NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 7 **Coordinate Geometry Class 10 NCERT Solutions Ex 7.3.**

- Coordinate Geometry Class 10 Ex 7.1
- Coordinate Geometry Class 10 Ex 7.2
- Coordinate Geometry Class 10 Ex 7.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Coordinate Geometry |

Exercise |
Ex 7.3 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3

NCERT Solutions for Class 10 Maths

Page No: 170

**Question 1.** Find the area of the triangle whose vertices are:

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

**Solution:
**

**Question 2.** In each of the following find the value of ‘*k*‘, for which the points are collinear.

(i) (7, -2), (5, 1), (3, –*k*)

(ii) (8, 1), (*k*, -4), (2, -5)

**Solution:
**

Concept Insight: Only three non collinear points can give a triangle.

**Question 3.** Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

**Solution:
**

**Let vertices of the triangle be A (0, –1), B (2, 1), C (0, 3)**

Let D, E, F are midpoints of the sides of this triangle. Coordinates of D, E, and F are given by –

**Question 4.** Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).

**Solution:
**

Let vertices of the quadrilateral be A (— 4, —2), 5 (—3, —5), C (3, —2) and D (2, 3). Join AC to form two triangles ∆ABC and ∆ACD

Concept Insight: Join either point A & C or B & D (not both) Compute Area of triangle separately & then add.

**Question 5. **A median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).

**Solution:
**

Let vertices of the triangle be A (4, — 6), B (3, — 2), C (5, 2)

Let D be the midpoint of side BC of ∆ABC. So AD is the median in ∆ABC,

But area can not be negative. So area of ∆ADC is 3 square units. Clearly median AD has divided ∆ABC in two triangles of equal areas.

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