Question 11: Please help. What is the equation of the line that is parallel to y−5=−13(x+2) and passes through the point (6,−1)?
The equation will be in slope-intercept form.

Question 11 Please help What is the equation of the line that is parallel to y513x2 and passes through the point 61 The equation will be in slopeintercept form class=

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Answer:

Step-by-step explanation:

Assume that you have two distinct and nonvertical lines, ℓ 1 {\ell _1} ℓ1 and ℓ 2 {\ell _2} ℓ2 in Slope-Intercept Form. They are parallel lines if their slopes are equal or the same. They are perpendicular lines if their slopes are opposite reciprocals of each other, or the product of their slopes equals −1.

Answer is y= -1

The equation will be y = -1/3x + 1 in slope-intercept form.

What is the equation of a line?

The general equation of a line is y = mx + c

where m is the slope of the line and c is the intercept.

A linear equation is defined as an equation in which the highest power of the variable is always one.

First, we need to determine the slope of the given line.

To do this, we convert the equation to the slope/intercept form:

y − 5 = -1/3(x+2)

y = -1/3x -2/3 + 5

y = -1/3x - 13/3

So the slope of the given line is -13.

The line passes through the point (6,−1)

Now we can use the point-slope form of a line to create the equation of the new line:

(y + 1) = (-1/3) (x - 6)

y = -1/3x + 2 -1

y = -1/3x + 1

Hence, the equation will be y = -1/3x + 1 in slope-intercept form.

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