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NCERT Exemplar · Class 10 Mathematics Coordinate Geometry

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EXERCISE 7.2 1–12 (part 3 of 6)

  1. State whether the following statements are true or false. Justify your answer.

    Exercise 1

    △ABC\displaystyle \triangle \mathrm{ABC} with vertices A(−2,0),B(2,0)\displaystyle \mathrm{A}(-2,0), \mathrm{B}(2,0) and C(0,2)\displaystyle \mathrm{C}(0,2) is similar to △DEF\displaystyle \triangle \mathrm{DEF} with vertices D(−4,0)\displaystyle \mathrm{D}(-4,0) E(4,0)\displaystyle \mathrm{E}(4,0) and F(0,4)\displaystyle \mathrm{F}(0,4).

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    NCERT’s answer
    True. Because all three sides of both triangles are proportional.
    True \[AB=\sqrt{(2-(-2))^2}=4,\quad BC=\sqrt{2^2+2^2}=2\sqrt2,\quad CA=\sqrt{2^2+2^2}=2\sqrt2 \] \[DE=\sqrt{(4-(-4))^2}=8,\quad EF=\sqrt{4^2+4^2}=4\sqrt2,\quad FD=\sqrt{4^2+4^2}=4\sqrt2 \] \[\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=\frac12 \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q1 Sides are proportional (SSS similarity), so the triangles are similar.

  2. Exercise 2

    Point P(−4,2)\displaystyle \mathrm{P}(-4,2) lies on the line segment joining the points A(−4,6)\displaystyle \mathrm{A}(-4,6) and B(−4,−6)\displaystyle \mathrm{B}(-4,-6).

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    NCERT’s answer
    True. The three points lie on the line \(\displaystyle x=-4\).
    True \[AP=\sqrt{(-4-(-4))^2+(6-2)^2}=4,\quad PB=\sqrt{(-4-(-4))^2+(2-(-6))^2}=8 \] \[AP+PB=4+8=12=AB \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q2 \(\displaystyle A, P, B\) all lie on \(\displaystyle x=-4\) and \(\displaystyle AP+PB=AB\), so \(\displaystyle P\) lies between \(\displaystyle A\) and \(\displaystyle B\).

  3. Exercise 3

    The points (0,5),(0,−9)\displaystyle (0,5),(0,-9) and (3,6)\displaystyle (3,6) are collinear.

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    NCERT’s answer
    False, since two points lie on the \(\displaystyle y\)-axis and one point lies in quadrant I.
    False \[\text{Area}=\tfrac12\big|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big| \] \[=\tfrac12\big|0(-9-6)+0(6-5)+3(5-(-9))\big|=\tfrac12(42)=21\neq0 \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q3 A nonzero area means the three points are not collinear.

  4. Exercise 4

    Point P(0,2)\displaystyle \mathrm{P}(0,2) is the point of intersection of y\displaystyle y-axis and perpendicular bisector of line segment joining the points A(−1,1)\displaystyle \mathrm{A}(-1,1) and B(3,3)\displaystyle \mathrm{B}(3,3).

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    NCERT’s answer
    False. \(\displaystyle \mathrm{PA}=\sqrt{2}\) and \(\displaystyle \mathrm{PB}=\sqrt{10}\), i.e., \(\displaystyle \mathrm{PA} \neq \mathrm{PB}\).
    False \[PA=\sqrt{(0+1)^2+(2-1)^2}=\sqrt2 \] \[PB=\sqrt{(0-3)^2+(2-3)^2}=\sqrt{10} \] \[PA\neq PB \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q4 Every point of the perpendicular bisector is equidistant from \(\displaystyle A\) and \(\displaystyle B\), so \(\displaystyle P(0,2)\) is not on it.

  5. Exercise 5

    Points A(3,1),B(12,−2)\displaystyle \mathrm{A}(3,1), \mathrm{B}(12,-2) and C(0,2)\displaystyle \mathrm{C}(0,2) cannot be the vertices of a triangle.

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    NCERT’s answer
    True, since ar \(\displaystyle (\Delta \mathrm{ABC})=0\).
    True \[\text{Area}=\tfrac12\big|3(-2-2)+12(2-1)+0(1-(-2))\big| \] \[=\tfrac12|-12+12|=0 \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q5 Zero area means A, B, C are collinear, so they cannot be the vertices of a triangle.

  6. Exercise 6

    Points A(4,3),B(6,4),C(5,−6)\displaystyle \mathrm{A}(4,3), \mathrm{B}(6,4), \mathrm{C}(5,-6) and D(−3,5)\displaystyle \mathrm{D}(-3,5) are the vertices of a parallelogram.

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    NCERT’s answer
    False, since the diagonals donot bisect each other.
    False \[\text{Midpoint of }AC=\left(\frac{4+5}{2},\frac{3-6}{2}\right)=(4.5,-1.5) \] \[\text{Midpoint of }BD=\left(\frac{6-3}{2},\frac{4+5}{2}\right)=(1.5,4.5) \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q6 Diagonals of a parallelogram bisect each other; these midpoints differ, so ABCD is not one.

