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.
