
    (B) Rectangle.\[OA = 3, \quad AB = 4, \quad BC = 3, \quad CO = 4 \] \[\angle O = \angle A = \angle B = \angle C = 90^\circ \]Opposite sides equal, every angle a right angle, but adjacent sides unequal — a rectangle, not a square.NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-1_Q13

  4. Exercise 14

    If P(−1,1),Q(3,−4),R(1,−1),S(−2,−3)\displaystyle \mathrm{P}(-1,1), \mathrm{Q}(3,-4), \mathrm{R}(1,-1), \mathrm{S}(-2,-3) and T(−4,4)\displaystyle \mathrm{T}(-4,4) are plotted on the graph paper, then the point(s) in the fourth quadrant are (A) P and T (B) Q and R (C) Only S (D) P\displaystyle \mathrm{P} and R

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    NCERT’s answer
    (B)
    (B) Q and R.\[P(-1,1),\ T(-4,4):\quad x<0,\ y>0\ \Rightarrow\ \text{II} \] \[Q(3,-4),\ R(1,-1):\quad x>0,\ y<0\ \Rightarrow\ \text{IV} \] \[S(-2,-3):\quad x<0,\ y<0\ \Rightarrow\ \text{III} \]Only Q and R carry a positive abscissa with a negative ordinate — quadrant IV's sign pattern.NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-1_Q14

  5. Exercise 15

    If the coordinates of the two points are P(−2,3)\displaystyle \mathrm{P}(-2,3) and Q(−3,5)\displaystyle \mathrm{Q}(-3,5), then (abscissa of P ) - (abscissa of Q) is (A) - 5\displaystyle 5 (B) 1\displaystyle 1 (C) - 1\displaystyle 1 (D) - 2\displaystyle 2

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 1\).\[(\text{abscissa of P}) - (\text{abscissa of Q}) = (-2) - (-3) \] \[= -2 + 3 = 1 \]The abscissa is a point's first coordinate: P's is \(\displaystyle -2\), Q's is \(\displaystyle -3\).

  6. Exercise 16

    If P(5,1),Q(8,0),R(0,4),S(0,5)\displaystyle \mathrm{P}(5,1), \mathrm{Q}(8,0), \mathrm{R}(0,4), \mathrm{S}(0,5) and O(0,0)\displaystyle \mathrm{O}(0,0) are plotted on the graph paper, then the point(s) on the x\displaystyle x-axis are (A) P and R (B) R and S (C) Only Q (D) Q and O

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    NCERT’s answer
    (D)
    (D) Q and O.\[y_P=1,\ y_Q=0,\ y_R=4,\ y_S=5,\ y_O=0 \]Only Q and O have zero ordinate, the condition for a point to lie on the \(\displaystyle x\)-axis.NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-1_Q16

  7. Exercise 17

    Abscissa of a point is positive in (A) I and II quadrants (B) I and IV quadrants (C) I quadrant only (D) II quadrant only

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    NCERT’s answer
    (B)
    (B) I and IV quadrants.\[\text{abscissa} > 0 \]Both quadrants I and IV lie to the right of the \(\displaystyle y\)-axis, where every abscissa is positive.

  8. Exercise 18

    The points whose abscissa and ordinate have different signs will lie in (A) I and II quadrants (B) II and III quadrants (C) I and III quadrants (D) II and IV quadrants

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    NCERT’s answer
    (D)
    (D) II and IV quadrants.\[\text{II}:\ x<0,\ y>0 \qquad \text{IV}:\ x>0,\ y<0 \]Only these two quadrants pair a negative coordinate with a positive one — opposite signs on abscissa and ordinate.

  9. Exercise 19

    In Fig. 3.1\displaystyle 3.1, coordinates of P are (A) (−4,2)\displaystyle (-4, 2) (B) (−2,4)\displaystyle (-2, 4) (C) (4,−2)\displaystyle (4,-2) (D) (2,−4)\displaystyle (2,-4) NCERT_Question_Class9_Maths_Exemplar_Ch3_Ex3-1_Q19

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    NCERT’s answer
    (B)
    (B) \(\displaystyle (-2,4)\).\[\text{Fig. 3.1: } P \text{ sits above } x=-2\text{, at height } y=4 \]NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-1_Q19

  10. Exercise 20

    In Fig. 3.2\displaystyle 3.2, the point identified by the coordinates (−5,3)\displaystyle (-5, 3) is (A) T (B) R (C) L (D) S NCERT_Question_Class9_Maths_Exemplar_Ch3_Ex3-1_Q20

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    NCERT’s answer
    (C)
    (C) L.\[\text{Fig. 3.2: } L \text{ is at } x=-5,\ y=3 \]NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-1_Q20