Given the equation of the line and point p not on the line, find the equation of a line parallel to the given line and a line perpendicular to the given line through the given point; y= -3x + 7; p(2,-4)

y = -3x + 2; y = ⅓x -14/3

Step-by-step explanation:

Step 1. Find the equation of the parallel line

Original line: y = -3x+ 7

Parallel line: slope = m₁ = -3

The line passes through (2, -4).

 y = m₁x + b      Insert the values

-4 = (-3)×2 + b

-4 = -6 + b        Add 6 to each side

b = 2

y = -3x + 2

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Step 2. Find the slope (m₂) of the perpendicular line

m₂ = -1/m₁     Substitute the value of m₁

m₂ = -1/(-3)

m₂ = ⅓

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Step 3. Find the equation for the perpendicular line

y = mx + b

y = ⅓x + b

The line passes through (2, -4).  

-4 = ⅓(2) + b      Substitute the values

-4 = ⅔ + b         Subtract ⅔ from each side

-4 - ⅔ = b

 b = -14/3

y = ⅓x - 14/3

In the image, below  the graph of your original equation is the red line.

The green line passing through (2, -4) is the parallel line.

The purple line passing through (2, -4) is the perpendicular to both lines.

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11