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NCERT Exemplar · Class 10 Mathematics Coordinate Geometry

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EXERCISE 7.3 1–10 (part 4 of 6)

  1. Exercise 1

    Name the type of triangle formed by the points A(−5,6),B(−4,−2)\displaystyle \mathrm{A}(-5,6), \mathrm{B}(-4,-2) and C(7,5)\displaystyle \mathrm{C}(7,5).

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    NCERT’s answer
    Scalene triangle
    NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q1 \[AB=\sqrt{(-4-(-5))^2+(-2-6)^2}=\sqrt{1+64}=\sqrt{65} \] \[BC=\sqrt{(7-(-4))^2+(5-(-2))^2}=\sqrt{121+49}=\sqrt{170} \] \[CA=\sqrt{(-5-7)^2+(6-5)^2}=\sqrt{144+1}=\sqrt{145} \] All three sides are unequal, and no pair of squared sides sums to the third, so no angle is \(\displaystyle 90^\circ\).Answer: Scalene triangle.

  2. Exercise 2

    Find the points on the x\displaystyle x-axis which are at a distance of 25\displaystyle 2 \sqrt{5} from the point (7,−4)\displaystyle (7, -4). How many such points are there?

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    NCERT’s answer
    $\displaystyle (9, 0)$, $\displaystyle (5, 0)$, $\displaystyle 2$ points
    \[(x-7)^2+(0-(-4))^2=(2\sqrt5)^2 \] \[(x-7)^2+16=20 \] \[(x-7)^2=4 \implies x=9 \text{ or } x=5 \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q2Answer: \(\displaystyle (5,0)\) and \(\displaystyle (9,0)\); two such points.

  3. Exercise 3

    What type of a quadrilateral do the points A (2,−2)\displaystyle (2, -2), B (7,3)\displaystyle (7, 3), C (11,−1)\displaystyle (11, -1) and D (6,−6)\displaystyle (6, -6) taken in that order, form?

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    NCERT’s answer
    Rectangle
    NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q3 \[AB=\sqrt{(7-2)^2+(3-(-2))^2}=\sqrt{25+25}=5\sqrt2 \] \[BC=\sqrt{(11-7)^2+(-1-3)^2}=\sqrt{16+16}=4\sqrt2 \] \[CD=\sqrt{(6-11)^2+(-6-(-1))^2}=\sqrt{25+25}=5\sqrt2 \] \[DA=\sqrt{(2-6)^2+(-2-(-6))^2}=\sqrt{16+16}=4\sqrt2 \] \[AC=\sqrt{(11-2)^2+(-1-(-2))^2}=\sqrt{81+1}=\sqrt{82} \] \[BD=\sqrt{(6-7)^2+(-6-3)^2}=\sqrt{1+81}=\sqrt{82} \] Opposite sides are equal (\(\displaystyle AB=CD\), \(\displaystyle BC=DA\)), so ABCD is a parallelogram; its diagonals are equal too (\(\displaystyle AC=BD\)), so it is a rectangle — not a square, since adjacent sides differ.Answer: Rectangle.

  4. Exercise 4

    Find the value of a\displaystyle a, if the distance between the points A(−3,−14)\displaystyle \mathrm{A}(-3,-14) and B(a,−5)\displaystyle \mathrm{B}(a,-5) is 9\displaystyle 9 units.

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    NCERT’s answer
    \(\displaystyle a=-3\)
    \[AB^2=(a-(-3))^2+(-5-(-14))^2=9^2 \] \[(a+3)^2+81=81 \] \[(a+3)^2=0 \implies a=-3 \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q4Answer: \(\displaystyle a=-3\)

  5. Exercise 5

    Find a point which is equidistant from the points A(−5,4)\displaystyle \mathrm{A}(-5,4) and B(−1,6)\displaystyle \mathrm{B}(-1,6)? How many such points are there?

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    NCERT’s answer
    \(\displaystyle (-3,5)\) the middle point of AB. Infinite number of points. In fact all points which are solutions of the equation \(\displaystyle 2 x+y+1=0\).
    NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q5 \[M=\left(\frac{-5+(-1)}{2},\ \frac{4+6}{2}\right)=(-3,\ 5) \] \[MA^2=(-3-(-5))^2+(5-4)^2=5, \qquad MB^2=(-3-(-1))^2+(5-6)^2=5 \] \[PA^2=PB^2 \implies (x+5)^2+(y-4)^2=(x+1)^2+(y-6)^2 \implies 2x+y+1=0 \] Every point \(\displaystyle P(x,y)\) of this line is equidistant from A and B, so there are infinitely many such points.Answer: \(\displaystyle M(-3,5)\) is one; infinitely many points lie on the line \(\displaystyle 2x+y+1=0\), the perpendicular bisector of AB.

