Question

Find the point P on the line yequals6x that is closest to the point left parenthesis 74 comma 0 right parenthesis. What is the least distance between P and left parenthesis 74 comma 0 right parenthesis​

Answer 1

Answer:

The least distance would be 73 units ( approx )

Step-by-step explanation:

Let the coordinates of P are $(x_1, y_1)$,

Since, point P lies on the line y = 6x

So, P will satisfy the above line,

$y_1=6x_1------(1)$,

Now, By the distance formula,

The distance between P and (74,0),

$D=\sqrt{(74-x_1)^2+(0-y_1)^2}$

$=\sqrt{(74-x_1)^2+(-6x_1)^2$   ( From equation (1) )

$=\sqrt{(74-x_1)^2 + 36x_1^2}$

Differentiating with respect to $x_1$,

$D'(x_1)=-\frac{1}{(74-x_1)^2 + 36x_1^2}[-2(74-x_1)+72x_1]$

Again differentiating with respect to x,

$D''(x_1)=\frac{394272}{(36x^2+(74-x)^2)^\frac{3}{2}}$

For maxima or minima,

$D'(x_1)=0$,

$-\frac{1}{(74-x_1)^2 + 36x_1^2}[-2(74-x_1)+72x_1]=0$

$-2(74-x_1)+72x_1=0$

$-148+2x_1+72x_1=0$

$74x_1=148$

$x_1=2$

At $x_1=2$ D''(x) = positive

Thus, for $x_1=2$ D is minimum,

And, minimum distance,

$D(2) = \sqrt{(74-2)^2 + 36(2)^2}=72.9931503636\approx 73$