Answered

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Write an equation for a line perpendicular to y=3x+1 and passing through the point (6,2)

Sagot :

MrRoyal

Answer:

[tex]y = -\frac{1}{3}x + 4[/tex]

Step-by-step explanation:

Required

Equation of line

passes through [tex](6,2)[/tex]

In an equation of the form [tex]y =mx + b[/tex]; the slope is [tex]m[/tex]

So, by comparison;

The slope of [tex]y = 3x + 1[/tex] is: [tex]m =3[/tex]

From the question, we understand that the required equation is perpendicular to [tex]y = 3x + 1[/tex]

This means that its slope is:

[tex]m_2 =-\frac{1}{m}[/tex]

So, we have:

[tex]m_2 =-\frac{1}{3}[/tex]

The line equation is:

[tex]y = m_2(x - x_1) + y_1[/tex]

Where:

[tex](x_1,y_1) = (6,2)[/tex]

So, we have:

[tex]y = -\frac{1}{3}(x - 6) + 2[/tex]

[tex]y = -\frac{1}{3}x + 2 + 2[/tex]

[tex]y = -\frac{1}{3}x + 4[/tex]

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