Find the point on the line y = 3x + 5 that is closest to the origin.

Answer 1

The shortest distance from a point to a line is the length of the perpendicular segment from a point to a line.

Since we are looking for a perpendicular line segment, we need to get its slope:

The given line is y = 3x + 5. Remember two lines are perpendicular when their slopes are opposite reciprocals. The line is in slope-intercept form: y = mx + b, where m is the slope.

The slope of the given line is 3: (m = 3)

Therefore, the slope of the perpendicular line segment is -13: (m_⊥ = -1/3)

and it's in the form of y = -1/3 x + b

This line has the point (0,0). Substitute this for x and y to get the value of b.

0 = -1/3 (0) + b

b = 0

Therefore the equation of the line is y = -1/3 x.

Now we can get the point of intersection of y = -1/3 x and y = 3x + 5 which is the closest to the origin by solution to System of Equations

-y = 1/3 x

y = 3x + 5

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0 = 10/3 x + 5

-10/3 x = 5

x = (-3/10)(5)

x = - 3/2

Using y = -1/3 x:

y = (-1/3) (-3/2)

y = 1/2

Therefore the point that we are looking for is (-3/2, 1/2)

Answer 2

Suppose (x,y) is on the line

Then if s is the distance of the point from the origin

s² = x² + y² = x² + (3x+5)²

We want to minimise s, so we differentiate and set ds/dx = 0

2s ds / dx = 2x + 2 (3x+5) (3) = 2x + 18x + 30 = 20x + 30

ds/dx = 0 so

20x + 30 = 0

so x = -1.5 , and y = 3 x -1.5 + 5 = 0.5

so (-1.5, 0.5)

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You can also solve this be finding where the perpendicular line through (0,0) cuts the line

The line has slope 3

perpendicular line has slope -1/3

perpendicular line through origin has equation y = (-1/3) x

-1/3 x = 3x + 5

gives x = -1.5