What part of the coordinate plane is equidistant from the points a(-3, 2) and b(3, 2).

Solution:

Given, the points are a(-3, 2) and b(3, 2).

We have to find the part of the coordinate plane that is equidistant from the given points.

Using distance formula,

Distance = \(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

Distance a² = \((x-(-3))^{2}+(y-2)^{2}=(x+3)^{2}+(y-2)^{2}\)

Distance b² = \((x-3)^{2}+(y-2)^{2}\)

Point P is equidistant from points a and b.

I.e. distance a = distance b

\((x-(-3))^{2}+(y-2)^{2}=(x+3)^{2}+(y-2)^{2}\) = \((x-3)^{2}+(y-2)^{2}\)

(x - 3)² = (x + 3)²

x² - 6x + 9 = x² + 6x + 9

By grouping,

x² + x² - 6x - 6x = 9 - 9

12x = 0

x = 0

This is the equation of a vertical line through the point x = 0.

In other words, it is the y-axis.

Therefore, the part of the plane equidistant from points a and b are all points along the y-axis.

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What part of the coordinate plane is equidistant from the points a(-3, 2) and b(3, 2).

Summary:

The part of the coordinate plane that is equidistant from the points a(-3 2) and b(3, 2) are all points along the y-axis.