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rjkz [21]
3 years ago
7

Write an equation of the line that is perpendicular to a line with the coordinates (-3,3) (0,1) that passes through the points (

0,-2).
Mathematics
1 answer:
ziro4ka [17]3 years ago
5 0

Hey there,

This is a two-step problem and two equations are involved in figuring out the problems.

Looking at the points you know the y-intercept of both equations but you would need to solve for the slope.

First Step:

  • \frac{0-(-3) }{1-3 } = \frac{3}{-2}

The slope of the first equation would be -\frac{3}{2} now we write it in slope intercept form and get y = -\frac{3}{2} x + 1.

Second Step:

Perpendicular lines have slopes that are the negative reciprocal of the original line. So the perpendicular line would have the slope as \frac{2}{3} .

The equation would be: y = \frac{2}{3} x - 2

Hope I helped,

Amna

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Answer:

The correct options are;

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3) Use the sum identity for cosine to rewrite the denominator

4) Divide both the numerator and denominator by cos(x)·cos(y)

5) Simplify fractions by dividing out common factors or using the tangent quotient identity

Step-by-step explanation:

Given that the required identity is Tangent (x + y) = (tangent (x) + tangent (y))/(1 - tangent(x) × tangent (y)), we have;

tan(x + y) = sin(x + y)/(cos(x + y))

sin(x + y)/(cos(x + y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y))

(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y))

(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y)) = (tan(x) + tan(y))(1 - tan(x)·tan(y)

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drek231 [11]

Answer:

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