Respuesta :
The line perpendicular to the line AB passes through the point [tex](0,2)[/tex] i.e., [tex]\fbox{\begin\\\\ \bf option 3\\\end{minispace}}[/tex].
Further explanation:
From the given figure in the question it is observed that the line AB passes through the points [tex](-2,4)[/tex] and [tex](2,-8)[/tex].
The coordinate for the point C is [tex](6,4)[/tex].
Step1: Obtain the slope of the line AB.
The slope of a line which passes through the points [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] is calculated as follows:
[tex]\fbox{\begin\\\ \math m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\end{minispace}}[/tex] (1)
It is given that the line AB passes through the points [tex](-2,4)[/tex] and [tex](2,-8)[/tex].
To obtain the slope for the line AB substitute [tex]-2[/tex] for [tex]x_{1}[/tex], [tex]2[/tex] for [tex]x_{2}[/tex], [tex]4[/tex] for [tex]y_{1}[/tex] and [tex]-8[/tex] for [tex]y_{2}[/tex] in equation (1).
[tex]\begin{aligned}m&=\frac{-8-4}{2+2}\\&=\frac{-12}{4}\\&=-3\end{aligned}[/tex]
Therefore, the slope of the line AB is [tex]-3[/tex].
Consider the slope of AB as, [tex]m_{1}[/tex] so, [tex]m_{1}=-3[/tex].
Step2: Obtain the slope of the perpendicular line.
The slope of line AB is [tex]m_{1}=-3[/tex].
Consider a line which is perpendicular to the line AB passing through the point C. Assume the slope of the perpendicular line as [tex]m_{2}[/tex].
The product of slope of two mutually perpendicular lines is always equal to [tex]-1[/tex].
The equation formed for the slope is as follows:
[tex]\fbox{\begin\\\ \math m_{1}\times m_{2}=-1\\\end{minispace}}[/tex]
Substitute the value of [tex]m_{1}[/tex] in the above equation.
[tex]\begin{aligned}-3\times m_{2}&=-1\\m_{2}&=\frac{1}{3}\end{aligned}[/tex]
Therefore, the slope of the perpendicular line is [tex]m_{2}=\frac{1}{3}[/tex].
Step3: Obtain the equation of the perpendicular line.
The slope for perpendicular line is [tex]\frac{1}{3}[/tex] and the line passes through the point C. The coordinate for the point C are [tex](6,4)[/tex].
The point slope form of a line is as follows:
[tex]\fbox{\begin\\\ \math (y-y_{1})=m(x-x_{1})\\\end{minispace}}[/tex]
Substitute [tex]\frac{1}{3}[/tex] for [tex]m[/tex], [tex]6[/tex] for [tex]x_{1}[/tex] and [tex]4[/tex] for [tex]y_{1}[/tex] in the above equation.
[tex]\begin{aligned}(y-4)&=\frac{1}{3}(x-6)\\3y-12&=x-6\\3y-x&=6\end{aligned}[/tex]
Therefore, the equation of the perpendicular line is [tex]3y-x=6[/tex].
Option 1:
In option 1 it is given that the line perpendicular to AB passes through the point [tex](-6,0)[/tex].
The equation of the line which is perpendicular to AB is [tex]3y-x=6[/tex].
Substitute [tex]-6[/tex] for [tex]x[/tex] in the above equation.
[tex]\begin{aligned}3y-(-6)&=6\\3y+12&=6\\3y&=-6\\y&=-2\end{aligned}[/tex]
From the above calculation it is concluded that the line passes through the point [tex](-6,-2)[/tex].
This implies that option 1 is incorrect.
Option 2:
In option 2 it is given that the line perpendicular to AB passes through the point [tex](0,-6)[/tex].
The equation of the line which is perpendicular to AB is [tex]3y-x=6[/tex].
Substitute [tex]0[/tex] for [tex]x[/tex] in the above equation.
[tex]\begin{aligned}3y-(0)&=6\\3y&=6\\y&=2\end{aligned}[/tex]
From the above calculation it is concluded that the line passes through the point [tex](0,2)[/tex].
This implies that option 2 is incorrect.
Option 3:
In option 3 it is given that the line perpendicular to AB passes through the point [tex](0,2)[/tex].
The equation of the line which is perpendicular to AB is [tex]3y-x=6[/tex].
Substitute [tex]0[/tex] for [tex]x[/tex] in the above equation.
[tex]\begin{aligned}3y-(0)&=6\\3y&=6\\y&=2\end{aligned}[/tex]
From the above calculation it is concluded that the line passes through the point [tex](0,2)[/tex].
This implies that option 3 is correct.
Option 4:
In option 4 it is given that the line perpendicular to AB passes through the point [tex](2,0)[/tex].
The equation of the line which is perpendicular to AB is [tex]3y-x=6[/tex].
