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If the abscissa of a point is y and the ordinate is x then the coordinates of the point are ……..
The distance of a point (-5, 6) from the y- axis is
the distance of the point (-5, 6) from y- axis is |-5| units.
In general, the distance of the point (x,y) from x-axis is y and from y-axis is x.
If (2, 2p + 2) is the mid-point of (3p, 4) and (-2, 2q), then the value of p and q are
2 = (3p - 2)/2 and 2p + 2 = (2q + 4)/2
3p = 6 and 2p = q
p = 2
and q = 2p = 4
Point P(x,y) is a point in the third quadrant of the Cartesian plane. Then
The distance of a point (2, 3) from the x- axis is
"The distance between the origin and the coordinates (2, 3) is
Square root of (2 – 0)^2 + (3 – 0)^2
Square root of 2^2 + 3^2
Square root of 4 + 9
Square root of 25
Hence 5 will be the total distance between the coordinates and origin.
In case of x-axis we will only consider the value of y coordinate that is 3.
For a line whose equation is 2x + y = 5, which one of the following points lies on it ?
Option C is correct answer.
For the given point (2,1)
x = 2 and y = 1
Putting value in 2x + y = 5
2 (2) +1 = 5
5 = 5
So point (2,1) lies on the line 2x + y = 5
The distance of a point (2, 3) from the y – axis is.
The distance of a point from the y -axis is always the value of it's x-coordinate
That is , the distance of point p(2,3 ) from y axis is 2 units.
The perpendicular distance of a point Q(4, 7) from y-axis is
Distance of point from y-axis is x -coordinate of given point,
So,since,value of x-coordinate is 4
so, distance = 4 unit
The coordinates of a point on the y- axis of the form are …….
This is because if it will be not 0 at then the point will fall somewhre else on the graph instead of y- axis.So the co-ordinatemust be (0,Y) to fall on y- axis.
If x = y then (x,y) = (y,x).