We are given the following information:

• Point $B(1, 2)$
• The x-coordinate of point $A$ is $-11$
• The distance between points $A$ and $B$ is 15 units

We need to find the possible coordinates of point $A$, given the distance between the two points. To do so, we can use the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Where:

• $(x_1, y_1)$ are the coordinates of point $A$
• $(x_2, y_2)$ are the coordinates of point $B$
• $d$ is the distance between the points

Let’s plug in the known values:

• $A(x_1, y_1) = (-11, y_1)$
• $B(1, 2)$
• Distance $d = 15$

So the equation becomes:

$15 = \sqrt{(1 - (-11))^2 + (2 - y_1)^2}$

Simplify the equation:

$15 = \sqrt{(1 + 11)^2 + (2 - y_1)^2}$ $15 = \sqrt{12^2 + (2 - y_1)^2}$ $15 = \sqrt{144 + (2 - y_1)^2}$

Now square both sides to eliminate the square root:

$15^2 = 144 + (2 - y_1)^2$ $225 = 144 + (2 - y_1)^2$ $225 - 144 = (2 - y_1)^2$ $81 = (2 - y_1)^2$

Now, solve for $y_1$ by taking the square root of both sides:

$2 - y_1 = \pm 9$

This gives us two possible equations:

1. $2 - y_1 = 9$ → $y_1 = -7$
2. $2 - y_1 = -9$ → $y_1 = 11$

So, the possible coordinates of point $A$ are:

$A(-11, -7) \quad \text{or} \quad A(-11, 11)$

Summary of the possible coordinates of point A:

• $A(-11, -7)$
• $A(-11, 11)$

Would you like further explanation on how this works? Or do you have any other questions?

Here are some related questions you might find helpful:

1. How do we use the distance formula to find the distance between two points?
2. What if the x-coordinate of point A was positive? How would that affect the solution?
3. How can we apply this method to find the coordinates of a point in 3D space?
4. What happens if the distance between the points is zero? What would the coordinates be in that case?
5. How do you solve quadratic equations that arise from the distance formula?

Tip: Always make sure to simplify and square both sides when dealing with square roots in distance problems!