Crawford Hardy
11/01/2023 · High School
Find the point on the line \( \frac{x}{5}+\frac{y}{10}=1 \) that is closest to \( (1,0) \). The point on the line \( \frac{x}{5}+\frac{y}{10}=1 \) that is closest to \( (1,0) \) is (Type an ordered pair, using integers or fractions.)
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Step-by-step Solution
To find the point on the line \( \frac{x}{5} + \frac{y}{10} = 1 \) that is closest to the point \( (1, 0) \), we can use the method of Lagrange multipliers or geometric considerations. Here, we will use the geometric approach.
First, we can rewrite the line equation in slope-intercept form. Starting from:
\[
\frac{x}{5} + \frac{y}{10} = 1
\]
we can multiply through by 10 to eliminate the fractions:
\[
2x + y = 10
\]
Now, we can express \( y \) in terms of \( x \):
\[
y = 10 - 2x
\]
Next, we want to minimize the distance from the point \( (1, 0) \) to a point \( (x, y) \) on the line. The distance \( d \) from \( (1, 0) \) to \( (x, y) \) is given by:
\[
d = \sqrt{(x - 1)^2 + (y - 0)^2}
\]
Substituting \( y = 10 - 2x \) into the distance formula gives:
\[
d = \sqrt{(x - 1)^2 + (10 - 2x)^2}
\]
To minimize \( d \), we can minimize \( d^2 \) instead, which is:
\[
d^2 = (x - 1)^2 + (10 - 2x)^2
\]
Expanding this:
\[
d^2 = (x - 1)^2 + (10 - 2x)^2 = (x - 1)^2 + (100 - 40x + 4x^2)
\]
Combining like terms:
\[
d^2 = (x^2 - 2x + 1) + (4x^2 - 40x + 100) = 5x^2 - 42x + 101
\]
To find the minimum, we take the derivative of \( d^2 \) with respect to \( x \) and set it to zero:
\[
\frac{d}{dx}(5x^2 - 42x + 101) = 10x - 42
\]
Setting the derivative equal to zero:
\[
10x - 42 = 0 \implies 10x = 42 \implies x = \frac{42}{10} = 4.2
\]
Now, substituting \( x = 4.2 \) back into the line equation to find \( y \):
\[
y = 10 - 2(4.2) = 10 - 8.4 = 1.6
\]
Thus, the point on the line that is closest to \( (1, 0) \) is:
\[
\boxed{\left( \frac{21}{5}, \frac{8}{5} \right)}
\]
Quick Answer
The point on the line closest to \( (1, 0) \) is \( \left( \frac{21}{5}, \frac{8}{5} \right) \).
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