What point on the line 3x – y = 4 is closest to the point A(–2, 3)?

Answer 1

What point on the line 3x – y = 4 is closest to the point A(–2, 3)?

Rewrite the equation for the line in slope-intercept form:

y=3x-4

The slope of this line is m = 3.

A line perpendicular to this line passing through point A(-2,3) has a slope m= -1/3.

The y-intercept for this perpendicular line [go down 2/3 and over 2 from A(-2,3) which is calculated from the slope to get to x-0] is (-2,3) + (2,-2/3) = (0,2 1/3), so b = 2 1/3.

The equation for this perpendicular line is

y=(-1/3)x+2 1/3 or

x+3y=7

Solving the system of two linear equations with two unknows substituting y=3x-4 into this equation::

x+3(3x-4)=7

Solve for x:

10x-12=7

x=19/10

19/10 + 3y = 7

30y=70-19

y = 51/30

check:

51/30 =? 3(19/10) - 4

51/30=?57/10-40/10

17/10=17/10

Answer is x=1.9 and y=1.7 or point (1.9,1.7)

Answer 2

Let D = distance from point on the line to A thus D= sqrt(x+2)²+(7-3x)²). Find dD/dx =(20x-40)/2sqrt(10x²-40x+53) and set equal to zero to find dD/dx=0 at x=2 and y=2. This yields D= sqrt(17).