Find the point on the line y = 4x + 5 that is closest to the origin.
Answers
Step 1
Find the equation of the line perpendicular to [tex] y = 4x + 5 [/tex] that pas through the origin
we know that
the slope of the equation [tex] y = 4x + 5 [/tex] is
[tex] m1=4 [/tex]
if two lines are perpendicular
then
the product of their slopes is equal to [tex] -1 [/tex]
[tex] m1*m2=-1 [/tex]
the slope of the line perpendicular is equal to
[tex] m2=-\frac{1}{m1} \\\\ m2=-\frac{1}{4} [/tex]
with m2 and the origin find the equation of the line
[tex] y-y1=m(x-x1)\\ y-0=(-\frac{1}{4} )*(x-0)\\ y=-\frac{1}{4} x [/tex]
Step 2
Solve the system
[tex] y = 4x + 5 [/tex]---> equation [tex] 1 [/tex]
[tex] y=-\frac{1}{4} x [/tex]-----> equation [tex] 2 [/tex]
Multiply equation [tex] 1 [/tex] by [tex] -1 [/tex]
Adds equation [tex] 1 [/tex] and equation [tex] 2 [/tex]
[tex] -y = -4x - 5 [/tex]
[tex] y=-\frac{1}{4} x \\ ----- [/tex]
[tex] 0=-4x-\frac{1}{4} x-5\\ \\ \frac{17}{4} x=-5\\ \\ x=-\frac{20}{17} [/tex]
[tex] y = 4x + 5 [/tex]
[tex] y = 4*(-\frac{20}{17}) + 5 \\ \\ y=\frac{5}{17} [/tex]
the solution is the point [tex] (-\frac{20}{17} ,\frac{5}{17} ) [/tex]
[tex] (-\frac{20}{17} ,\frac{5}{17} ) [/tex]=[tex] (-1.176 ,0.294 ) [/tex]
therefore
the answer is
the point is [tex] (-\frac{20}{17} ,\frac{5}{17} ) [/tex]
see the attached figure
The point [tex]\boxed{(-1.176,0.294)}[/tex] on the line [tex]y=4x+5[/tex] is the closest point to the origin.
Further explanation:
The general form of linear function is as follows:
[tex]\boxed{y=mx+c}[/tex]
A linear function has one independent variable and one dependent variable. The independent variable is [tex]x[/tex] and the dependent variable is [tex]y[/tex].
Here, [tex]c[/tex] is the constant term and [tex]m[/tex] is the slope and gives the rate of change of dependent variable.
The point slope form of a line is given as follows:
[tex]\boxed{y-y_{1}=m(x-x_{1})}[/tex]
where [tex](x_{1},y_{1})[/tex] is the point on the line and [tex]m[/tex] is the slope of the line.
It is given that the equation of the line is as follows:
[tex]y=4x+5[/tex]
where [tex]4[/tex] is the slope of the line.
Consider the slope of the line [tex]y=4x+5[/tex] as [tex]m_{1}=4[/tex].
We first find the line perpendicular to the line [tex]y=4x+5[/tex] that passes through the origin as shown in Figure 1 in the attachment below.
If two lines are perpendicular then the product of their slopes is [tex]-1[/tex] that is [tex]m_{1}\tiimes m_{2}=-1[/tex].
And [tex]m_{2}[/tex] is the slope of the perpendicular line.
Calculated the value of [tex]m_{2}[/tex] as follows:
[tex]\begin{aligned}m_{1}\times m_{2}&=-1\\m_{2}&=-\dfrac{-1}{m_{1}}\\&=\dfrac{-1}{4}\\&=-0.25\end{aligned}[/tex]
Therefore, the slope of perpendicular line is [tex]-0.25[/tex].
We have the point [tex](0,0)[/tex] on the line and the slope of the line.
Thus, the equation of line is,
[tex]\begin{aligned}y-y_{1}&=m_{2}(x-x_{1})\\y-0&=(-0.25)(x-0)\\y&=-0.25x\end{aligned}[/tex]
The equation of the perpendicular line is as follows:
[tex]\boxed{y=-0.25x}[/tex] .......(2)
Substitute [tex]y=-0.25x[/tex] in equation (1).
[tex]\begin{aligned}-\dfrac{x}{4}&=4x+5\\-\dfrac{x}{4}-4x&=5\\-\dfrac{17x}{4}&=5\\-17x&=5\times 4\\x&=-\dfrac{20}{17}\\x&\approx-1.176\end{aligned}[/tex]
Now, put the value of [tex]x[/tex] in equation (2) to get the value of [tex]y[/tex] as,
[tex]\begin{aligned}y&=-0.25\times (-1.176)\\&=0.294\end{aligned}[/tex]
Therefore, the closest point is [tex]\boxed{(-1.176,0.294)}[/tex].
Learn more
1. A Problem on the slope-intercept form https://brainly.com/question/1473992.
2. A Problem on the center and radius of an equation https://brainly.com/question/9510228
3. A Problem on general form of the equation of the circle https://brainly.com/question/1506955
Answer details:
Grade: High school
Subject: Mathematics
Chapter: Linear equations
Keywords: Linear equations, slope of a line, equation of the line, function, real numbers, ordinates, abscissa, interval, open interval, closed intervals, semi-closed intervals, semi-open intervals, sets, range domain, codomain.
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I hope this helped!
Comment if you have any questions!!
Glad to help!
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To find the probability that a randomly selected single parent in the U.S. military is a woman in the Air Force, we need to divide the number of women in the Air Force by the total number of single parents in the military.
1. First, find the number of women in the Air Force from the table. This is given as 6 thousand.
2. Next, find the total number of single parents in the military. This is given as 97 thousand.
3. Finally, divide the number of women in the Air Force by the total number of single parents in the military and express the result as a fraction or decimal to represent the probability.
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[tex]\[ P = \frac{6}{97} \][/tex]
[tex]\[ P ≈ 0.0619 \][/tex]
Therefore, the probability that a randomly selected single parent in the U.S. military is a woman in the Air Force is approximately 0.0619 or 6.19%.
Complete question:
This problem involves empirical probability. The table shows the breakdown of 97 thousand single parents on active duty in the U.S. military in a certain year. All numbers are in thousands and rounded to the nearest thousand. Use the data in the table to find the probability that a randomly selected single parent in the U.S. military is a woman in the Air Force.
Army Navy Marine Corps Air Force Total
Male 27 26 4 14 71
Female 11 8 1 6 26
Total 38 34 5 20 97
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aprox
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