To answer the question, first figure out the equaton of the line. To do that first find the slope. The formula for the slope of a line is

\(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\) where m = slope, \({y}_{2}\) = y-coordinate in the second point, \({y}_{1}\) = y-coordinate in the first point, \({x}_{2}\) = x-coordinate in the second point, and \({x}_{1}\) = x-coordintate in the first point.

\(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\)

m = ?

\({y}_{2}\) = 10

\({y}_{1}\) = -2

\({x}_{2}\) = 5

\({x}_{1}\) = 1

\(m=\frac{10-(-2)}{5-1}\)

\(m=\frac{10+2}{5-1}\)

\(m=\frac{12}{5-1}\)

\(m=\frac{12}{4}\)

\(m=\frac{3}{1}\)

\(m=3\)

Now that we know that the slope of the line is 3, put that in the equation for a line. The equation for a line is

\(y=mx+b\) where \(y\) = y-coordinate, \(m\) = slope, \(x\) = x-coordinate, and \(b\) = y intercept (where line crosses the y axis). Take one of the points and susitute \(x\) and \(y\) in the equation so we can solve for b.

\(y=mx+b\)

\(y\) = 10

\(m\) = 3

\(x\) = 5

\(b\) = ?

\(10=3\times5+b\)

\(10=15+b\)

\(10-15=15+b-15\)

\(-5=15+b-15\)

\(-5=b-0\)

\(-5=b\)

\(b=-5\)

Now fill in what you know leaving \(y\) and \(x\) as \(y\) and \(x\).

\(y=3x-5\)

To figure out at which point the line crosses the y-axis, subsitute \(x\) for 0 and solve for \(x\).

\(y=3x-5\)

\(y=3\times0-5\)

\(y=0-5\)

\(y=-5\)

The point that the line crosses the y-axis is at point \((0,-5)\) which means that the answer is neither a, b, c, or d.