# Question #efd88

One of my favorite forms of a linear equation is called "Point-Slope form". It looks like: $y - {y}_{1} = m \left(x - {x}_{1}\right)$ where your point is the ordered pair $\left({x}_{1} , {y}_{1}\right)$ and the slope is m.
1: Substitute in m = $- \frac{2}{7}$ and (-15,10) for the point.
y - 10 = $- \frac{2}{7}$(x -(-15)) or more simply, y - 10 = $- \frac{2}{7}$(x+15).
If the instructor wishes for the equation to be in a different form, then use algebraic techniques to manipulate it. For example, add 10 to both sides to obtain: y = $- \frac{2}{7}$(x+15) + 10. Continue by distributing the slope and simplifying if necessary.
2: Perpendicular lines have slopes whose product is -1. The given slope is -5, so the "opposite reciprocal" would be $\frac{1}{5}$. That is the desired slope. A point at the y-axis of -10 is considered to be the y-intercept, or (0,-10). In this case, one could write the equation in slope-intercept form, y = mx + b. You would get: y = $\frac{1}{5}$x - 10.