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Write an equation that goes through (8,1) and is perpendicular to 2y + 4x =12

Sagot :

VirαtKσhli

Answer:

2y -x + 6 = 0

Step-by-step explanation:

Here a equation and a point is given to us and we are interested in finding a equation which is perpendicular to the given equation and passes through the given point .

The given equation is ,

[tex]\sf \longrightarrow 2y + 4x = 12 \\ [/tex]

[tex]\sf \longrightarrow 4x + 2y = 12 [/tex]

Firstly convert this into slope intercept form , to find out the slope of the line.

[tex]\sf \longrightarrow 2y = -4x +12\\ [/tex]

[tex]\sf \longrightarrow y =\dfrac{-4x+12}{2}\\ [/tex]

[tex]\sf \longrightarrow y =\dfrac{-4x}{2}+\dfrac{12}{2}\\ [/tex]

[tex]\sf \longrightarrow \red{ y = -2x + 6 } [/tex]

Now on comparing it to slope intercept form which is y = mx + c , we have ,

  • m = -2

And as we know that the product of slopes of two perpendicular lines is -1 . So the slope of the perpendicular line will be negative reciprocal of the slope of the given line. Therefore ,

[tex]\sf \longrightarrow m_{\perp}=\dfrac{-1}{-2}=\red{\dfrac{1}{2}} [/tex]

Now we may use the point slope form of the line to find out the equation of the line using the given point . The point slope form is,

[tex]\sf \longrightarrow y - y_1 = m(x - x_1) [/tex]

Now on substituting the respective values we have,

[tex]\sf \longrightarrow y - 1 = \dfrac{1}{2}(x-8)\\ [/tex]

[tex]\sf \longrightarrow 2(y-1 )= x -8 \\[/tex]

[tex]\sf \longrightarrow 2y -2=x-8\\ [/tex]

[tex]\sf \longrightarrow \underline{\boxed{\bf 2y - x + 6=0}} [/tex]

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