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(-5,5) and is perpendicular to y=-3x+1 in point slope form

User Shequana
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Final Answer:

The equation of the line that passes through the point (-5,5) and is perpendicular to y = -3x + 1 in point slope form is y = (1/3)x + 20/3.

Step-by-step explanation:

Sure, let's find the equation of the line that passes through the point (-5,5) and is perpendicular to the line with the equation y = -3x + 1.

When two lines are perpendicular, the slope of one line is the negative reciprocal of the slope of the other. The slope (m) of the given line y = -3x + 1 is -3. The slope of the line perpendicular to it will therefore be the negative reciprocal of -3, which is 1/3.

The point-slope form of the equation of a line is given by:

y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a point on the line.

Here, we have the slope m = 1/3 and the point (x₁, y₁) = (-5,5).

Now place these values into the point-slope form:

y - 5 = (1/3)(x - (-5))

Simplify the equation:

y - 5 = (1/3)(x + 5)

Distribute the 1/3:

y - 5 = (1/3)x + (1/3) * 5

y - 5 = (1/3)x + 5/3

Now we can add 5 to both sides to solve for y:

y = (1/3)x + 5/3 + 15/3

y = (1/3)x + 20/3

Thus, the equation of the line that passes through the point (-5,5) and is perpendicular to the line y = -3x + 1 in point-slope form is:

y = (1/3)x + 20/3

User Roman Pfneudl
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