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Sagot :
The equation of a line that is also perpendicular to the line [tex]x = 11[/tex] and contains the point [tex](7,6)[/tex] is [tex]y=6[/tex]
As per the question statement, we are supposed to find the equation of a line that is also perpendicular to the line [tex]x = 11[/tex] and contains the point [tex](7,6)[/tex].
Before solving that, we need to know that the given line [tex]x = 11[/tex] is parallel to the y-axis as no point is passing through y axis hence the slope [tex]m_{1}[/tex] of the given line will be [tex]\frac{1}{0}[/tex].
The required line is perpendicular to the given line, so it will be parallel to the x-axis and hence it's slope would be slope [tex]m_{2}[/tex] = [tex]0[/tex]
We know the equation of the line passing through a some point [tex]x_{1}[/tex] and [tex]y_{1}[/tex] having slope m is given by:
[tex]y-y_{1} =m_{2} (x-x_{1})[/tex]
Here the required line is passing through (7,6), hence substituting the values of [tex]x_{1}, y_{1} ,m_{2}[/tex], we get
[tex]y-6=0(x-7)[/tex]
or [tex]y-6=0[/tex]
or [tex]y=6[/tex]
- Equation of a line: The equation of line is the algebraic method of representing the set of points which together forms a line in a coordinate system.
To learn more about equation of a line, click on the link given below:
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