If the relationship is parallel, you find the gradient/slope (m) in the equation y=mx+c. Then substitute the point and the gradient/slope into the equation y-y1=m(x-x1), after that you work it out algebraically making y the subject.
If the relationship is perpendicular then you find the gradient/slope (m) from the equation y=mx+c, after doing that you sub m into the equation m1*m2= -1 (solve algebraically to make m2 the subject). Then you substitute the answer of m2 and the points into the equation y-y1=m2(x-x1).
1) Use point-slope formula to find the slope-intercept form (or y = mx + b form) of the equation. m represents the slope, and and represent the x and y values of a point the line intersects. So, substitute for m, -4 for , and 7 for . Then, simplify and isolate y on the left side like so: