The equation for line A is shown below: Y= -2/3x-4 Line A and Line B are perpendicular, and the point (-2 , 1) lies on Line B. Which equation Below represents line B?

Answer:

[tex]y = \frac{2}{3}x + \frac{7}{3}[/tex]

Step-by-step explanation:

Given

Line A:

[tex]y = -\frac{2}{3}x - 4[/tex]

Line B:

[tex](-2,1)[/tex]

Required

Determine the equation of Line B

First, we need to determine the slope of A using

[tex]y =mx + b[/tex]

Where:

[tex]m = slope[/tex]

By comparison:

[tex]y = -\frac{2}{3}x - 4[/tex]

[tex]m = -\frac{2}{3}[/tex]

Since A and B are perpendicular:

The slope of B (m2) is:

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

[tex]m_2 = -1/\frac{-2}{3}[/tex]

[tex]m_2 = -1 * \frac{-3}{2}[/tex]

[tex]m_2 = \frac{3}{2}[/tex]

The equation of B is calculated as thus:

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

Where

[tex](x_1,y_1) = (-2,1)[/tex]

[tex]m = \frac{2}{3}[/tex]

So, the equation becomes

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

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

Open bracket

[tex]y - 1 = \frac{2}{3}x + \frac{4}{3}[/tex]

Make y the subject

[tex]y = \frac{2}{3}x + \frac{4}{3} + 1[/tex]

[tex]y = \frac{2}{3}x + \frac{4 + 3}{3}[/tex]

[tex]y = \frac{2}{3}x + \frac{7}{3}[/tex]

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