Explanation
We know, for the point (x,y),
in I quadrant both x and y are positive, in II quadrant x is negative and y is positive, in III quadrant both x and y are negative and in IV quadrant x is positive and y is negative.
Given, the point (8,7).
That is, x is 8, which is positive and y is 7, which is also positive.
Hence, the given point (8,7) lies in the Ist quadrant.
Therefore, the statement is true.
Hence, option A is correct.
Step - 1: Plot and identify points in the Cartesian plane.
The given equation is x=y.
So, if x=1 and y=1.
If x=−5 and y=−5.
If x=0 and y=0.
Step - 2: By Induction.
x=a and y=a.
So, any point P on the line x=y will be of the form P(a,a).
Hence the correct option (A) (a, a).
We know, in the xy-plane, any point on the y-axis will have x=0. Hence, any point on the y-axis is of the form (0,y), where y can be any non-zero number.
But here, given the point (a,b) and b=0.
Then, the point will be (a,0).
Clearly, the point lies on the x-axis.
Hence, the given statement is false.
Therefore, option B is correct
Given, P(0,−4).
Here, the x-coordinate is 0 and y-coordinate is −4.
We know, a point on the x-axis has the form (a,0), where a is any number.
But the given point is in the form (0,a) and therefore lies on the y-axis.
Therefore, the point P(0,4) lies on the y-axis and so, the given statement is false.
Hence, option B is correct.
Let the required point be P.
Given, the point lies on x-axis.
Then, its ordinate is 0 or y=0.
Also, given the point is at a distance of 7 units to the right of origin.
So, its abscissa will be +7. Hence, the co-ordinates of the required point are (7,0).
Given, the point lies on y-axis.
Then, its abscissa is 0 or x=0.
Also, given the point is at a distance of 6 units below the x-axis,
So, its ordinate will be −6. Hence, the co-ordinates of the required point are (0,−6).
Also, given the point is at a distance of 3 units to the left of the origin.
So, its abscissa will be −3. Hence, the co-ordinates of the required point are (−3,0).
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