Find the point P on the line yequals=66x that is closest to the point (74 comma 0 )(74,0). What is the least distance between P and (74 comma 0 )(74,0)?

The point P that is closest to the point (74, 0) is P(7.4, 22.2), and the least distance between P and (74, 0) is 70.2

Step-by-step explanation:

Given

y = 66x

And a point A(74, 0)

We are required to find the point on the line that is closest to A, and then find the distance between these points.

Let the point be P(s, t)

Because it's coordinate satisfies the given equation

y = 3x

Let t = 3s, then we have

p(s, t) = P(s, 3s)

Next, we find the distance between P and A

Let the distance be

D = √[(x1 - y1)² + (x2 - y2)²]

Here x1 = 74, x2 = 0, y1 = s, y2 = 3s.

D = √[(74 - s)² + (0 - 3s)²]

= √(s² - 148s + 5476 + 9s²)

= √(10s² - 148s + 5479)

To maximize the distance, we find the derivative of D with respect to s, and set it to 0.

That is dD/ds = 0

dD/ds =

(20s - 148)/2√(10s² - 148s 5479) = 0

=> 20s - 148 = 0

20s = 148

s = 148/20

s = 7.4

Since t = 3s, we have

t = 22.2

So, the point is P(7.4, 22.2)

Finally, we find the distance D.

D = √[(74 - s)² + (0 - 3s)²]

D = √[(74 - 7.4)² + (0 - 3×7.4)²]

D = √(4435.56 + 492.84)

= √4928.4

D = 70.2

Therefore, the distance is 70.2

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°