The two vertices that form the base of an isosceles triangle are (-5, 3) and (2, 3). What are the coordinates of the other vertex?

To find the coordinates of the third vertex of the isosceles triangle, we need to use symmetry. Since the triangle is isosceles, the two sides that are equal in length are the sides formed by the base, which is the line segment joining the two given vertices.

First, let's find the midpoint of the base. The midpoint formula is given by:

Midpoint = ((x1 + x2) / 2, (y1 + y2) / 2)

So, the midpoint of the base is:

Midpoint = ((-5 + 2) / 2, (3 + 3) / 2)
= (-3/2, 3)

Now, since the triangle is isosceles, the third vertex will form a line segment that is perpendicular to the base and passes through the midpoint.

To find the slope of the line segment joining the two given vertices (base), we use the slope formula:

Slope = (y2 - y1) / (x2 - x1)

where (x1, y1) and (x2, y2) are the coordinates of the two given vertices of the base.

So, the slope of the base is:

Slope = (3 - 3) / (2 - (-5))
= 0 / 7
= 0

Now, since the line segment joining the third vertex and the midpoint is perpendicular to the base, the slope of this line segment will be the negative reciprocal of the slope of the base.

So, the negative reciprocal of 0 is undefined.

Therefore, the line segment joining the third vertex and the midpoint is a vertical line. Since the midpoint is (-3/2, 3), the x-coordinate of the third vertex will be the same as the x-coordinate of the midpoint, which is -3/2.

Therefore, the coordinates of the third vertex are (-3/2, y), where y can be any value.

In summary, the coordinates of the other vertex of the isosceles triangle are (-3/2, y), where y can be any real number.

notice the points given form a horizontal line, so the length of the side must be 7

the x coordinate of the third point must be midway between the two points (simple geometry, look at your sketch)
let the third point be P
P is (-3/2, y)
so the distance from P to (2,3) must be 7
√(2 + 3/2)^2 + (y-3)^2 = 7
√( 49/4 + (y-3)^2) = 7
square both sides
49/4 + (y-3)^2 = 49
(y-3)^2 = 147/4
y-3 = ± √147/2
y = 3 ± √147/2 or (6 ± √147)/2

There is a point above the base and it is (-3/2 , (6+√147)/2 )
and one below which is (-3/2 , 6 -√147)/2 )