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Question

Find the coordinates of the foot of the perpendicular from the point A with coordinates (-5,5) to the line 10x +2y -3= 0

(i) What is the x coordinate of the foot of the perpendicular?

(ii) What is the y coordinate of the foot of the perpendicular?

(iii) Find the distance from the point A to the foot of the perpendicular.

Answer 1

(i)  x = – 45/52

(ii)  y = 303/52

(iii)  distance = 4.22

Start by placing the line's equation in point-slope form so we can use the fact that its slope times the perpendicular's slope equals -1:

10x + 2y - 3 = 0....subtract 10x and -3 from each side

2y = -10x + 3....divide each side by 2

y = -5x + 1.5

So, this slope, -5, times the slope we need equals -1:

(-5) * m = -1....divide each side by -5

m = 1/5

So now we know the perpendicular's equation is:  y = (1/5)x + b. What is "b"? We know the perpendicular goes through the point (-5, 5) so let us substitute:

y = (1/5)x + b....rearrange to get "b" by itself

b = y - (1/5)x....substitute for (-5, 5)

b = 5 - (1/5)*(-5)

b = 5 - (-1) = 5 + 1 = 6

So the perpendicular's equation is:  y = (1/5)x + 6.

Now we have two equations that yield the same values at one point:  the point the cross which is also the "foot of the perpendicular." If we set the point-slope sides equal to each other we can solve for x, then substitute that value into either to get y. That will answer (i) and (ii):

(1/5)x + 6 = -5x + 1.5....add 5x to each side; subtract 6 from each side

(5 + 1/5)x = 1.5 - 6

(26/5)x = -9/2....multiply each side by 5/26 (the inverse of 26/5)

x = (-9/2) * (5/26) = –45/52 which is (i)

Substitute back into either equation:

y = (-5) * (-45/52) + 1.5

y = 225/52 + 78/52 = 303/52 which is (ii)

Now we have two points on the perpendicular, so we just use the distance formula to find the distance:

d = {(x1 - x2)^2 + (y1 - y2)^2}^0.5

d = {(-5 - -45/52)^2 + (5 - 303/52)^2}^0.5

just use a calculator or spreadsheet to do the arithmetic — there's no extra value here to the fractions involved!

d = 4.22 which is (iii).

Answer 2

http://search.yahoo.com/search?p=What+is...