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

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 (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given as,            

Here, x₁ = -6, x₂ = -4, x₃ = 3 and y₁ = 10, y₂ = 6, y₃ = -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.

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