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
[tex]\boxed{ \ y - y_1 = m(x - x_1) \ }[/tex]
The equation of the line goes through points (2, 5) and has a gradient m.
[tex]\boxed{ \ y - 5 = m(x - 2) \ }[/tex]
Step-2: the x-intercept and the y-intercept
For y = 0, we get the x-intercept.
[tex]\boxed{ \ 0 - 5 = m(x - 2) \ }[/tex]
[tex]\boxed{ \ - 5 = mx - 2m \ }[/tex]
[tex]\boxed{ \ mx = 2m - 5 \ }[/tex]
Divide both sides by m.
Hence, [tex]\boxed{\boxed{ \ x = 2 - \frac{5}{m} \ }}[/tex]
For x = 0, we get the y-intercept.
[tex]\boxed{ \ y - 5 = m(0 - 2) \ }[/tex]
[tex]\boxed{ \ y - 5 = - 2m \ }[/tex]
Add 5 form both sides.
Hence, [tex]\boxed{\boxed{ \ y = 5 - 2m \ }}[/tex]
Step-3: prepare the area cut off in the first quadrant
We see a right triangle with:
- [tex]\boxed{ \ base \ (x-intercept) = 2 - \frac{5}{m} \ }[/tex]
- [tex]\boxed{ \ height \ (y-intercept) = 5 - 2m \ }[/tex]
Area cut off in the first quadrant = [tex]\boxed{ \ \frac{1}{2} \times base \times height \ }[/tex]
[tex]\boxed{ \ A(m) = \frac{1}{2} \times (2 - \frac{5}{m}) \times (5 - 2m) \ }[/tex]
[tex]\boxed{ \ A(m) = \frac{1}{2} \times (10 - 4m - \frac{25}{m} + 10) \ }[/tex]
[tex]\boxed{ \ A(m) = \frac{1}{2} \times (20 - 4m - 25m^{-1}) \ }[/tex]
Thus, [tex]\boxed{\boxed{ \ A(m) = 10 - 2m - \frac{25}{2}m^{-1} \ }}[/tex]
Step-4: the stationary state for the least area
Let us differentiate for A(m).
[tex]\boxed{ \ A'(m) = -2 + \frac{25}{2}m^{-2} \ }[/tex]
[tex]\boxed{ \ A'(m) = -2 + \frac{25}{2m^2} \ }[/tex]
The stationary state for A(m).
[tex]\boxed{ \ A'(m) = 0 \ }[/tex]
[tex]\boxed{ \ -2 + \frac{25}{2m^2} = 0 \ }[/tex]
Multiply both sides by 2m².
[tex]\boxed{ \ -4m^2 + 25 = 0 \ }[/tex]
[tex]\boxed{ \ -(4m^2 - 25) = 0 \ }[/tex]
[tex]\boxed{ \ -(2m - 5)(2m + 5)) = 0 \ }[/tex]
We get [tex]\boxed{m = \frac{5}{2}}[/tex] and [tex]\boxed{m = -\frac{5}{2}}.[/tex]
Because the line equation is in the first quadrant, with the line slopes downwards, this line has a negative gradien, i.e., [tex]\boxed{m = -\frac{5}{2}}.[/tex]
Final step: the equation of the line in the point-slope form
Let us recall the equation in step-1 above.
[tex]\boxed{ \ y - 5 = m(x - 2) \ }[/tex]
Substitute the value of m, so that the form y - A = B (x - C) is obtained, namely:
[tex]\boxed{\boxed{ \ y - 5 = -\frac{5}{2}(x - 2) \ }}[/tex]
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Notes:
Let us find out the least area from the first quadrant.
[tex]\boxed{ \ m = -\frac{5}{2} \rightarrow A = 10 - 2\bigg(-\frac{5}{2}\bigg) - \frac{25}{2}\bigg(-\frac{2}{5}\bigg)} \ }[/tex]
A = 10 + 5 + 5
Thus, the least area is 20 square units.
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The x-intercept
[tex]x = 2 - \frac{5}{-\frac{5}{2}} \rightarrow \boxed{ \ x = 4 \ }[/tex]
The y-intercept
[tex]y = 5 - 2(-\frac{5}{2}) \rightarrow \boxed{ \ y = 10 \ }[/tex]
Learn more
- Finding the equation, in slope-intercept form, of the line that is parallel to the given line and passes through a point https://brainly.com/question/1473992
- Determine the equation represents Nolan’s line https://brainly.com/question/2657284
- The derivatives of the composite function https://brainly.com/question/6013189#
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
