NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry
Exercise 7.3

Coordinate geometry short problems with worked answers.

Question. 1

Name the type of triangle formed by the points \(A(-5,6),\ B(-4,-2),\ C(7,5)\).

Answer:

Scalene.

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Question. 2

Find the point(s) on the x–axis at a distance \(2\sqrt5\) from \((7,-4)\). How many such points are there?

Answer:

(5, 0) and (9, 0) — two points.

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Question. 3

What type of quadrilateral is formed by the points \(A(2,-2),\ B(7,3),\ C(11,-1),\ D(6,-6)\) in order?

Answer:

Rectangle.

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Question. 4

Find \(a\) if the distance between \(A(-3,-14)\) and \(B(a,-5)\) is \(9\).

Answer:

\(a=-3\)

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Question. 5

Find a point equidistant from \(A(-5,4)\) and \(B(-1,6)\). How many such points are there?

Answer:

One example is (-3, 5). There are infinitely many such points (on the perpendicular bisector).

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Question. 6

Find the coordinates of the point \(Q\) on the x–axis which lies on the perpendicular bisector of the segment joining \(A(-5,-2)\) and \(B(4,-2)\). Also, name the type of triangle formed by \(Q, A, B\).

Answer:

\(Q = \left(-\dfrac{1}{2}, 0\right)\)

Triangle \(QAB\) is isosceles (\(QA=QB\)).

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Question. 7

Find \(m\) if the points \((5,1),\ (-2,-3),\ (8,2m)\) are collinear.

Answer:

\(m=\dfrac{19}{14}\)

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Question. 8

If \(A(2,-4)\) is equidistant from \(P(3,8)\) and \(Q(-10,y)\), find \(y\). Also find \(PQ\).

Answer:

\(y=-3\) or \(y=-5\)

\(PQ=\sqrt{290}\) when \(y=-3\); and \(PQ=13\sqrt{2}\) when \(y=-5\).

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Question. 9

Find the area of the triangle with vertices \((-8,4),\ (-6,6),\ (-3,9)\).

Answer:

0

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Question. 10

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

Answer:

Ratio \(6:7\) (internally), point \(\big(-\dfrac{34}{13},\,0\big)\).

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Question. 11

Find the ratio in which the point \(P\big(\dfrac{3}{4},\dfrac{5}{12}\big)\) divides the segment joining \(A\big(\dfrac{1}{2},\dfrac{3}{2}\big)\) and \(B(2,-5)\).

Answer:

\(1:5\) (\(AP:PB\)).

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Question. 12

If \(P(9a-2,-b)\) divides the segment joining \(A(3a+1,-3)\) and \(B(8a,5)\) in the ratio \(3:1\), find \(a\) and \(b\).

Answer:

\(a=1,\ b=-3\)

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Question. 13

If \((a,b)\) is the midpoint of the segment joining \(A(10,-6)\) and \(B(k,4)\) and \(a-2b=18\), find \(k\) and \(|AB|\).

Answer:

\(k=22\), \(|AB|=2\sqrt{61}\)

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Question. 14

The centre of a circle is \((2a,\,a-7)\). If it passes through \((11,-9)\) and has diameter \(10\sqrt2\), find \(a\).

Answer:

\(a=3\) or \(a=5\)

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Question. 15

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

Answer:

\(k=19\)

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Question. 16

If \(D\big(-\dfrac{1}{2},\dfrac{5}{2}\big),\ E(7,3),\ F\big(\dfrac{7}{2},\dfrac{7}{2}\big)\) are the midpoints of the sides of \(\triangle ABC\), find \(\text{area}(\triangle ABC)\).

Answer:

\(11\) square units.

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Question. 17

The points \(A(2,9),\ B(a,5),\ C(5,5)\) are vertices of a triangle right–angled at \(B\). Find \(a\) and the area of \(\triangle ABC\).

Answer:

\(a=2\), area \(=6\).

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Question. 18

Find the coordinates of the point \(R\) on the segment joining \(P(-1,3)\) and \(Q(2,5)\) such that \(PR=\dfrac{3}{5}PQ\).

Answer:

\(R\big(\dfrac{4}{5},\dfrac{21}{5}\big)\)

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Question. 19

Find \(k\) if the points \(A(k+1,2k),\ B(3k,2k+3),\ C(5k-1,5k)\) are collinear.

Answer:

\(k=2\)

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Question. 20

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

Answer:

Ratio \(8:1\) (internally, from \(B:C\)). Point \(\big(\dfrac{8}{3},\,-\dfrac{1}{9}\big)\).

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NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 7: Coordinate Geometry – Exercise 7.3 | Detailed Answers