Question

Find the equation of the line through the point (2, 5) that cuts off the least area from the first quadrant. Give your answer using the form below. y − A = B(x − C)

Answer 1

y - 5 = - ⁵/₂ (x - 2)

Further explanation

This case asking the end result in the form of a point-slope form. Before that what has to be done is the differential process for the stationary state.

Step-1: prepare the point-slope form

$\boxed{ \ y - y_1 = m(x - x_1) \ }$

The equation of the line goes through points (2, 5) and has a gradient m.

$\boxed{ \ y - 5 = m(x - 2) \ }$

Step-2: the x-intercept and the y-intercept

For y = 0, we get the x-intercept.

$\boxed{ \ 0 - 5 = m(x - 2) \ }$

$\boxed{ \ - 5 = mx - 2m \ }$

$\boxed{ \ mx = 2m - 5 \ }$

Divide both sides by m.

Hence, $\boxed{\boxed{ \ x = 2 - \frac{5}{m} \ }}$

For x = 0, we get the y-intercept.

$\boxed{ \ y - 5 = m(0 - 2) \ }$

$\boxed{ \ y - 5 = - 2m \ }$

Add 5 form both sides.

Hence, $\boxed{\boxed{ \ y = 5 - 2m \ }}$

Step-3: prepare the area cut off in the first quadrant

We see a right triangle with:

• $\boxed{ \ base \ (x-intercept) = 2 - \frac{5}{m} \ }$
• $\boxed{ \ height \ (y-intercept) = 5 - 2m \ }$

Area cut off in the first quadrant = $\boxed{ \ \frac{1}{2} \times base \times height \ }$

$\boxed{ \ A(m) = \frac{1}{2} \times (2 - \frac{5}{m}) \times (5 - 2m) \ }$

$\boxed{ \ A(m) = \frac{1}{2} \times (10 - 4m - \frac{25}{m} + 10) \ }$

$\boxed{ \ A(m) = \frac{1}{2} \times (20 - 4m - 25m^{-1}) \ }$

Thus, $\boxed{\boxed{ \ A(m) = 10 - 2m - \frac{25}{2}m^{-1} \ }}$

Step-4: the stationary state for the least area

Let us differentiate for A(m).

$\boxed{ \ A'(m) = -2 + \frac{25}{2}m^{-2} \ }$

$\boxed{ \ A'(m) = -2 + \frac{25}{2m^2} \ }$

The stationary state for A(m).

$\boxed{ \ A'(m) = 0 \ }$

$\boxed{ \ -2 + \frac{25}{2m^2} = 0 \ }$

Multiply both sides by 2m².

$\boxed{ \ -4m^2 + 25 = 0 \ }$

$\boxed{ \ -(4m^2 - 25) = 0 \ }$

$\boxed{ \ -(2m - 5)(2m + 5)) = 0 \ }$

We get $\boxed{m = \frac{5}{2}}$ and $\boxed{m = -\frac{5}{2}}.$

Because the line equation is in the first quadrant, with the line slopes downwards, this line has a negative gradien, i.e., $\boxed{m = -\frac{5}{2}}.$

Final step: the equation of the line in the point-slope form

Let us recall the equation in step-1 above.

$\boxed{ \ y - 5 = m(x - 2) \ }$

Substitute the value of m, so that the form y - A = B (x - C) is obtained, namely:

$\boxed{\boxed{ \ y - 5 = -\frac{5}{2}(x - 2) \ }}$

- - - - - - - - - -

Notes:

Let us find out the least area from the first quadrant.

$\boxed{ \ m = -\frac{5}{2} \rightarrow A = 10 - 2\bigg(-\frac{5}{2}\bigg) - \frac{25}{2}\bigg(-\frac{2}{5}\bigg)} \ }$

A = 10 + 5 + 5

Thus, the least area is 20 square units.

- - - - - - - - - -

The x-intercept

$x = 2 - \frac{5}{-\frac{5}{2}} \rightarrow \boxed{ \ x = 4 \ }$

The y-intercept

$y = 5 - 2(-\frac{5}{2}) \rightarrow \boxed{ \ y = 10 \ }$

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Keywords: the equation of the line, through the point, (2, 5), cuts off, the least area, the first quadrant, y − A = B(x − C), point-slop, gradient, slope

Answer 2

The equation of the line would be

y = mx+b 

where m is the slope, b is the y intercept. 

The line will form a triangular region in the first quadrant. Its area would be 1/2 base times height. The height is the y intercept and the base is y intercept divided by slope. Therefore,

A = b^2/2m

At point (2,5)

5 = 2m+b

Substitute that in the area

A = b^2/5-b to find the least area, differentiate the area with respect to the height and equate it to 0 

dA/db = 0

find b and use that to find m. Then, you can have the equation of the line.