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

In this article you will get the Class 10 Mathematics chapter 7, Coordinate Geometry: NCERT Exemplar Problems and Solutions (Part-IIA). In this part answers to Q. No. 1-6 from exercise 7.2 of class 10 Maths Exampler for chapter Coordinate Geometry, are given. Practice thsese questions to prepare for CBSE Class 10 Maths Board Exam 2018.

Created On: Jul 25, 2017 11:44 IST
Modified On: Jul 25, 2017 12:13 IST

Coordinate Geometry Short Answer Questions, Coordinate Geometry Class 10 NCERT ExemplarHere you get the CBSE Class 10 Mathematics chapter 7, Coordinate Geometry: NCERT Exemplar Problems and Solutions (Part-IIA). This part of the chapter includes solutions of Question Number 1 to 6 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.

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 7.2

Very Short Answer Type Questions (Q. No. 1-6)

Write whether True or False and justify your answer

Question. 1 ΔABC with vertices A (– 2, 0), B (2, 0) and C (0, 2) is similar to ΔDEF with vertices D (-4, 0), E (4, 0) and F (0, 4).

Solution. True

Explanation:

Here, we see that sides of ΔABC and ΔFDE are proportional

Therefore, by SSS similarity rule both the triangles are similar.        

Question. 2 The point P (-4, 2) lies on the line segment joining the points A (-4, 6) and B (-4, -6).

Solution. True

Explanation:

If we plot all the points P (-4, 2), A (-4, 6) and B (-4, -6) on the graph paper then, we will find, point P (-4, 2) lies on the line segment joining the points A (-4, 6) and B (-4, -6),

 

Question. 3 The points (0, 5), (0, -9) and (3, 6) are collinear.

Solution. False

Given, x1 = 0, x2 = 0 , x3 = 3 and  y1 = 5, y2 = −9, y3 = 6

Three points will be collinear if the area of triangle formed by these three points is zero.

Now, area of triangle formed by three points (x1, y1), (x2, y2) and (x3, y3) is given as,            

As the area of triangle formed by the points (0, 5), (0, ‒ 9) and (3, 6) is non-zero, hence the points are non-collinear.

Question. 4 Point P (0, 2) is the point of intersection of Y-axis and perpendicular bisector of line segment joining the points A (‒1, 1) and B (3, 3).

Solution. False

For P(0, 2) to be the point of intersection of y–axis and perpendicular bisector of the line joining the points A(–1, 1) and B(3, 3), P must be equidistant from A and B.

i.e., PA = PB

Question. 5 The points A (3, 1), B (12, -2) and C (0, 2) cannot be vertices of a triangle.

Solution. True

For any three points to be the vertices of a triangle, the area of that triangke must be non-zero.

Now, area of triangle formed by three points (x1, y1), (x2, y2) and (x3, y3) is given as,            

     

Since, area of ΔABC = 0, therefore, the points, A (3, 1), B (12, ‒2) and C (0, 2) are collinear and can’t be the vertices of a triangle.


Question. 6 The points A (4, 3), B (6, 4), C (5, - 6) and D (-3, 5) are vertices of a parallelogram.

Solution. False

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

By using distance formula, distance between A (4, 3) and B (6, 4), 

In parallelogram, opposite sides are equal. But, in this case all sides AB, BC, CD and DA are different.

Therefore, given coordinates are not the vertices of a parallelogram.

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