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NCERT Exemplar · Class 10 Mathematics Coordinate Geometry

58 questions · 58 still being checked

EXERCISE 7.3 11–20 (part 5 of 6)

  1. Exercise 11

    Find the ratio in which the point P(34,512)\displaystyle \mathrm{P}\left(\frac{3}{4}, \frac{5}{12}\right) divides the line segment joining the points A(12,32)\displaystyle \mathrm{A}\left(\frac{1}{2}, \frac{3}{2}\right) and B(2,−5)\displaystyle \mathrm{B}(2,-5).

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    NCERT’s answer
    $\displaystyle 1$:$\displaystyle 5$
    Let \(\displaystyle P\) divide \(\displaystyle AB\) in ratio \(\displaystyle k:1\). \[\dfrac34=\dfrac{2k+\tfrac12}{k+1} \] \[3(k+1)=4\left(2k+\tfrac12\right) \] \[3k+3=8k+2 \] \[1=5k \quad\Rightarrow\quad k=\dfrac15 \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q11 Answer: \(\displaystyle 1:5\).

  2. Exercise 12

    If P(9a−2,−b)\displaystyle \mathrm{P}(9 a-2,-b) divides line segment joining A(3a+1,−3)\displaystyle \mathrm{A}(3 a+1,-3) and B(8a,5)\displaystyle \mathrm{B}(8 a, 5) in the ratio 3\displaystyle 3 : 1\displaystyle 1, find the values of a\displaystyle a and b\displaystyle b.

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    NCERT’s answer
    \(\displaystyle a=1 \quad b=-3\)
    Section formula, ratio \(\displaystyle 3:1\) from \(\displaystyle A\) to \(\displaystyle B\): \[9a-2=\dfrac{3(8a)+1(3a+1)}{3+1}=\dfrac{27a+1}{4} \] \[36a-8=27a+1 \] \[9a=9 \quad\Rightarrow\quad a=1 \] \[-b=\dfrac{3(5)+1(-3)}{4}=\dfrac{12}{4}=3 \quad\Rightarrow\quad b=-3 \] \[A(4,-3),\quad P(7,3),\quad B(8,5) \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q12 Answer: \(\displaystyle a=1,\ b=-3\).

  3. Exercise 13

    If (a,b)\displaystyle (a, b) is the mid-point of the line segment joining the points A(10,−6)\displaystyle \mathrm{A}(10,-6) and B (k,4)\displaystyle (k, 4) and a−2b=18\displaystyle a-2 b=18, find the value of k\displaystyle k and the distance AB.

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    NCERT’s answer
    \(\displaystyle k=22, \mathrm{AB}=2 \sqrt{61}\)
    \[a=\dfrac{10+k}{2},\quad b=\dfrac{-6+4}{2}=-1 \] \[a-2b=18 \] \[\dfrac{10+k}{2}-2(-1)=18 \] \[\dfrac{10+k}{2}=16 \] \[k=22 \] \[(a,b)=(16,-1) \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q13 \[AB=\sqrt{(22-10)^2+(4+6)^2}=\sqrt{144+100}=\sqrt{244}=2\sqrt{61} \] Answer: \(\displaystyle k=22,\ AB=2\sqrt{61}\).

  4. Exercise 14

    The centre of a circle is (2a,a−7)\displaystyle (2 a, a-7). Find the values of a\displaystyle a if the circle passes through the point (11,−9)\displaystyle (11, -9) and has diameter 102\displaystyle 10 \sqrt{2} units.

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    NCERT’s answer
    \(\displaystyle a=5,3\)
    \[r=\dfrac{10\sqrt2}{2}=5\sqrt2 \] \[(2a-11)^2+(a-7+9)^2=(5\sqrt2)^2 \] \[(2a-11)^2+(a+2)^2=50 \] \[4a^2-44a+121+a^2+4a+4=50 \] \[5a^2-40a+75=0 \] \[a^2-8a+15=0 \] \[(a-3)(a-5)=0 \] \[a=3 \ \text{or}\ a=5 \] \[a=3:\ O_1(6,-4),\qquad a=5:\ O_2(10,-2) \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q14 Answer: \(\displaystyle a=3\) or \(\displaystyle a=5\).

  5. Exercise 15

    The line segment joining the points A(3,2)\displaystyle \mathrm{A}(3,2) and B(5,1)\displaystyle \mathrm{B}(5,1) is divided at the point P in the ratio 1\displaystyle 1:2\displaystyle 2 and it lies on the line 3x−18y+k=0\displaystyle 3 x-18 y+k=0. Find the value of k\displaystyle k.

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    NCERT’s answer
    $\displaystyle 19$
    By the section formula, with \(\displaystyle P\) dividing \(\displaystyle AB\) in the ratio \(\displaystyle 1:2\): \[P = \left( \frac{1(5)+2(3)}{1+2}, \frac{1(1)+2(2)}{1+2} \right) = \left( \frac{11}{3}, \frac{5}{3} \right) \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q15 Since \(\displaystyle P\) lies on \(\displaystyle 3x-18y+k=0\): \[3\left(\frac{11}{3}\right) - 18\left(\frac{5}{3}\right) + k = 0 \] \[11 - 30 + k = 0 \] \[k = 19 \] Answer: \(\displaystyle k = 19\)

  6. Exercise 16

    If D(−12,52),E(7,3)\displaystyle \mathrm{D}\left(\frac{-1}{2}, \frac{5}{2}\right), \mathrm{E}(7,3) and F(72,72)\displaystyle \mathrm{F}\left(\frac{7}{2}, \frac{7}{2}\right) are the midpoints of sides of ΔABC\displaystyle \Delta \mathrm{ABC}, find the area of the ΔABC\displaystyle \Delta \mathrm{ABC}.

