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.
