This article brings the Part-IB of NCERT Exemplar Problems and Solutions for class 10 Mathematics chapter 7, Coordinate Geometry. This part is in continuation with NCERT Exemplar Class 10 Msths chapter 7, Part-IA. In this part answers to Q. No. 9-18 are given with a detailed explanation.
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Coordinate Geometry:
Exercise 7.1
Multiple Choice Questions (Q. No. 10-18)
Question. 10 The point which lies on the perpendicular bisector of the line segment joining the points A (-2, -5) and B (2, 5) is
(a) (0, 0)
(b) (0, 2)
(c) (2, 0)
(d) (-2, 0)
Answer. (a)
Explanation:
The perpendicular bisector of any line segment always passes through the mid-point of that line segment.
Hence, (0, 0) is the required point which lies on the perpendicular bisector of the line segment joining the points A (-2, -5) and B (2, 5).
Question. 11 The fourth vertex D of a parallelogram ABCD whose three vertices are A (-2, 3), B (6, 7) and C (8, 3) is
(a) (0, 1)
(b) (0, -1)
(c) (-1, 0)
(d) (1, 0)
Answer. (b)
Explanation:
Suppose the co-ordinate of fourth vertex or point D of parallelogram be (x4, y4) and P, Q be the middle points of AC and BD, respectively.
Question. 12 If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then
Answer. (d)
Explanation:
A rough diagram of the point P (2, 1) on the line segment joining the points A (4, 2) and B (8, 4) is shown in the figure given below:
Question. 13 If P (a/3, 4) is the mid-point of the line segment joining the points Q (-6, 5) and R (-2, 3), then the value of a is
(a) - 4
(b) -12
(c) 12
(d) -6
Answer. (b)
Explanation:
Given that, P (a/3, 4) is the mid-point of the line segment joining the points Q(-6, 5) and R (-2, 3).
Question. 14 The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the Y-axis at
(a) (0, 13)
(b) (0, -13)
(c) (0, 12)
(d) (13, 0)
Answer. (a)
Explanation:
It is given that the perpendicular bisector of the line segment is perpendicular on the line segment.
If two lines having slope m1 and m2 are perpendicular to each other, then,
Putting x = 0 in above equation we get:
3 × 0 + y = 13
⟹ y = 13.
Hence, the required point is (0, 13).
Question. 15 The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is
Answer. (a)
Explanation:
Form the given figure, coordinates of three vertices of a triangle are: O (0, 0), A (0, 2y), B (2x, 0).
Suppose the required point be P whose coordinates are (h, k)
Now, P is equidistant from the three vertices of DAOB, therefore, PO = PA = PB or (PO)2 = (PA)2 = (PB)2
By using distance formula, we get:
Question. 16 If a circle drawn with origin as the centre passes through (13/2, 0), then the point which does not lie in the interior of the circle is
Question. 17 A line intersects the Y-axis and X-axis at the points P and Q, respectively. If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
(a) (0, -5) and (2, 0)
(b) (0, 10) and (-4, 0)
(c) (0, 4) and (-10, 0)
(d) (0, -10) and (4, 0)
Answer. (d)
Explanation:
Suppose the coordinates of P are (0, y) and Q are (x, 0).
The given situation can be represented by the following diagram:
Question. 18 The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is
(a) (a + b + c)2
(b) 0
(c) (a + b + c)
(d) abc
Answer. (b)
Explanation:
