Find the point P on the line y=4x that is closes to the point (68,0). What is the least distance between P and (68,0)
line from line to point is perpendicular to original line :)
m' = -1/4
in
y = -x/4 + b
now through point
0 = -68/4 + b
b = 17
so
line to point is y = -x/4 + 17
where does it hit original line?
-x/4 + 17 = 4 x
-x + 68 = 16 x
17 x = 68
x = 4
so intersection at (4 , 16)
now how far is it from (4,16) to (68,0)
d^2 = 64^2 + 16^2
and check to see that the answer agrees with the formula giving the distance from a point (x,y) to the line Ax+By+C=0
d = |Ax+By+C|/√(A^2+B^2)
Well, let's clown around with some math!
The line y = 4x is in the form of y = mx, where m represents the slope. In this case, the slope is 4.
Now, we can find the equation of a line perpendicular to this line by taking the negative reciprocal of the slope. So, the slope of the perpendicular line will be -1/4.
Using the point (68, 0), we can write an equation for the perpendicular line in the form y = mx + b, where b represents the y-intercept. Substituting the values, we get 0 = (-1/4) * 68 + b. Solving for b, we find b = 17.
So, the equation of the perpendicular line is y = (-1/4)x + 17.
To find the point of intersection between the two lines, we can set the equations equal to each other. Thus, 4x = (-1/4)x + 17. Solving for x, we get x = 4.
Substituting x = 4 into the equation of the original line, y = 4(4), we find that y = 16.
Therefore, the point of intersection is P(4, 16).
The distance between P(4, 16) and (68, 0) can be found using the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance = sqrt((68 - 4)^2 + (0 - 16)^2)
Distance = sqrt(64^2 + (-16)^2)
Distance = sqrt(4096 + 256)
Distance = sqrt(4352)
Distance ≈ 65.93
So, the least distance between P and (68, 0) is approximately 65.93 units.
To find the point P on the line y = 4x that is closest to the point (68, 0), we need to find the perpendicular distance between the line and the point.
The line y = 4x can be expressed in the form of y = mx + c, where m is the slope of the line and c is the y-intercept. In this case, the slope of the line is 4 and the y-intercept is 0.
To find the perpendicular distance between the line and the point (68, 0), we will use the formula d = |Ax + By + C| / sqrt(A^2 + B^2), where (x, y) is a point on the line and A, B, and C are the coefficients of the equation of the line.
Let's start by finding the coefficients A, B, and C for our line y = 4x.
Since the equation of the line is y = 4x, we can rearrange it to get -4x + y = 0. Comparing this equation with the standard form Ax + By + C = 0, we can see that A = -4, B = 1, and C = 0.
Now, we can substitute these values into the formula for the perpendicular distance:
d = |(-4)(68) + (1)(0) + (0)| / sqrt((-4)^2 + 1^2)
= |-272 + 0 + 0| / sqrt(16 + 1)
= |272| / sqrt(17)
= 272 / sqrt(17)
≈ 66.184
Therefore, the least distance between the point P on the line y = 4x and the point (68, 0) is approximately 66.184 units.