Question

What is the point of intersection of a line parallel to the X-axis at a distance of 4 units above it with the Y-axis?​

Answer 1

Answer:

Correct is

y=2

We know that the equation of a straight line parallel to the x-axis at a distance b from it is y=b.

Therefore, the equation of a straight line parallel to the x-axis at a distance of 2 units above the x-axis is y=2.

Answer 2

The point of intersection of the given line with the y-axis will be (0, 4).

Given,

A line || to the x-axis, at a distance of 4 units above it.

To find,

The point of intersection of the line with the y-axis.

Solution,

It can be seen that the given line is || to the x-axis.

Now, when a line is || to the x-axis, then its equation is in the form of

$y=\pm a$

Where a is the distance above or below the x-axis, and

$+a$ or just $a$ when above,

$-a$ when below the x-axis.

Here, the distance is 4 units above the x-axis. So,

$y=4$

⇒ the distance along the y-axis or y-coordinate = 4 units.

Now, on the y-axis, the abscissa is 0 for all the points.

So, the distance along the x-axis or,

x-coordinate = 0.

⇒ coordinates of the point of intersection of the line = (0, 4).

Therefore, the point of intersection of the given line with the y-axis will be (0, 4).