The distance between and arbitrary point #(x,y)=(x,4x+7)# on this line and the point #(0,-3)# is #\sqrt{(x-0)^2+(4x+7-(-3))^2}=\sqrt{17x^{2}+80x+100}#.

Minimizing the squared distance will occur at the same value of #x# where the distance is minimized, therefore, we can focus on minimizing the function #f(x)=17x^{2}+80x+100#. Its derivative is #f'(x)=34x+80#, which has one root at #x=-80/34=-40/17\approx -2.353#.

That this value of #x# gives a minimum distance is clear since the graph of #f(x)# is a parabola opening upward, though you can also note that the second derivative is #f''(x)=34>0# for all #x#, making the graph of #f# concave up.

When #x=-40/17#, #y=4\cdot(-40/17)+7=\frac{119-160}{17}=-41/17#. Hence, the point #(x,y)=(-40/17,-41/17)# is the point on the line #y=4x+7# that is closest to the point #(0,-3)#. The minimum distance itself is #\sqrt{f(-40/17)}=\sqrt{100/17}=10/sqrt(17)=\frac{10\sqrt{17)}{17} \approx 2.425#.

Here's a picture for this situation. The line segment intersects the given line at a right angle.