In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 16 Co-ordinate Geometry Ex 16 A Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 16 Co-ordinate Geometry Ex 16 A Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16 A Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 16 |

Chapter Name | Coordinate Geometry |

Exercise | 16 A |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry** **Ex 16 A**

**Question 1:**

(i) The given points are A(9,3) and B(15,11).

(ii) The given points are A(7,4) and B(-5,1).

(iii) The given points are A(-6, -4) and B(9,-12).

(iv) The given points are A(1, -3) and B(4, -6).

(v) The given points are P(a + b, a – b) and Q(a – b, a + b).

(vi) The given points are P(a sin a, a cos a) and Q(a cos a, – a sina).

**Question 2:**

(i) The given point is A(5, -12) and let O(0,0) be the origin.

(ii) The given point is B(-5, 5) and let O(0,0) be the origin.

(iii) The given point is C(-4, -6) and let O(0,0) be the origin.

**Question 3:**

The given points are A(a, -1) and B(5,3).

**Question 4:**

Let R(10,y) be the point at a distance of 10 units from P(2, -3).

**Question 5:**

Let A(6, -1), B(1,3) and C(k,8) are the given points.

**Question 6:**

Let A(a, 2), B(8, -2) and C(2,-2) be the given points. Then first we find:

Therefore, a = 5

**Question 7:**

Let any point P on x – axis is (x,0) which is equidistant from A(-2, 5) and B(-2, 9).

This is not admissible.

Hence, there is no point on x – axis which is equidistant from A(-2, 5) and B(-2, 9).

**Question 8:**

Let any point P on x – axis is (0,y) which is equidistant from A(5, -2) and B(-3, 2)

Thus, the point on y – axis is (0, -2).

**Question 9:**

he point A(4,3) and B(x,5) lie on a circle. Its centre is O(2,3)

**Question 10:**

Let A(6, -1) and B(2,3) be the given point and P(x,y) be the required point, we get

**Question 11:**

Let A(11, -8) be the given point and let P(x,0) be the required point on x – axis

Then,

Hence, the required points are (17,0) and (5,0).

**Question 12:**

Let the required points be P(x,y), then

PA = PB = PC. The points A, B, C are (5,3), (5, -5) and (1, -5) respectively.

Hence, the point P is (3, -1).

**Question 13:**

Let A(1, -1), B(5, 2) and C(9, 5) are the given points. Then

Hence the given points A, B, C are collinear.

**Question 14:**

(i) Let A (6,9), B(0,1) and C(-6, -7) be the given points. Then

Hence the given A, B, C are collinear.

(ii) Let A(-1, -1), B(2,3) and C(8,11) be the given points. Then

Hence the given A, B, C are collinear.

(iii) Let P(1,1), Q(-2,7) and R(3, -3) be the given points, then

(iv) Let P(2,0), Q(11,6) and R(-4,-4) be the given points

Then,

Hence the given P, Q, R are collinear.

**Question 15:**

Let A(3,0), B(6,4) and C(-1,3) are the given points. Then

∴ ∆ ABC is an isosceles right – angled triangle.

This shows that ∆ ABC is right angled at A.

**Question 16:**

Vertices of triangle ABC are A(7, 10), B(-2, 5) and C(3, -4)

∴ ∆ ABC is a right angled triangle.

Hence ∆ ABC is an isosceles right triangle.

**Question 17:**

Let A(-5,6), B(3,0) and C(9,8) be the given points. Then

**Question 18:**

Let O(0,0), A(3,√3) and B(3,-√3) are the given points.

Hence, DABC is equilateral and each of its sides being 2√3 units.

**Question 19:**

Let A(2,1), B(5,2), C(6,4) and D(3,3) are the angular points of a parallelogram ABCD. Then

Diagonal AC ≠ Diagonal BD.

Thus ABCD is not a rectangle but it is a parallelogram because its opposite sides are equal and diagonals are not equal.

**Question 20:**

(i) Let A(0, -4), B(6,2), C(3,5) and D(-3,-1) are the vertices of quad. ABCD. Then

Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal.

Hence, quad. ABCD is a rectangle.

(ii) Let A(2, -2), B(14, 10), C(11, 13) and D(-1, 1) be the angular points of quad. ABCD, then

Thus, ABCD is a quadrilateral whose opposite sides are equal and diagonals are equal.

Hence, quad. ABCD is rectangle.

(iii) Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) are the vertices of quad. ABCD. Then

Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal

Hence, quad. ABCD is a rectangle.

**Question 21:**

(i) Let A(6,2), B(2,1), C(1,5) and D(5,6) be the angular points of quad. ABCD. Join AC and BD

Thus, ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.

Hence, quad ABCD is a square.

(ii) Let P(0, -2), Q(3,1), R(0,4) and S(-3,1) be the angular points of quad. ABCD

Thus, PQRS is a quadrilateral in which all sides are equal and the diagonals are equal.

Hence, quad. PQRS is a square.

(iii) The angular points of quadrilateral ABCD are A(3,2), B(0,5), C(-3,2) and D(0,-1)

Thus, all sides of quad. ABCD are equal and diagonals are also equal.

Quad. ABCD is a square.

**Question 22:**

Let A(-3,2), B(-5, -5), C(2, -3) and D(4,4) be the angular point of quad ABCD. Join AC and BD.

Thus, ABCD is a quadrilateral having all sides equal but diagonals are unequal.

Hence, ABCD is a rhombus.

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