Find the point on the line − 5 x + 7 y − 3 = 0 which is closest to the point ( 5 , 1 ) .

Question

I am not sure how to approach this one!

Answer 1

This is actually easy in geometry. The point on a line closest to another point is the point on the line that intersects the line perpendicular to the line going through that point. The line -5x + 7y - 3 = 0 is also y = 5/7x + 3/7 meaning the slope is 5/7. The perpendicular would have a slope of -7/5 and the line with a slope of -7/5 going through (5, 1) is y - 1 = -7/5(x - 5) or y = -7/5x + 8. To find the intersection (the point on the line closest to (5, 1) just set them equal to each other:

5/7x + 3/7 = -7/5x + 8

35(5/7x + 3/7) = 35(-7/5x + 8)

25x + 15 = -49x + 280

74x = 265

x = 265/74

Then plugging that in: y = 5/7x + 3/7 we get

y = 5/7(265/74) + 3/7 = 221/74

So the point is (265/74, 221/74)

To do this using calculus is a little harder. Let the point on the line -5x + 7y - 3 = 0 be (x₁, y₁) since y = 5/7x + 3/7 then y₁ = 5/7x₁ + 3/7 so the point is (x₁, (5/7x₁ + 3/7))

The distance between (x₁, (5/7x₁ + 3/7)) and (5, 1) can be found using the distance formula:

d = √((x₂ - x₁)² + (y₂ - y₁)²) or, in this case with the points being (x₁, (5/7x₁ + 3/7)) and (5, 1):

d(x₁) = √((5 - x₁)² + (1 - (5/7x₁ + 3/7))²)

d(x₁) = √((5 - x₁)² + (4/7 - 5/7x₁))²)

d(x₁) = √((25 - 10x₁ + x₁²) + (16/49 - 40/49x₁ + 25/49x₁²))

d(x₁) = √(1241/49 - 530/49x₁ + 74/49x₁²)

d(x₁) = 1/7√(1241 - 530x₁ + 74x₁²)

d(x₁) = 1/7(1241 - 530x₁ + 74x₁²)^1/2

To minimize it, we take the derivative and set it equal to zero. To take the derivative, use the power rule and the chain rule.

d'(x₁) = (1/2)1/7(1241 - 530x₁ + 74x₁²)^-1/2(148x₁ - 530)

d'(x₁) = (148x₁ - 530)/(14(1241 - 530x₁ + 74x₁²)^-1/2)

d'(x₁) = (74x₁ - 265)/(7(1241 - 530x₁ + 74x₁²)^1/2)

Setting this equal to zero we will only get a solution if the numerator equals zero so:

74x₁ - 265 = 0

x₁ = 265/74 which is the same value we got using geometry.

Plugging this into y = 5/7x₁ + 3/7 we get y₁ = 221/74

Again, the point closest to (5, 1) is (265/74, 221/74)

Answer 2

One way is to write y as f(x) : y = 5/7 x + 3/7

You want to minimize the distance from point (5,1) to the line:

d = sqrt((5-x)² + (1-(5x/7 +3/7))²) where (x,y) is on the line

which will be a minimum when the quantity under the square root sign is a min: (make life easier)

d = (5-x)² + (4/7 - 5x/7)²

Take the derivative with respect to x and equate to zero and solve for critical x value.

I get x = 530/148 (checked)

Good luck.!

Answer 3

Equation of perpendicular

y = -7x / 5 + b

To pass through (5, 1)

1 = -7(5)/ 5 + b

1 = -7 + b

8 = b

y = -7x/5 + 8

Point of intersection

-7x/5 + 8 = 5x/7 + 3/7

-49x + 280 = 25x + 15 LCD 35

265 = 74x

265 / 74 = x

221 / 74 = y