One of my favorite forms of a linear equation is called "Point-Slope form". It looks like: #y-y_1=m(x-x_1)# where your point is the ordered pair #(x_1,y_1)# and the slope is m.
1: Substitute in m = #-2/7# and (-15,10) for the point.
y - 10 = #-2/7#(x -(-15)) or more simply, y - 10 = #-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 = #-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 #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 = #1/5#x - 10.