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NCERT Exemplar · Class 10 Mathematics Coordinate Geometry

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EXERCISE 7.4 1–6 (part 6 of 6)

  1. Exercise 1

    If (−4,3)\displaystyle (-4,3) and (4,3)\displaystyle (4,3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

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    NCERT’s answer
    \(\displaystyle (0, 3-4 \sqrt{3})\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-4_Q1 \[A(-4,3),\ B(4,3): \qquad AB = 4-(-4) = 8, \qquad M = \left(\frac{-4+4}{2},\,3\right) = (0,3) \] Third vertex \(\displaystyle C\) lies on the perpendicular bisector of \(\displaystyle AB\), the line \(\displaystyle x=0\). \[h = \frac{\sqrt3}{2}\cdot AB = \frac{\sqrt3}{2}\cdot 8 = 4\sqrt3 \] \[C = (0,\ 3 \pm 4\sqrt3) \] The origin is below \(\displaystyle AB\) (\(\displaystyle y=3\)), so \(\displaystyle C\) must be below \(\displaystyle AB\) too. Answer: \(\displaystyle (0,\ 3-4\sqrt3) \)

  2. Exercise 2

    A (6,1)\displaystyle (6, 1), B (8,2)\displaystyle (8, 2) and C (9,4)\displaystyle (9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ΔADE\displaystyle \Delta \mathrm{ADE}.

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    NCERT’s answer
    \(\displaystyle \frac{3}{4}\) sq. units.
    NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-4_Q2 Diagonals of a parallelogram bisect each other: \[\text{mid}(AC) = \left(\frac{6+9}{2},\,\frac{1+4}{2}\right) = (7.5,\,2.5) = \text{mid}(BD) \] \[B=(8,2) \ \Rightarrow\ D = \big(2(7.5)-8,\ 2(2.5)-2\big) = (7,3) \] \[E = \text{mid}(D,C) = \left(\frac{7+9}{2},\,\frac{3+4}{2}\right) = (8,\,3.5) \] \[\text{Area}(\triangle ADE) = \tfrac12\big|\,x_A(y_D-y_E)+x_D(y_E-y_A)+x_E(y_A-y_D)\,\big| \] \[= \tfrac12\big|\,6(-0.5)+7(2.5)+8(-2)\,\big| = \tfrac12(1.5) \] Answer: \(\displaystyle \dfrac{3}{4} \) sq units

  3. Exercise 3

    The points A(x1,y1),B(x2,y2)\displaystyle \mathrm{A}\left(x_1, y_1\right), \mathrm{B}\left(x_2, y_2\right) and C(x3y3)\displaystyle \mathrm{C}\left(x_3 y_3\right) are the vertices of ΔABC\displaystyle \Delta \mathrm{ABC}.
    (i)
    The median from A meets BC at D. Find the coordinates of the point D.
    (ii)
    Find the coordinates of the point P on AD such that AP:PD=2:1\displaystyle \mathrm{AP}: \mathrm{PD}=2: 1
    (iii)
    Find the coordinates of points Q and R on medians BE and CF, respectively such that BQ:QE=2:1\displaystyle \mathrm{BQ}: \mathrm{QE}=2: 1 and CR:RF=2:1\displaystyle \mathrm{CR}: \mathrm{RF}=2: 1
    (iv)
    What are the coordinates of the centroid of the triangle ABC?

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    NCERT’s answer
    (i)
    \(\displaystyle \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)\)
    (ii)
    \(\displaystyle \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
    (iii)
    same as (ii)
    (iv)
    same as (ii)
    NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-4_Q3
    (i)
    \(\displaystyle D\) is the midpoint of \(\displaystyle BC\):
    \[D = \left(\frac{x_2+x_3}{2},\ \frac{y_2+y_3}{2}\right) \]
    (ii)
    Section formula, \(\displaystyle P\) divides \(\displaystyle AD\) in ratio \(\displaystyle 2:1\) from \(\displaystyle A\):
    \[P = \left(\frac{2\cdot\dfrac{x_2+x_3}{2}+x_1}{3},\ \frac{2\cdot\dfrac{y_2+y_3}{2}+y_1}{3}\right) = \left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right) \]
    (iii)
    With \(\displaystyle E=\text{mid}(A,C)\) and \(\displaystyle F=\text{mid}(A,B)\), the same section formula on \(\displaystyle BQ:QE=2:1\) and \(\displaystyle CR:RF=2:1\) gives
    \[Q = \left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right) = R \]
    (iv)
    Since \(\displaystyle P=Q=R\), all three medians meet at
    \[\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right) \]
    Answer: centroid \(\displaystyle = \left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3}\right) \)

  4. Exercise 4

    If the points A(1,−2),B(2,3)C(a,2)\displaystyle \mathrm{A}(1,-2), \mathrm{B}(2,3) \mathrm{C}(a, 2) and D(−4,−3)\displaystyle \mathrm{D}(-4,-3) form a parallelogram, find the value of a\displaystyle a and height of the parallelogram taking AB as base.