  6. Exercise 6

    Find the coordinates of the point Q on the x\displaystyle x-axis which lies on the perpendicular bisector of the line segment joining the points A(−5,−2)\displaystyle \mathrm{A}(-5,-2) and B(4,−2)\displaystyle \mathrm{B}(4,-2). Name the type of triangle formed by the points Q, A and B.

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    NCERT’s answer
    \(\displaystyle \left(\frac{-1}{2}, 0\right)\), isosceles triangle
    \[\text{Midpoint of } AB=\left(\frac{-5+4}{2},\ \frac{-2-2}{2}\right)=\left(-\tfrac12,\ -2\right) \] AB is horizontal; its perpendicular bisector is the vertical line \(\displaystyle x=-\tfrac12\), meeting the x-axis at: \[Q=\left(-\tfrac12,\ 0\right) \] \[QA^2=\left(\tfrac92\right)^2+2^2=\tfrac{97}{4}, \qquad QB^2=\left(\tfrac92\right)^2+2^2=\tfrac{97}{4} \] \(\displaystyle QA=QB\), so \(\displaystyle \triangle QAB\) is isosceles. NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q6Answer: \(\displaystyle Q\left(-\tfrac12,0\right)\); \(\displaystyle \triangle QAB\) is isosceles.

  7. Exercise 7

    Find the value of m\displaystyle m if the points (5,1),(−2,−3)\displaystyle (5,1),(-2,-3) and (8,2m)\displaystyle (8,2 m) are collinear.

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    NCERT’s answer
    \(\displaystyle \frac{19}{14}\)
    Collinear points enclose zero area: \[\tfrac12\left|5(-3-2m)+(-2)(2m-1)+8(1-(-3))\right|=0 \] \[-15-10m-4m+2+32=0 \] \[19-14m=0 \implies m=\tfrac{19}{14} \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q7Answer: \(\displaystyle m=\dfrac{19}{14}\)

  8. Exercise 8

    If the point A(2,−4)\displaystyle \mathrm{A}(2,-4) is equidistant from P(3,8)\displaystyle \mathrm{P}(3,8) and Q(−10,y)\displaystyle \mathrm{Q}(-10, y), find the values of y\displaystyle y. Also find distance PQ.

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    NCERT’s answer
    \(\displaystyle y=-3,-5, \mathrm{PQ}=\sqrt{290}, 13 \sqrt{2}\)
    Equidistant: \(\displaystyle AP = AQ\). \[AP^2=(3-2)^2+(8+4)^2=145 \] \[AQ^2=(-10-2)^2+(y+4)^2=144+(y+4)^2 \] \[145=144+(y+4)^2 \] \[(y+4)^2=1 \] \[y=-3\ (Q)\quad\text{or}\quad y=-5\ (Q') \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q8 \[PQ=\sqrt{(-10-3)^2+(y-8)^2} \] \[y=-3:\ PQ=\sqrt{169+121}=\sqrt{290} \] \[y=-5:\ PQ'=\sqrt{169+169}=13\sqrt{2} \] Answer: \(\displaystyle y=-3,\ PQ=\sqrt{290}\) or \(\displaystyle y=-5,\ PQ=13\sqrt2\).

  9. Exercise 9

    Find the area of the triangle whose vertices are (−8,4)\displaystyle (-8, 4), (−6,6)\displaystyle (-6, 6) and (−3,9)\displaystyle (-3, 9).

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    NCERT’s answer
    $\displaystyle 0$
    \[\text{Area}=\tfrac12\big|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big| \] \[=\tfrac12\big|(-8)(6-9)+(-6)(9-4)+(-3)(4-6)\big| \] \[=\tfrac12|24-30+6| \] \[=\tfrac12(0)=0 \] Area \(\displaystyle 0\): the points are collinear. NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q9 Answer: \(\displaystyle 0\); the points are collinear.

  10. Exercise 10

    In what ratio does the x\displaystyle x-axis divide the line segment joining the points (−4,−6)\displaystyle (-4, -6) and (−1,7)\displaystyle (-1, 7)? Find the coordinates of the point of division.

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    NCERT’s answer
    $\displaystyle 6$:$\displaystyle 7$, \(\displaystyle \left(\frac{-34}{13}, 0\right)\)
    Let the \(\displaystyle x\)-axis divide \(\displaystyle A(-4,-6)\), \(\displaystyle B(-1,7)\) at \(\displaystyle P\) in ratio \(\displaystyle m:n\). \[0=\dfrac{7m-6n}{m+n} \] \[7m=6n \quad\Rightarrow\quad m:n=6:7 \] \[x=\dfrac{6(-1)+7(-4)}{6+7}=\dfrac{-34}{13} \] NCERT_Solution_Class10_Maths_Exemplar_Ch7_Ex7-3_Q10 Answer: ratio \(\displaystyle 6:7\), point \(\displaystyle \left(-\tfrac{34}{13}, 0\right)\).