Respuesta :
Answer:
[tex]\displaystyle \left(\frac{278}{113},\, -\frac{215}{113}\right)[/tex].
Step-by-step explanation:
The question is asking for the point on the line where the distance from the given point is minimized. Since distance is non-negative, minimizing distance is equivalent to minimizing the square of the distance.
To find the coordinates of the requested point using calculus, apply the following steps:
- Find the general expression (in terms of the [tex]x[/tex]-coordinate) of a point on the given line.
- Express the square of the distance between the two points in terms of the [tex]x[/tex]-coordinate of the point on the line.
- Differentiate the expression for distance to find the [tex]x[/tex]-coordinate of the point where this distance is minimized.
- Substitute the [tex]x[/tex]-coordinate of the point back into the general expression for the point to find the [tex]y[/tex]-coordinate.
Rearrange the equation of the line [tex]7\, x + 8\, y - 2[/tex] to find the [tex]y[/tex]-coordinate in terms of [tex]x[/tex]:
[tex]\displaystyle y = \frac{1}{8}\, \left(-7\, x + 2\right)[/tex].
In other words, if the [tex]x[/tex]-coordinate of a point on this line is [tex]x[/tex], the [tex]y[/tex]-coordinate of this point would be [tex](1/8)\, (-7\,x + 2)[/tex]. Hence, [tex](x,\, (1/8)\, (-7\, x + 2))[/tex] is a general expression for all points on this line.
The distance between two points [tex](x_{0},\, y_{0})[/tex] and [tex](x_{1},\, y_{1})[/tex] on a cartesian plane is:
[tex]\displaystyle \sqrt{(x_{1} - x_{0})^{2} + (y_{1} - y_{0})^{2}}[/tex].
The square of the distance is [tex](x_{1} - x_{0})^{2} + (y_{1} - y_{0})^{2}[/tex]. Hence, in this question, the square of the distance between [tex](-2,\, -7)[/tex] and a point [tex](x,\, (1/8)\, (-7\, x + 2))[/tex] on the given line would be:
[tex]\begin{aligned} & \left(x - (-2)\right)^{2} + \left(\frac{1}{8}\, (-7\, x + 2) - (-7)\right)^{2} \\ =\; & \left(x + 2\right)^{2} + \left(\left(-\frac{7}{8}\right)\, x + \frac{29}{4} \right)\right)^{2}\end{aligned}[/tex].
Apply the chain rule and the power rule to differentiate the squared distance with respect to [tex]x[/tex]:
[tex]\begin{aligned} & 2\, (x + 2) + \left(-\frac{7}{8}\right)\, (2)\, \left(\left(-\frac{7}{8}\right)\, x + \frac{29}{4}\right) \\ =\; & 2\, x + \frac{49}{32}\, x + 4 - \frac{203}{16}\\ =\; & \frac{113}{32}\, x - \frac{139}{16}\end{aligned}[/tex].
Differentiate again to find the second derivative: [tex](113/32)[/tex]. Since the second derivative of this expression is positive for all [tex]x[/tex], the zero of the first derivative would indeed be a minimum of the distance.
Set the first derivative of the squared distance to [tex]0[/tex] and solve for [tex]x[/tex]:
[tex]\displaystyle x = \frac{278}{113}[/tex].
Substitute [tex]x = (278/113)[/tex] back into the general expression for points on this line to find the [tex]y[/tex]-coordinate of this point:
[tex]\begin{aligned} y &= \frac{1}{8}\, \left(-7\, x + 2\right) \\ &= \frac{1}{8}\, \left((-7)\, \left(\frac{278}{113}\right) + 2\right) \\ &= -\frac{215}{113}\end{aligned}[/tex].
