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NCERT Exemplar · Class 9 Mathematics Coordinate Geometry

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EXERCISE 3.4 1–5 (part 6 of 6)

  1. Exercise 1

    Points A(5,3),B(−2,3)\displaystyle \mathrm{A}(5,3), \mathrm{B}(-2,3) and D(5,−4)\displaystyle \mathrm{D}(5,-4) are three vertices of a square ABCD . Plot these points on a graph paper and hence find the coordinates of the vertex C.

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    NCERT’s answer
    C$\displaystyle (-2, -4)$
    AB lies along \(\displaystyle y=3\) and AD along \(\displaystyle x=5\), giving a right angle at A: \[AB = |5-(-2)| = 7, \qquad AD = |3-(-4)| = 7 \] BC continues from B, vertical and equal to AD: \[C = (-2,\ 3-7) = (-2,-4) \] \[CD = |5-(-2)| = 7 = AB \quad \text{(closes the square)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-4_Q1 Answer: \(\displaystyle C(-2,-4)\).

  2. Exercise 2

    Write the coordinates of the vertices of a rectangle whose length and breadth are 5\displaystyle 5 and 3\displaystyle 3 units respectively, one vertex at the origin, the longer side lies on the x\displaystyle x-axis and one of the vertices lies in the third quadrant.

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    NCERT’s answer
    $\displaystyle (0, 0)$, $\displaystyle (-5, 0)$, $\displaystyle (0, -3)$
    The longer side ($\displaystyle 5$) lies on the \(\displaystyle x\)-axis from the origin, and the breadth ($\displaystyle 3$) must reach into the third quadrant, so both run negative: \[O(0,0),\quad A(-5,0),\quad B(-5,-3),\quad C(0,-3) \] NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-4_Q2 Answer: \(\displaystyle O(0,0),\ A(-5,0),\ B(-5,-3),\ C(0,-3)\).

  3. Exercise 3

    Plot the points P(1,0),Q(4,0)\displaystyle \mathrm{P}(1,0), \mathrm{Q}(4,0) and S(1,3)\displaystyle \mathrm{S}(1,3). Find the coordinates of the point R such that PQRS is a square.

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    NCERT’s answer
    $\displaystyle (4,3)$
    PQ is horizontal and PS is vertical, both of length \(\displaystyle 3\), giving a right angle at P: \[PQ = |4-1| = 3, \qquad PS = |3-0| = 3 \] For PQRS to close as a square, R takes Q's abscissa and S's ordinate: \[R = (4,3) \] \[QR = |3-0| = 3 = PS, \qquad RS = |4-1| = 3 = PQ \] NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-4_Q3Answer: \(\displaystyle R(4,3)\)

  4. Exercise 4

    From the Fig. 3.8\displaystyle 3.8, answer the following: (i) Write the points whose abscissa is 0\displaystyle 0 . (ii) Write the points whose ordinate is 0. (iii) Write the points whose abscissa is -5. NCERT_Question_Class9_Maths_Exemplar_Ch3_Ex3-4_Q4

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    NCERT’s answer
    (i)
    A, L and O (ii) G, I and O (iii) D and H
    NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-4_Q4
    (i)
    Points with abscissa \(\displaystyle 0\) lie on the \(\displaystyle y\)-axis: \(\displaystyle A(0,3)\), \(\displaystyle L(0,-4)\), \(\displaystyle O(0,0)\).
    (ii)
    Points with ordinate \(\displaystyle 0\) lie on the \(\displaystyle x\)-axis: \(\displaystyle I(-2,0)\), \(\displaystyle G(5,0)\), \(\displaystyle O(0,0)\).
    (iii)
    Points with abscissa \(\displaystyle -5\): \(\displaystyle D(-5,1)\), \(\displaystyle H(-5,-3)\).
    Answer: (i) A, L, O (ii) G, I, O (iii) D, H

  5. Exercise 5

    Plot the points A(1,−1)\displaystyle \mathrm{A}(1,-1) and B (4,5)\displaystyle (4,5) (i) Draw a line segment joining these points. Write the coordinates of a point on this line segment between the points A and B. (ii) Extend this line segment and write the coordinates of a point on this line which lies outside the line segment AB.

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    NCERT’s answer
    (i)
    $\displaystyle (2,1), \quad$ (ii) $\displaystyle (5,7)$
    \[4-1=3, \qquad 5-(-1)=6 \]
    \[\text{each unit step in } x \text{ raises } y \text{ by } \dfrac{6}{3}=2 \]
    (i)
    \[M=(1+1,\,-1+2)=(2,1) \] lies between A and B: \(\displaystyle 1<2<4\).
    (ii)
    \[N=(4+1,\,5+2)=(5,7) \] lies beyond B, outside segment AB: \(\displaystyle 5>4\).
    NCERT_Solution_Class9_Maths_Exemplar_Ch3_Ex3-4_Q5
    Answer: (i) \(\displaystyle M(2,1)\) (ii) \(\displaystyle N(5,7)\)