The equation of line q is y = 4/3*x - 100. What is the smallest possible distance in the xy-plane from the point with coordinates (0, 0) to any point on line q?
Options:
A) 48
B) 50
C) 60
D) 75
E) 100
This can be solve in two steps and without any complex calculation.   Given : equation of line q as y=(4/3)x -100. So the line intercept the axes at (0,-100) and (75,0).   This can be considered a right angle triangle with right angle at (0,0) . So Base=100 , Height=75 and Hypotenuse =125 (By Pythagoras triplet)   So a perpendicular from the (0,0) to hypotenuse will be the answer.   Area of triangle= 0.5*100*75=0.5*125* x   =&gt; x=60;   SO answer is 60=60
<result>C</result>
The equation of line q is y = 4/3*x - 100. What is the smallest possible distance in the xy-plane from the point with coordinates (0, 0) to any point on line q?