Find a point on the y-axis that is at the same distance from the points $(-5, 2)$ and $(9, -2)$.

Step 1: Define the point on the y-axis

A point on the $y$ -axis always has its $x$ -coordinate equal to 0. Therefore, we can represent the required point as P(0, y).

Step 2: Set up the distance condition

The problem states that the point $P$ is equidistant from points $A(-5, 2)$ and $B(9, -2)$. This means the distance from $P$ to $A$ is equal to the distance from $P$ to $B$.

Step 3: Apply the distance formula

To simplify calculations, we square both sides of the distance equality, $PA = PB$, to remove the square roots. We then apply the distance formula, which states that the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step 4: Substitute the coordinates and simplify

Substitute the coordinates of P(0, y), $A(-5, 2)$, and $B(9, -2)$ into the squared distance formula. Then, simplify the squared terms and expand the binomials using the formula $(a \pm b)^2 = a^2 \pm 2ab + b^2$.

Step 5: Solve for y

First, combine the constant terms on both sides. Then, subtract $y^2$ from both sides. Collect the $y$ terms on one side and the constant terms on the other side. Finally, divide to find the value of $y$.

Step 6: State the coordinates of the point

Since we found $y = -7$ and the point is on the $y$ -axis, its coordinates are $(0, -7)$.

More from this chapter

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