Substitute [tex]2[/tex] for [tex]x[/tex] in the above equation.
[tex]\begin{aligned}3y-(2)&=6\\3y&=8\\y&=\frac{8}{3}\end{aligned}[/tex]
From the above calculation it is concluded that the line passes through the point [tex](2,\frac{8}{3})[/tex].
This implies that option 4 is incorrect.
Therefore, the line perpendicular to the line AB passes through the point [tex](0,2)[/tex] i.e., [tex]\fbox{\begin\\\ \bf option 3\\\end{minispace}}[/tex].
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Answer details:
Grade: High school
Subject: Mathematics
Chapter: Lines
Keywords: Geometry, coordinate geometry, lines, equation, graph, curve, slope, perpendicular, point slope form, slope intercept form.
The line perpendicular to the line AB passes through the point[tex](0,2)[/tex] i.e., option [tex]3[/tex]
From the given figure in the question it is observed that the line AB passes through the points and .
The coordinate of point C is .
Obtain the slope of the line AB.
The slope of a line which passes through the points [tex](x_1 ,y_1)[/tex] and [tex](x_2,y_2)[/tex] is as follows:
[tex]m=\dfrac{{y_2}-{y_1}}{{x_2}-{x_1}}[/tex]-----------------1
It is given that the line AB passes through the points [tex](-2,4)[/tex] and [tex](2,-8)[/tex] .
To obtain the slope for the line AB substitute for , for , for , for , in equation (1).
[tex]m=\dfrac{-12}{4}[/tex]
[tex]m=-3[/tex]
Therefore, the slope of the line AB is [tex]-3[/tex] .
Consider the slope of AB as [tex]m_1[/tex],so [tex]m_1=-3[/tex].
Now the slope of the perpendicular line.
The slope of line AB is [tex]m_1=-3.[/tex]
Consider a line which is perpendicular to the line AB passing through the point C. Assume the slope of the perpendicular line as [tex]m_2[/tex] .
The product of slope of two mutually perpendicular lines is always equal to [tex]-1[/tex] .
The equation formed for the slope is as follows:
Substitute the value of in the above equation.
Therefore, the slope of the perpendicular line is [tex]m_2=\dfrac{1}{3}[/tex] .
Now find the equation of the perpendicular line.
The slope for perpendicular line is [tex]\dfrac{1}{3}[/tex] and the line passes through the point C. The coordinate for the point C are [tex](6,4)[/tex] .
The point slope form of a line is as follows:
[tex](y-y_1)=m(x-x_1)[/tex]
Substitute [tex]\dfrac{1}{3}[/tex] for [tex]m[/tex] , [tex]6[/tex] for [tex]x_1[/tex]. and [tex]4[/tex] for [tex]y_1[/tex] in the above equation.
[tex](y-3)=\dfrac{1}{3}{(x-6)}[/tex]
[tex]3y-12=x-6[/tex]
[tex]3y-x=6[/tex]
Therefore, the equation of the perpendicular line is [tex]3y-x=6[/tex] .
Option 1:
In option 1 it is given that the line perpendicular to AB passes through the point .
The equation of the line which is perpendicular to AB is [tex]3y-x=6[/tex] .
Substitute[tex]-6[/tex] for [tex]x[/tex] in the above equation.
[tex]3y-(-6)=6\\[/tex]
[tex]3y+12=6[/tex]
[tex]3y=-6[/tex]
[tex]y=-2[/tex]
From the above calculation it is concluded that the line passes through the point [tex](-6,-2)[/tex] .
This implies that option 1 is incorrect.
Option 2:
In option 2 it is given that the line perpendicular to AB passes through the point [tex](0,-6)[/tex] .
The equation of the line which is perpendicular to AB is .
Substitute for in the above equation.
[tex]3y=6[/tex]
[tex]y=2[/tex]
From the above calculation it is concluded that the line passes through the point .
This implies that option 2 is incorrect.
Option 3:
In option 3 it is given that the line perpendicular to AB passes through the point .
The equation of the line which is perpendicular to AB is .
Substitute [tex]0[/tex] for [tex]x[/tex] in the above equation.
[tex]3y-(0)=6[/tex]
[tex]3y=6[/tex]
[tex]y=2[/tex]
From the above calculation it is concluded that the line passes through the point [tex](0,2)[/tex] .
This implies that option 3 is correct.
Option 4:
In option 4 it is given that the line perpendicular to AB passes through the point .
The equation of the line which is perpendicular to AB is [tex]3y-x=6[/tex] .
Substitute for in the above equation.
From the above calculation it is concluded that the line passes through the point [tex](2,\dfrac{8}{3})[/tex] .
This implies that option 4 is incorrect.
Therefore, the line perpendicular to the line AB passes through the point [tex](0,2)[/tex] i.e., Option [tex]3[/tex]
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