The point of intersection of X and Y axes is called :
The point (2, 3) is at a distance of _______________ units from xaxis :
The point (3, 2) is at a distance of _______________ units from yaxis :
The point (–3, 2) belongs to Quadrant _______________ :
The point (2, –3) belongs to quadrant _______________ :
The x value is POSITIVE which means that the point is to the right of the origin.
The y value is NEGATIVE which means that the point is below the origin.
The 4th quadrant is to the right and below the origin,(The bottom right quadrant)
All points in the 4th quadrant have the signs as (+,−)
The point (3, 2) belongs to quadrant _______________ :
The point (–2, –3) belongs to Quadrant :
The point (–2, 0) lies on :
The point (0, –2) lies on :
The point (3, 0) lies on :
The point (0, 3) lies on :
The distance between the points (–4, 7) and (1, –5) is :
Use the distance formula to determine the distance between two points.
√{1(4)}^{2} +{(5)7}^{2}
=√25+144
=√169
=13
The distance of the point (–2, –2) from the origin is :
If the distance between points (p, –5), (2, 7) is 13 units, then p is _____________ :
The correct answer is a.
Distance between two points (x1,y1) and (x2,y2) can be calculated using the formula −
Distance between the points (p,−5) and (2,7)=
(2−p)2 +(7−(−5)) 2 =13
(2p)2 + 144 = 13
(2p)2 + 144 = 169
(2p)2 = 25
2p = 5 or 2p = 5
p = 3 or p = 7
The points (a, a) (–a, a) and (– (√3) a, (√3)a) form the vertices of an :
Find the ratio in which the line joining the points (6, 4) and (1, –7) is divided by xaxis.
The points (2, –1), (3, 4), (–2, 3) and (–3, –2) taken in the order from the vertices of _________________:
In rhombus, all sides are equal and diagonals are not equal.
Distance between two points = [(x2−x1)2 + (y2−y1)2]1/2
AB = [(3−2)2 + (4+1)2]1/2 =(26)1/2
BC = [(3+2)2 + (3−4)2]1/2 = (26)1/2
CD = [(3+2)2 + (23)2]1/2 = (26)1/2
DA = [(32)2 + (2+1)2]1/2 = (26)1/2
AC = [(2+2)2 + (4)2]1/2 = 4(2)1/2
BD = [(33)2 + (4+2)2]1/2 = 6(2)1/2
AB=BC=CD=DA All sides are equal
AC and BD Diagonals are not equal
The points (0, 0), (–2, 0) and (3, 0) _________________:
The point on Yaxis equidistant from (–3, 4) and (7, 6) is _________________:
The point on Xaxis equidistant from (5, 4) and (–2, 3) is _________________:
The points on Xaxis at a distance of 10 units from (11, –8) are :
Let P(x, 0) be the point on the xaxis. Then as per the question we have
AP = 10
Hence, the points on the xaxis are (5, 0) and (17,0) .
The points on Yaxis at a distance of 13 units from point (–5, 7) are :
The coordinates of the centre of a circle passing through (1, 2), (3, –4) and (5, –6) is _________________:
Given three points (1,2),(3,−4)and(5,−6)
Let the coordinates o centre be O(h,k)
The distance from the centre to a point on the circumference is equal to radius.
∴ OA=OB=OC
From distance formula
OA^{2}=OB^{2}
The points (–5, 6), (3, 0) and (9, 8) form the vertices of :
The points (4, 4), (3, 5) and (–1, –1) from the vertices of :
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 








