NCERT Solutions 10th Maths Chapter 7 Coordinate Geometry Exercise 7.3

NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 we will restart our exploration of the world of Coordinate Geometry. Thus, we are providing you Chapter 7 Coordinate Geometry NCERT Class 10 Maths Solutions that will help in achieving more marks. You don't have to wander and waste your precious time in finding best CBSE NCERT Solutions.

Exercise 7.3
1. Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
Solution:


2. In each of the following find the value of ‘k’ for which the points are collinear.
(i) (7, -2), (5, 1), (3, k)
(ii) (8, 1), (k, -4), (2, -5)
Solution:
(i) Since the given points are collinear, the area of the triangle formed by them must be 0, i.e.,

(ii) Since the given points are collinear, the area of the triangle formed by them must be 0, i.e.,

Answer
DOB is a straight line.
Therefore, ?DOC + ? COB = 180°
? ?DOC = 180° - 125°
= 55°
In ?DOC,
?DCO + ? CDO + ? DOC = 180°
(Sum of the measures of the angles of a triangle is 180º.)
? ?DCO + 70º + 55º = 180°
? ?DCO = 55°
It is given that ?ODC ~ ?OBA.
? ?OAB = ?OCD [Corresponding angles are equal in similar triangles.]
? ? OAB = 55°
? ?OAB = ?OCD [Corresponding angles are equal in similar triangles.]
? ?OAB = 55°
3.E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ?ABE ~ ?CFB.
Answer

In ?ABE and ?CFB,
?A = ?C (Opposite angles of a parallelogram)
?AEB = ?CBF (Alternate interior angles as AE || BC)
? ?ABE ~ ?CFB (By AA similarity criterion)