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Q.
We divided the plane of the paper into four equal parts. by drawing two mutually perpendicular lines, X'OX and YOY'. These lines are called the axes. Here X'OX is called x-axis and YOY' is called y-axis. There axes divide the plane of the paper into four parts, called quadrants.
The position of a point in a plane is denoted by an ordered pair (a,b), where a is called the x co-ordinate and y is called y co-ordinate.
In which quadrant does the point(-4, -7) lie?
The point (-4, -7) lies in 3rd quadrant.
The point (1, 5) lies in 1st quadrant.
The point (9, -2) lies in 4th quadrant.
The point (-7, 6) lies in 2nd quadrant.
The point (0, 9) lies in y-axis.
The point (9, 0) lies in x-axis.
Find the coordinates of the point equidistant from the points A(1, 2), B(3, –4) and C(5, –6).
Given three points A(1,2) B(3,-4) and C(5,-6).
To find the perpendicular bisectors of AB:-
To find the perpendicular bisectors of AC:-
Now, solve the above two equations:-
⇒ 1 / 3(x - 2) - 1 = 1 / 2(x - 3) - 2
⇒ 2(x - 2) - 6 = 3(x - 3) - 12
⇒ x = 11
⇒ y = 1 / 2(x - 3) - 2 = 1 / 2(11 - 3) - 2 = 2
The coordinates of the points equidistant from the point A(1, 2), B(3, -4) and C(5, -6) are (11, 2)
Find the distance of the point A(3, -3) from the origin.
OA = √32+(-3)2 = √9+9 = √18 = 3√2
P is a point on x-axis at a distance of 4 units from y-axis to its right. The co-ordinates of P are:
The co-ordinates of P are A(4, 0)
A is a point on y-axis at a distance of 5 units from x-axis lying below x-axis. The co-ordinates of A are:
The co-ordinates of A are A(0, -5)
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