dhairya275 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

I'll post the answer later. Thanks

I've corrected a typo: the equation of the line should be y = (4/3)x - 100.

The x-intercept of a line occurs when y=0.
Substituting y=0 into y = (4/3)x - 100, we get:
0 = (4/3)x - 100
x = 75.
Thus, the x-intercept of y = (4/3)x - 100 is (75,0).

DRAW the figure:


The SHORTEST DISTANCE between a point and a line must form a RIGHT ANGLE with the line.
Thus:
d = the shortest distance between the origin and y = (4/3)x - 100.

∆ABC is a multiple of a 3-4-5 triangle:
25*(3:4:5) = 75125.
Thus, BC = 125.

Any side of a triangle can be considered the base.
Each base has a corresponding height.
A = 1/2(bh).
Since the area must be the same no matter which base and height are used, bh must always yield the same product.
In the triangle above:
If AB=75 is considered the base, the corresponding height is AC=100.
If BC=125 is considered the base, the corresponding height is d.
Since bh must yield the same product in each case, we get:
75*100 = 125d
d = 60.

The correct answer is C.

If we have the equation Ax + By + C = 0 and the point as (x1, y1) then the PERPENDICULAR DISTANCE onto the line from the point (x1, y1) is :



So I transformed the equation as 4x - 3y - 300 = 0 and the point is (0, 0).
So the perpendicular distance is
|4(0) - 3(0) - 300|
------------------------
√ (3^2 + 4^2)

|300|
--------
5

Answer : 60