  7. Exercise 7

    A circle has its centre at the origin and a point P(5,0)\displaystyle \mathrm{P}(5,0) lies on it. The point Q(6,8)\displaystyle \mathrm{Q}(6,8) lies outside the circle.

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    NCERT’s answer
    True, radius of the circle \(\displaystyle =5\) and \(\displaystyle \mathrm{OP}>5\)
    True. \(\displaystyle OP\) is the circle's radius. \[OP=\sqrt{5^2+0^2}=5 \] \[OQ=\sqrt{6^2+8^2}=\sqrt{100}=10 \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q7 \(\displaystyle OQ=10>5=OP\), so Q lies outside the circle.

  8. Exercise 8

    The point A(2,7)\displaystyle \mathrm{A}(2,7) lies on the perpendicular bisector of line segment joining the points P(6,5)\displaystyle \mathrm{P}(6,5) and Q(0,−4)\displaystyle \mathrm{Q}(0,-4).

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    NCERT’s answer
    False, since \(\displaystyle \mathrm{AP} \neq \mathrm{AQ}\)
    False. A point on the perpendicular bisector of \(\displaystyle PQ\) is equidistant from \(\displaystyle P\) and \(\displaystyle Q\). \[AP=\sqrt{(6-2)^2+(5-7)^2}=\sqrt{20} \] \[AQ=\sqrt{(0-2)^2+(-4-7)^2}=\sqrt{125} \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q8 \(\displaystyle AP\neq AQ\), so A does not lie on the bisector.

  9. Exercise 9

    Point P(5,−3)\displaystyle \mathrm{P}(5,-3) is one of the two points of trisection of the line segment joining the points A(7,−2)\displaystyle \mathrm{A}(7,-2) and B(1,−5)\displaystyle \mathrm{B}(1,-5).

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    NCERT’s answer
    True, since P divides AB in the ratio \(\displaystyle 1: 2\)
    True. The trisection point nearer \(\displaystyle A\) divides \(\displaystyle AB\) in ratio \(\displaystyle 1:2\). \[\left(\frac{1(1)+2(7)}{1+2},\ \frac{1(-5)+2(-2)}{1+2}\right)=(5,-3) \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q9 This equals \(\displaystyle P\), so \(\displaystyle P\) is a point of trisection of \(\displaystyle AB\).

  10. Exercise 10

    Points A(−6,10),B(−4,6)\displaystyle \mathrm{A}(-6,10), \mathrm{B}(-4,6) and C(3,−8)\displaystyle \mathrm{C}(3,-8) are collinear such that AB=29AC\displaystyle \mathrm{AB}=\frac{2}{9} \mathrm{AC}.

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    NCERT’s answer
    True, since B divides AC in the ratio \(\displaystyle 2: 7\)
    True. \[AB=\sqrt{(-4+6)^2+(6-10)^2}=\sqrt{20}=2\sqrt5 \] \[BC=\sqrt{(3+4)^2+(-8-6)^2}=\sqrt{245}=7\sqrt5 \] \[AC=\sqrt{(3+6)^2+(-8-10)^2}=\sqrt{405}=9\sqrt5 \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q10 \[AB+BC=2\sqrt5+7\sqrt5=9\sqrt5=AC \] So \(\displaystyle B\) lies on \(\displaystyle AC\), between \(\displaystyle A\) and \(\displaystyle C\). \[\frac{AB}{AC}=\frac{2\sqrt5}{9\sqrt5}=\frac{2}{9} \]

  11. Exercise 11

    The point P(−2,4)\displaystyle \mathrm{P}(-2,4) lies on a circle of radius 6\displaystyle 6 and centre C (3,5)\displaystyle (3, 5).

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    NCERT’s answer
    False, since \(\displaystyle \mathrm{PC}=\sqrt{26}<6\), P will lie inside the circle.
    False. \(\displaystyle PC\) is the distance from \(\displaystyle P\) to the centre. \[PC=\sqrt{(3-(-2))^2+(5-4)^2}=\sqrt{25+1}=\sqrt{26} \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q11 \(\displaystyle \sqrt{26}\neq 6\); since \(\displaystyle \sqrt{26}<6\), P lies inside the circle, not on it.

  12. Exercise 12

    The points A (−1,−2)\displaystyle (-1, -2), B (4,3)\displaystyle (4, 3), C (2,5)\displaystyle (2, 5) and D (−3,0)\displaystyle (-3, 0) in that order form a rectangle.

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    NCERT’s answer
    True, Mid-points of both the diagonals are the same and the diagonals are of equal length.
    True. \[AB=\sqrt{5^2+5^2}=\sqrt{50},\quad BC=\sqrt{2^2+2^2}=\sqrt{8} \] \[CD=\sqrt{5^2+5^2}=\sqrt{50},\quad DA=\sqrt{2^2+2^2}=\sqrt{8} \] \(\displaystyle AB=CD\) and \(\displaystyle BC=DA\) make \(\displaystyle ABCD\) a parallelogram. \[AC=\sqrt{3^2+7^2}=\sqrt{58},\quad BD=\sqrt{7^2+3^2}=\sqrt{58} \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-2_Q12 Equal diagonals in a parallelogram make it a rectangle.