Here you get the CBSE Class 10 Mathematics chapter 7, Coordinate Geometry: NCERT Exemplar Problems and Solutions (Part-IIB). This part is a continuation of Part-I and contains solutions of Question Number 7 to 12 from Exercise 7.2 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Coordinate Geometry. This exercise comprises only the Very Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

**NCERT Exemplar Solution for CBSE Class 10 Mathematics: Coordinate Geometry (Part-IIA)**

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.

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

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

**Exercise 7.2**

**Very Short Answer Type Questions (Q. No. 7-12)**

**Write whether True or False and justify your answer**

**Question. 7** A circle has its centre at the origin and a point *P *(5, 0) lies on it. The point *Q *(6, 8) lies outside the circle.

**Solution. True**

Point *Q *(6, 8) will lie outside the circle if its distance from the centre of circle is greater than the radius of that circle.

**Question. 8** The point *A *(2, 7) lies on the perpendicular bisector of the line segment joining the points *P *(6, 5) and *Q* (0, -4).

**Solution. False**

** **

**Question. 9 **The point *P*(5, -3) is one of the two points of trisection of line segment joining the points *A*(7*, *-2) and *B*(1, -5).

**Solution. ***True*

Suppose, *P *(5, -3) divides the line segment joining the points *A *(7, -2) and *B *(1, -5) in the ratio *m* : 1* *internally.

So the point *P* divides the line segment *AB* in ratio 1 : 2. Therefore, point *P* in the point of trisection of *AB*.

**Question. 10** The points *A*(*-*6,10), *B*(-4, 6) and *C*(3, -8) are collinear such that *AB =* *AC.*

**Solution. ***True*

For points *A*(*-*6,10), *B*(-4, 6) and *C*(3, -8) to be collinear, the area of triangle formed by these points must be zero.

Now, area of triangle formed by three points (*x*_{1}, *y*_{1}), (*x*_{2}, *y*_{2}) and (*x*_{3}, *y*_{3}) is given as,

Here, *x*_{1} = -6, *x*_{2} = -4, *x*_{3} = 3 and *y*_{1} = 10, *y*_{2} = 6, *y*_{3} = -8

**Question. 11** The point *P*(-2, 4) lies on a circle of radius 6 and centre (3, 5).

**Solution. ***False*

The point *P*(-2, 4) will lie on given circle if its distance from centre (3, 5) equal to the radius of this circle.

**Question. 12** The points *A *(-1, -2), *B *(4, 3), *C *(2, 5) and *D *(-3, 0) in that order form a rectangle.

**Solution. True**

Given points will be the vertices of a parallelogram, if the opposite sides and diagonals of that rectangle come out to be equal.

As, diagonals *AC *and SD are equal, therefore, the points *A* (-1, -2), *B *(4, 3), *C *(2, 5) and *D *(-3, 0) form a rectangle.

**CBSE Class 10 NCERT Textbooks & NCERT Solutions**

**NCERT Solutions for CBSE Class 10 Maths**

**NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters**