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)

y - 5 = - ⁵/₂ (x - 2)

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

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

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

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

Divide both sides by m.

Hence,

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

Add 5 form both sides.

Hence,

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

We see a right triangle with:

Area cut off in the first quadrant =

Thus,

Step-4: the stationary state for the least area

Let us differentiate for A(m).

The stationary state for A(m).

Multiply both sides by 2m².

We get and

Because the line equation is in the first quadrant, with the line slopes downwards, this line has a negative gradien, i.e.,

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

Let us recall the equation in step-1 above.

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

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Notes:

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

A = 10 + 5 + 5

Thus, the least area is 20 square units.

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The x-intercept

The y-intercept

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

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