To find the equation of a line with a given slope and a point through which it passes, we can use the point-slope form of a line equation:

$y - y_1 = m(x - x_1)$

Where:

• $m$ is the slope of the line,
• $(x_1, y_1)$ is a point on the line.

Given:

• Slope $m = 10$,
• Point $(x_1, y_1) = (5, -6)$.

Substitute these values into the point-slope formula:

$y - (-6) = 10(x - 5)$

Simplifying:

$y + 6 = 10(x - 5)$

Now, distribute the 10:

$y + 6 = 10x - 50$

Finally, solve for $y$ (subtract 6 from both sides):

$y = 10x - 50 - 6$

$y = 10x - 56$

Thus, the equation of the line is:

$y = 10x - 56$

Let me know if you'd like more details or have any questions!

Here are 5 related questions to expand on this concept:

1. How would the equation change if the slope were negative?
2. Can you find the equation of a line parallel to this one but passing through a different point?
3. How do you convert the equation of a line from point-slope form to slope-intercept form?
4. What is the equation of a line perpendicular to this one that passes through $(5, -6)$?
5. How can you find the x-intercept and y-intercept from the equation $y = 10x - 56$?

Tip: Always double-check if the point provided satisfies the final equation to verify the accuracy of your solution!