Respuesta :
Answer:Option D
Step-by-step explanation:
The given equation is [tex]y=\frac{-5}{3}x+11 \frac{1}{3}[/tex]
The slope m of the line is [tex]-\frac{5}{3}[/tex]
Slope of line perpendicular is reciprocal with opposite sign.
Slope of perpendicular m=[tex]\frac{3}{5}[/tex]
Equation of line passing through point P(-2,3) and having a slope m=[tex]\frac{3}{5}[/tex] is
y-3=[tex]\frac{3}{5}(x+2)[/tex]
y-3=[tex]\frac{3}{5}x+ \frac{6}{5}[/tex]
Adding 3 both sides
y=[tex]\frac{3}{5}x+4 \frac{1}{5}[/tex]
Option D
The equation of the line perpendicular to [tex]y=-\frac{5}{3}x+11\frac{1}{3}[/tex] and passes through a point [tex]P(-2,3)[/tex] is [tex]\boxed{\dfrac{3}{5}x+4\dfrac{1}{5}}[/tex].
Further explanation:
The linear equation of the line is [tex]y=mx+b[/tex] where [tex]m[/tex] is the slope of the line and [tex]b[/tex] is the [tex]y[/tex]-intercept of the line.
If the slope of the line [tex]1[/tex] is [tex]m_{1}[/tex] and the slope of the line [tex]2[/tex] is [tex]m_{2}[/tex] which is perpendicular to line [tex]1[/tex]. Then the relation between two slopes is expressed as follows:
[tex]\boxed{m_{1}\times m_{2}=-1}[/tex]
Therefore, we can say that the perpendicular slopes of the line are negative reciprocals of each other.
The point slope form equation of the line passing through a point [tex](x_{1},y_{1})[/tex] is as follows:
[tex]\boxed{y-y_{1}=m(x-x_{1})}[/tex]
Given:
The given equation is [tex]y=-\frac{5}{3}x+11\frac{1}{3}[/tex] and the point is [tex]P(-2,3)[/tex].
Step 1:
First we find the slope of the given equation.
The given equation is [tex]y=-\frac{5}{3}x+11\frac{1}{3}[/tex].
Compare the given equation with standard linear equation. It is obnserved that the slope and the [tex]y[/tex]-intercept of the equation is as follows:
[tex]m=-\dfrac{5}{3}\\ b=11\dfrac{1}{3}[/tex]
Step 2:
Now find the slope of the line that is perpendicular to the given equation of the line.
Consider [tex]m_{1}[/tex] as the slope of the given equation of the line and [tex]m_{2}[/tex] as the slope of the line which is perpendicular to the given line.
Substitute the value of [tex]m_{1}[/tex] in the relation between two slopes.
[tex]\begin{aligned}m_{1}\times m_{2}&=-1\\-\dfrac{5}{3}\times m_{2}&=-1\\m_{2}&=\dfrac{3}{5}\end{aligned}[/tex]
Step 3:
Now, we find the equation of the line containing the given point [tex]P(-2,3)[/tex].
Substitute [tex]x_{1}=-2[/tex], [tex]y_{1}=3[/tex] and the slope [tex]m_{2}=\frac{3}{5}[/tex] in the point slope form equation of the line.
[tex]\begin{aligned}y-y_{1}&=m(x-x_{1})\\y-3&=\dfrac{3}{5}(x-(-2))\\y&=3+\dfrac{3}{5}(x-(-2))\\y&=3+\dfrac{3}{5}x+\dfrac{6}{5}\\y&=\dfrac{3}{5}x+\dfrac{6}{5}+3\end{aligned}[/tex]
Further solution of the above equation is calculated as,
[tex]\begin{aligned}y&=\dfrac{3}{5}x+\dfrac{6+15}{5}\\y&=\dfrac{3}{5}x+\dfrac{21}{5}\\y&=\dfrac{3}{5}x+4\dfrac{1}{5}\end{aligned}[/tex]
Therefore, the equation of the line perpendicular to [tex]y=-\frac{5}{3}x+11\frac{1}{3}[/tex] and passes through a point [tex]P(-2,3)[/tex] is [tex]\boxed{\dfrac{3}{5}x+4\dfrac{1}{5}}[/tex].
Learn more:
1. Learn more about the equation in the slope intercept form https://brainly.com/question/1473992
2. Learn more about the equation of the circle https://brainly.com/question/1506955
3. Learn more about the line segment https://brainly.com/question/909890
Answer details:
Grade: High school
Subject: Mathematics
Chapter: Straight lines
Keywords: Slope, intercept, equation, perpendicular normal, angle between two line, point, coordinate, passes, parallel, cartesian plane, quadrilateral, distance of a point from a line, two point form, slope-intercept form.
