kannn wrote:
The equation of line n 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 n?
A. 48
B. 50
C. 60
D. 75
E. 100
The answer is C as follows.
Given the line y = 4/3*x - 100 ==> 4x-3y-300=0
As per the formula distance of a point (x1,y1) from line ax+by+c=0 is |ax1+by1+c|/\sqrt{(\((a^2)\)+\((b^2)\))}
Putting the value in the formula |a*0+b*0-300|/\sqrt{(\(4^2\)+\(3^2\))} ==> \(\frac{300}{5}\) ==>60
Hence answer is C
_________________
-------------------------------------------------------------------------------
Kudos are the only way to tell whether my post is useful.
GMAT PREP 1: Q50 V34 - 700
Veritas Test 1: Q43 V34 - 630
Veritas Test 2: Q46 V30 - 620
Veritas Test 3: Q45 V29 - 610
Veritas Test 4: Q49 V30 - 650
GMAT PREP 2: Q50 V34 - 700
Veritas Test 5: Q47 V33 - 650
Veritas Test 5: Q46 V33 - 650