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    NCERT’s answer
    \(\displaystyle a=-3, h=\frac{12 \sqrt{26}}{13}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-4_Q4 Diagonals of parallelogram \(\displaystyle ABCD\) bisect each other: \[\text{mid}(AC) = \text{mid}(BD) \ \Rightarrow\ \left(\frac{1+a}{2},\,\frac{-2+2}{2}\right) = \left(\frac{2-4}{2},\,\frac{3-3}{2}\right) = (-1,0) \] \[1+a = -2 \ \Rightarrow\ a = -3 \] So \(\displaystyle C=(-3,2)\). \[\text{ar}(\triangle ABC) = \tfrac12\big|\,1(3-2)+2(2+2)+(-3)(-2-3)\,\big| = \tfrac12(1+8+15) = 12 \] \[\text{ar}(ABCD) = 2\,\text{ar}(\triangle ABC) = 24 \] \[AB = \sqrt{(2-1)^2+(3+2)^2} = \sqrt{26} \] \[\text{height} = \frac{\text{ar}(ABCD)}{AB} = \frac{24}{\sqrt{26}} = \frac{12\sqrt{26}}{13} \] Answer: \(\displaystyle a=-3,\quad \text{height} = \dfrac{12\sqrt{26}}{13} \) units

  5. Exercise 5

    Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in figure 7.4. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students A, B, C and D? If so, what should be his position? NCERT_Question_Class10_Maths_Exemplar_Ch7_Ex7-4_Q5

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    NCERT’s answer
    Yes, Jaspal should be placed at the point \(\displaystyle (7,5)\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-4_Q5 From the figure: \(\displaystyle A(3,5),\ B(7,9),\ C(11,5),\ D(7,1)\). \[\text{mid}(AC) = \left(\frac{3+11}{2},\frac{5+5}{2}\right) = (7,5), \qquad \text{mid}(BD) = \left(\frac{7+7}{2},\frac{9+1}{2}\right) = (7,5) \] \[AC = 11-3 = 8, \qquad BD = 9-1 = 8 \] Both diagonals share the midpoint \(\displaystyle J(7,5)\) and have equal length \(\displaystyle 8\), so \[JA=JC=\frac{AC}{2}=4, \qquad JB=JD=\frac{BD}{2}=4 \] Answer: Yes, at \(\displaystyle J(7,5)\): column $\displaystyle 7$, row 5.

  6. Exercise 6

    Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter's school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines). If the house is situated at (2,4)\displaystyle (2, 4), bank at (5,8)\displaystyle (5, 8), school at (13,14)\displaystyle (13, 14) and office at (13,26)\displaystyle (13,26) and coordinates are in km.

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    NCERT’s answer
    House to Bank \(\displaystyle =5 \mathrm{~km}\) Bank to school = $\displaystyle 10$ km School to Office \(\displaystyle =12 \mathrm{~km}\) Total distance travelled \(\displaystyle =27 \mathrm{~km}\) Distance from house to office \(\displaystyle =24.6 \mathrm{~km}\) Extra distance \(\displaystyle =2.4 \mathrm{~km}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-4_Q6 House \(\displaystyle H(2,4)\), bank \(\displaystyle B(5,8)\), school \(\displaystyle S(13,14)\), office \(\displaystyle F(13,26)\). \[HB = \sqrt{(5-2)^2+(8-4)^2} = 5 \] \[BS = \sqrt{(13-5)^2+(14-8)^2} = 10 \] \[SF = \sqrt{(13-13)^2+(26-14)^2} = 12 \] \[\text{route} = HB+BS+SF = 5+10+12 = 27 \] \[HF = \sqrt{(13-2)^2+(26-4)^2} = \sqrt{605} = 11\sqrt5 \] \[\text{extra} = 27-11\sqrt5 \] Answer: \(\displaystyle (27-11\sqrt5)\ \text{km} \approx 2.4\ \text{km} \)