# NCERT Exemplar Solution for CBSE Class 10 Mathematics: Coordinate Geometry (Part-IIIA)

NCERT Exemplar Problems and Solutions (Part-IIIA) for Class 10 Mathematics chapter 7, Coordinate Geometry, is available here. In this part you will get simple and apt solutions to Q. No. 1-7 from exercise 7.3 of class 10 Maths Exampler for chapter Coordinate Geometry, are given.

Here you get the CBSE Class 10 Mathematics chapter 7, Coordinate Geometry: NCERT Exemplar Problems and Solutions (Part-IIIA). This part of the chapter includes solutions of Question Number 1 to 7 from Exercise 7.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Coordinate Geometry. This exercise comprises only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

**CBSE Class 10 Mathematics Syllabus 2017-2018**

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

**Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Coordinate Geometry:**

**Exercise 7.3**

**Short Answer Type Questions (Q. No. 1-7)**

**Question. 1** Name the type of triangle formed by the points *A *(-5, 6), *B *(-4, -2) and *C *(7, 5).

**Solution.**

**Question. 3 **What type of quadrilateral do the points *A *(2, -2), *B *(7, 3), *C *(11, -1) and *D *(6, -6)

taken in that order form?

**Solution.**

To find the type of quadrilateral, we will first find the length of all four sides and diagonals of

the quadrilateral formed by the points *A *(2, -2), *B *(7, 3), *C *(11, -1) and *D *(6, -6).

**Question. 4** Find the value of *a*, if the distance between the points *A *(-3, -14) and *B (a, *-5) is 9 units.

**Solution.**

Given, distance between *A *(-3, -14) and *B *(*a*, -5) is, *AB* = 9 units

By distance formula,

**Question. 5** Find a point which is equidistant from the points *A *(- 5, 4) and *B *(- 1, 6). How

many such points are there?

**Solution.**

Let, *M* (*h*, *k) *be the point which is equidistant from the points *A *(-5, 4) and *B* (-1, 6).

Point (-3, 5) satisfies equation (i).

So, all points which are solution of the equation 2*h* + *k *+ 1 = 0, are equidistant from the points *A *and *B*.

**Question. 6** Find the coordinates of the point *Q *on the *X*-axis which lies on the perpendicular bisector of the line segment joining the points *A *(- 5, -2) and *B *(4, -2). Name the type of triangle formed by the point *Q*,* A *and *B*.

**Solution.**

Situation given in question is shown in the figure given below,

As, the perpendicular bisector of the line segment *AB *bisects the segment *AB*, hence it will pass through the mid-point of AB i.e., R

Clearly *BQ = QA *≠ *AB, *which means that the triangle formed by the points *Q*, *A *and *B* is an isosceles.

**Question. 7** Find the value of *m, *if the points (5, 1), (- 2, - 3) and (8, 2*m) *are collinear.

**Solution. **

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