The point \(\boxed{(-1.176,0.294)}\) on the line \(y=4x+5\) is the closest point to the origin.
Further explanation:
The general form of linear function is as follows:
\(\boxed{y=mx+c}\)
A linear function has one independent variable and one dependent variable. The independent variable is \(x\) and the dependent variable is \(y\).
Here, \(c\) is the constant term and \(m\) is the slope and gives the rate of change of dependent variable.
The point slope form of a line is given as follows:
\(\boxed{y-y_{1}=m(x-x_{1})}\)
where \((x_{1},y_{1})\) is the point on the line and \(m\) is the slope of the line.
It is given that the equation of the line is as follows:
\(y=4x+5\)
where \(4\) is the slope of the line.
Consider the slope of the line \(y=4x+5\) as \(m_{1}=4\).
We first find the line perpendicular to the line \(y=4x+5\) 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 \(-1\) that is \(m_{1}\tiimes m_{2}=-1\).
And \(m_{2}\) is the slope of the perpendicular line.
Calculated the value of \(m_{2}\) as follows:
\(\begin{aligned}m_{1}\times m_{2}&=-1\\m_{2}&=-\dfrac{-1}{m_{1}}\\&=\dfrac{-1}{4}\\&=-0.25\end{aligned}\)
Therefore, the slope of perpendicular line is \(-0.25\).
We have the point \((0,0)\) on the line and the slope of the line.
Thus, the equation of line is,
\(\begin{aligned}y-y_{1}&=m_{2}(x-x_{1})\\y-0&=(-0.25)(x-0)\\y&=-0.25x\end{aligned}\)
The equation of the perpendicular line is as follows:
\(\boxed{y=-0.25x}\) .......(2)
Substitute \(y=-0.25x\) in equation (1).
\(\begin{aligned}-\dfrac{x}{4}&=4x+5\\-\dfrac{x}{4}-4x&=5\\-\dfrac{17x}{4}&=5\\-17x&=5\times 4\\x&=-\dfrac{20}{17}\\x&\approx-1.176\end{aligned}\)
Now, put the value of \(x\) in equation (2) to get the value of \(y\) as,
\(\begin{aligned}y&=-0.25\times (-1.176)\\&=0.294\end{aligned}\)
Therefore, the closest point is \(\boxed{(-1.176,0.294)}\).
Learn more
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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.
