7.Co ordinate geometry
The abscissa and ordinate of a given point are the distances of the point from y-axis and x-axis
respectively.
The coordinates of any point on x-axis are of the form (x, 0).
The coordinates of any point on y-axis are of the form (0,y)
Distance of a point P(x , y) from the orgin O(0,0) is
For Example: Find the coordinates of the point which is divided the line segment joining (-1,3)
and (4,-7) internally in the ratio 3 : 4.
Solution: Let the end points of AB be A (-1, 3) and B (4, -7).
(x1 = −1, y1 = 3) and (x2 = 4, y2 = −7)
Also, m = 3 and n = 4
Let P(x , y) be the required point, then by section formula,
Question 1.
Find the distance between the following pair of points,
(i) (2, 3) and (4, 1)
Answer:
Question 2.
Find the distance between the points (0, 0) and (36, 15).
Answer:
Given: Origin O (0, 0) and a point P (36, 15).
Distance between any point and origin =
∴ Distance =
=
=
= 39 units
Question 3.
Verify that the points (1, 5), (2, 3) and (-2, -1) are collinear or not.
Let us assume
A = (1, 5) B = (2, 3) and C = (-2, -11)
Now let us find the distance between points.
Let us consider AB, BC and CA
AB= √[(2 – 1)2 + (3 – 5)2]
= √[(1)2 + (-2)2]
=√(1 + 4)
=√5
BC= √[(-2 – 2)2 + (-11 – 3)2]
= √[(-4)2 + (-14)2]
=√(16 + 196)
=√212
CA= √[(-2 – 1)2 + (-11 – 5)2]
= √[(-3)2 + (-16)2]
=√(9+ 256)
=√265
Since AB + BC ≠ CA
Hence, the points (1, 5), (2, 3), and ( – 2, – 11) are not collinear.
Question 4.
Check whether (5, -2), (6, 4) and (7,-2) are the vertices of an isosceles triangle.
Answer:
Let A = (5, – 2); B = (6, 4) and C = (7, – 2).
Let the points (5, – 2), (6, 4), and (7, – 2) are representing the vertices A, B, and C respectively.
This implies, whether given points are vertices of an isosceles triangle
Question 5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Solution:
From figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6,1).
Find distance between points using distance formula, we get
All sides are of equal length. Therefore, ABCD is a square and hence, Champa was correct.
Question 6.
Show that the following points form an equilateral triangle A(a, 0), B(- a, 0), C(0, a√3).
Answer:
Given: A (a, 0), B (- a, 0), C (0, a√3).
Question 7
Show that the points (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order are the vertices of a rhombus. And find its area.
(Hint: Area of rhombus =
Answer:
Given in ▱ ABCD , A(-4, – 7), B (- 1, 2), C (8, 5) and D (5,-4)
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Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
i) (-1,-2), (1,0), (-1,2), (-3,0)
Answer:
Let A (- 1, -2), B (1, 0), C (- 1, 2), D (- 3, 0) be the given points. Distance formula
