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 shortest line is a perpendicular, which goes through \((0;0)\)
We know that lines (\(y=kx+b\)) are perpendicular when \(k1*k2=-1\).
So, \((4/3)*k2=-1\)
k2=3/4
Line has point (0;0) also.
So, \(y=(3/4)x\)
This line intersects with \(y=\frac{4}{3}x-100\).
Let's find the intersection point.
\(\frac{4}{3}x-100=-\frac{3}{4}\)
\(x=48\)
\(y=\frac{4}{3}*48-100\)
\(y=36\)
\((48;36)\)
The shortest distance from (0;0) to above mentioned point is
\(\sqrt{48*48+36*36}\)=\(60\)
(C)

1. Draw your line first
2. Note the pythagorean triple, that is divide by 25 to get your 3,4,5
3. Given proportions Hypothenuse is 125
4. Now area is either b*h or leg*leg (both are divided by 2 but in this case we can cancel them out)
5. 75(100) = 125 (x). 'x' represents the height which is perpendicular to the base and is the smallest distance from the origin
6. x = 60

Answer is C

Bonus method: If y'all want to save some time in there's a formula for distance from point to line that has been mentioned in the posts above but anyways this is how it would work for this problem

d = 100 / sqrt (1^2 + (4/3)^2) = 100 / 5/3 = 60

Sanity check, answer still 60

Hope this gives a hand

Hope it helps
Cheers!

J

Can u please explain this in more detail. Thank you

Is the goal as described as the attachment.

I follow the area of the triangle is (100*75)/2, but why are you setting the area of the triangle to (125*H)/2. I dont understand the second formula logic. Please explain.

This can be solve in two steps and without any complex calculation.

Given : equation of line 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
=> x=60;

SO answer is 60

agreed, nice approach jlgdr
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I know calculus isn't something expected to be known on the GMAT, but you can also use calculus to quickly solve this problem:

\(D^2 = x^2 + ((4x/3)-100)^2\)

\(D^2 = x^2 + ((16/9)x^2 - (8/3)(100)x +100^2)\)

\((D^2)' = 2x + 2*(16/9)*x - (8/3)(100) = 0\)

to find the minimum value we then solve for x when \((D^2)' = 0\)

\(0 = 2x + 32x/9 - (8/3)*100

(8/3)*(100) = 50x/9

x = (8*100*9)/(3*50)\)


Canceling out terms from the numerator and denominator we get \(x = 48\)

This x value will give us the minimum distance. Plugging this back in, we get:

\(D^2 = 48^2 + ((4/3)(48) - 100)^2\)

\(D^2 = 48^2 + (64 - 100)^2 = 48^2 + 36^2\)

Noticing that 6 is a factor of both 48 and 36;

\(D^2 = 6^2*(8^2 + 6^2)\)

\(D^2 = 36*(100)^2\)

\(D = 6*10 = 60.\)

Answer: C

This isn't math that will be tested on the GMAT, but maybe it could still provide shortcuts in problems like these.

For one line to be perpendicular to another, the relationship between their slopes has to be negative reciprocal \(-\frac{1}{m}\). In other words, the two lines are perpendicular if and only if the product of their slopes is \(-1\).

So, yes, you are right, the slope of a line which is perpendicular to y = 4/3*x - 100 is -3/4 (-1/(4/3) = -3/4).



Hope this helps.
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There is a formula to solve such questions.
The min distance between points (p,q) and the line segment ax+by+c=0 is given by
(|pa+bq+c|)/[sqrt(a^2+b^2)]




- Rajat