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NCERT Exemplar · Class 9 Mathematics Linear Equations in Two Variables

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EXERCISE 4.1 11–19 (part 2 of 5)

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

    Exercise 11

    x=5,y=2\displaystyle x=5, y=2 is a solution of the linear equation (A) x+2y=7\displaystyle x+2 y=7 (B) 5x+2y=7\displaystyle 5 x+2 y=7 (C) x+y=7\displaystyle x+y=7 (D) 5x+y=7\displaystyle 5 x+y=7

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    NCERT’s answer
    (C)
    (C) \(\displaystyle x+y=7\). \[5+2=7 \] satisfies this equation; substituting \(\displaystyle x=5,\ y=2\) into the other three gives \(\displaystyle 9\), \(\displaystyle 29\), and \(\displaystyle 27\), none equal to \(\displaystyle 7\).

  2. Exercise 12

    If a linear equation has solutions (−2,2)\displaystyle (-2, 2), (0,0)\displaystyle (0, 0) and (2,−2)\displaystyle (2, -2), then it is of the form (A) y−x=0\displaystyle y-x=0 (B) x+y=0\displaystyle x+y=0 (C) −2x+y=0\displaystyle -2 x+y=0 (D) −x+2y=0\displaystyle -x+2 y=0

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    NCERT’s answer
    (B)
    (B) \(\displaystyle x+y=0\). \[-2+2=0,\quad 0+0=0,\quad 2+(-2)=0 \] all three given points satisfy this equation, while each of the other options fails at least one of them.

  3. Exercise 13

    The positive solutions of the equation ax+by+c=0\displaystyle a x+b y+c=0 always lie in the (A) 1st quadrant (B) 2nd quadrant (C) 3rd quadrant (D) 4th quadrant

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    NCERT’s answer
    (A)
    (A) 1st quadrantA positive solution satisfies \[x>0, \quad y>0 \] for any \(\displaystyle a,b,c\) -- exactly the defining condition of the first quadrant.

  4. Exercise 14

    The graph of the linear equation 2x+3y=6\displaystyle 2 x+3 y=6 is a line which meets the x\displaystyle x-axis at the point (A) (0,2)\displaystyle (0,2) (B) (2,0)\displaystyle (2,0) (C) (3,0)\displaystyle (3,0) (D) (0,3)\displaystyle (0,3)

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    NCERT’s answer
    (C)
    (C) \(\displaystyle (3,0)\)\[2x+3y=6 \] On the \(\displaystyle x\)-axis, \(\displaystyle y=0\): \[2x=6 \implies x=3 \]

  5. Exercise 15

    The graph of the linear equation y=x\displaystyle y=x passes through the point (A) (32,−32)\displaystyle \left(\frac{3}{2}, \frac{-3}{2}\right) (B) (0,32)\displaystyle \left(0, \frac{3}{2}\right) (C) (1,1)\displaystyle (1,1) (D) (−12,12)\displaystyle \left(\frac{-1}{2}, \frac{1}{2}\right)

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    NCERT’s answer
    (C)
    (C) \(\displaystyle (1,1)\)\[y=x \] Testing \(\displaystyle (1,1)\): \[1=1 \quad \text{(true)} \] The other three points fail this check.

  6. Exercise 16

    If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation : (A) Changes (B) Remains the same (C) Changes in case of multiplication only (D) Changes in case of division only

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    NCERT’s answer
    (B)
    (B) Remains the same\[ax+by+c=0 \iff k(ax+by+c)=0, \quad k\neq 0 \] Multiplying or dividing by a nonzero constant gives an equivalent equation, true for exactly the same pairs \(\displaystyle (x,y)\).

  7. Exercise 17

    How many linear equations in x\displaystyle x and y\displaystyle y can be satisfied by x=1\displaystyle x=1 and y=2\displaystyle y=2 ? (A) Only one (B) Two (C) Infinitely many (D) Three

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    NCERT’s answer
    (C)
    (C) Infinitely many\[a(1)+b(2)+c=0 \] One condition on three free coefficients \(\displaystyle a,b,c\) leaves infinitely many equations satisfied by \(\displaystyle (1,2)\).

  8. Exercise 18

    The point of the form (a,a)\displaystyle (a, a) always lies on : (A) x\displaystyle x-axis (B) y\displaystyle y-axis (C) On the line y=x\displaystyle y=x (D) On the line x+y=0\displaystyle x+y=0

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    NCERT’s answer
    (C)
    (C) on the line \(\displaystyle y=x\) \[y=x:\quad a=a \] This holds for every value of \(\displaystyle a\); the point \(\displaystyle (a,a)\) fails the \(\displaystyle x\)-axis, \(\displaystyle y\)-axis and \(\displaystyle x+y=0\) tests unless \(\displaystyle a=0\).

  9. Exercise 19

    The point of the form (a,−a)\displaystyle (a,-a) always lies on the line (A) x=a\displaystyle x=a (B) y=−a\displaystyle y=-a (C) y=x\displaystyle y=x (D) x+y=0\displaystyle x+y=0

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    NCERT’s answer
    (D)
    (D) \(\displaystyle x+y=0\) \[x+y=0:\quad a+(-a)=0 \] This holds for every \(\displaystyle a\), so the point \(\displaystyle (a,-a)\) always satisfies \(\displaystyle x+y=0\).