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    NCERT’s answer
    $\displaystyle 11$
    By the shoelace formula on the midpoints: \[\text{ar}(\triangle DEF) = \frac12\left| x_D(y_E-y_F)+x_E(y_F-y_D)+x_F(y_D-y_E) \right| \] \[= \frac12\left| \left(-\frac12\right)\left(-\frac12\right) + 7(1) + \frac72\left(-\frac12\right) \right| = \frac12 \cdot \frac{11}{2} = \frac{11}{4} \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q16 The midpoints of the sides of a triangle form a triangle similar to it in ratio \(\displaystyle 1:2\), so \[\text{ar}(\triangle ABC) = 4\,\text{ar}(\triangle DEF) = 4 \times \frac{11}{4} = 11 \] Answer: \(\displaystyle 11\) square units

  7. Exercise 17

    The points A(2,9),B(a,5)\displaystyle \mathrm{A}(2,9), \mathrm{B}(a, 5) and C(5,5)\displaystyle \mathrm{C}(5,5) are the vertices of a triangle ABC right angled at B. Find the values of a\displaystyle a and hence the area of ΔABC\displaystyle \Delta \mathrm{ABC}.

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    NCERT’s answer
    \(\displaystyle a=2\), Area \(\displaystyle =6\) sq. unit
    Since \(\displaystyle B(a,5)\) and \(\displaystyle C(5,5)\) share \(\displaystyle y=5\), \(\displaystyle BC\) is horizontal; a right angle at \(\displaystyle B\) forces \(\displaystyle BA\) vertical, so \(\displaystyle a=2\). NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q17 \[AB = \sqrt{(2-2)^2+(9-5)^2} = 4, \quad BC = \sqrt{(5-2)^2+(5-5)^2} = 3 \] \[\text{ar}(\triangle ABC) = \frac12 \cdot BC \cdot AB = \frac12 \cdot 3 \cdot 4 = 6 \] Answer: \(\displaystyle a = 2\), area \(\displaystyle = 6\) square units

  8. Exercise 18

    Find the coordinates of the point R on the line segment joining the points P(−1,3)\displaystyle \mathrm{P}(-1,3) and Q(2,5)\displaystyle \mathrm{Q}(2,5) such that PR=35PQ\displaystyle \mathrm{PR}=\frac{3}{5} \mathrm{PQ}.

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    NCERT’s answer
    \(\displaystyle \left(\frac{4}{5}, \frac{21}{5}\right)\)
    \(\displaystyle PR=\frac35PQ\) means \(\displaystyle R\) divides \(\displaystyle PQ\) in the ratio \(\displaystyle 3:2\). By the section formula: \[R = \left( \frac{3(2)+2(-1)}{3+2}, \frac{3(5)+2(3)}{3+2} \right) = \left( \frac{4}{5}, \frac{21}{5} \right) \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q18 Answer: \(\displaystyle R = \left(\dfrac{4}{5}, \dfrac{21}{5}\right)\)

  9. Exercise 19

    Find the values of k\displaystyle k if the points A(k+1,2k),B(3k,2k+3)\displaystyle \mathrm{A}(k+1,2 k), \mathrm{B}(3 k, 2 k+3) and C(5k−1,5k)\displaystyle \mathrm{C}(5 k-1,5 k) are collinear.

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    NCERT’s answer
    \(\displaystyle 2, \frac{1}{2}\)
    Collinearity of \(\displaystyle A, B, C\) requires \(\displaystyle \text{ar}(\triangle ABC) = 0\): \[(k+1)(3-3k) + 3k(3k) + (5k-1)(-3) = 0 \] \[3 - 3k^2 + 9k^2 - 15k + 3 = 0 \] \[2k^2 - 5k + 2 = 0 \] \[(2k-1)(k-2) = 0 \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q19 Answer: \(\displaystyle k = \dfrac{1}{2}\) or \(\displaystyle k = 2\)

  10. Exercise 20

    Find the ratio in which the line 2x+3y−5=0\displaystyle 2 x+3 y-5=0 divides the line segment joining the points (8,−9)\displaystyle (8, -9) and (2,1)\displaystyle (2, 1). Also find the coordinates of the point of division.

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    NCERT’s answer
    $\displaystyle 8$:$\displaystyle 1$, \(\displaystyle \left(\frac{8}{3}, \frac{-1}{9}\right)\)
    Let the line meet \(\displaystyle P(8,-9)Q(2,1)\) at the point dividing it in ratio \(\displaystyle k:1\): \[\left( \frac{2k+8}{k+1}, \frac{k-9}{k+1} \right) \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q20 On \(\displaystyle 2x+3y-5=0\): \[2(2k+8) + 3(k-9) - 5(k+1) = 0 \] \[4k+16+3k-27-5k-5 = 0 \] \[2k - 16 = 0 \implies k = 8 \] \[\text{Point} = \left( \frac{2(8)+8}{9}, \frac{8-9}{9} \right) = \left( \frac{8}{3}, -\frac{1}{9} \right) \] Answer: ratio \(\displaystyle 8:1\), point \(\displaystyle \left(\dfrac{8}{3}, -\dfrac{1}{9}\right)\)