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find the equation of a straight line perpendicular to y=3x+2, which goes through the point (6,0)

Sagot :

chibabykalu16

Answer:

y = -1/3x + 2

Step-by-step explanation:

Given:

  • Original line equation: y = 3x + 2
  • Point through which the perpendicular line passes: (6, 0)

To find the equation of a straight line perpendicular to y = 3x + 2 and passing through (6, 0):

1. Determine the slope of the given line:

The slope (m) of y = 3x + 2 is 3.

2. Find the slope of the perpendicular line:

[tex]m_{\text{perpendicular}} = -\frac{1}{3}[/tex]

3. Use the point-slope form of the equation of a line:

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

Where:

  • m = -1/3
  • (x₁, y₁) = (6, 0)

4. Substitute the values into the point-slope form:

[tex]y - 0 = -\frac{1}{3}(x - 6)[/tex]

Simplify:

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

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

Therefore, the equation of the line perpendicular to y = 3x + 2 and passing through (6, 0) is:

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

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