The abscissa of a point is -6 and its distance from the point (1,3) is square root 74. Find the ordinate of the point.

To find the ordinate of the point, we can use the distance formula between two points in a plane:

Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Given that the abscissa of the point is -6 and the distance from the point (1,3) is sqrt(74), we can plug in the values into the distance formula:

sqrt(74) = sqrt((-6 - 1)^2 + (y - 3)^2)
sqrt(74) = sqrt(49 + (y - 3)^2)
74 = 49 + (y - 3)^2
25 = (y - 3)^2

Taking the square root of both sides, we get:

5 = y - 3

Adding 3 to both sides, we find:

y = 8

Therefore, the ordinate of the point is 8.

Let the ordinate of the point be y.

Using the distance formula, we have:

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

where $$
(x_1, y_1) = (-6, y)\\
(x_2, y_2) = (1,3)
$$

Substituting these values into the formula, we get:

$$
\sqrt{74} = \sqrt{(-6 - 1)^2 + (y - 3)^2}\\
\sqrt{74} = \sqrt{(-7)^2 + (y - 3)^2}\\
\sqrt{74} = \sqrt{49 + (y - 3)^2}\\
74 = 49 + (y - 3)^2\\
(y - 3)^2 = 74 - 49\\
(y - 3)^2 = 25\\
y - 3 = \pm 5\\
y = 3 \pm 5\\
y = 3 + 5 \quad \text{or} \quad y = 3 - 5\\
y = 8 \quad \text{or} \quad y = -2
$$

Therefore, the ordinate of the point is either 8 or -2.