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What is equation of a line that passes through the point (3,a) that has a slope of -2 and it is perpendicular to the line ay = 2x + 4

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Answer:

To find the equation of a line that passes through the point (3, a) and is perpendicular to the line ay = 2x + 4, we need to determine the slope of the given line and then use it to find the slope of the perpendicular line. Once we have the slope and a point on the line, we can use the point-slope form of a linear equation to find the equation of the perpendicular line.

First, let's determine the slope of the given line ay = 2x + 4. We can rewrite this equation in slope-intercept form (y = mx + b), where m represents the slope:

ay = 2x + 4

Dividing both sides by a:

y = (2/a)x + 4/a

Comparing this equation with y = mx + b, we can see that the slope of the given line is 2/a.

Since we want to find a line perpendicular to this one, we know that the product of their slopes must be -1. Therefore, the slope of our perpendicular line will be -1 divided by 2/a:

m_perpendicular = -1 / (2/a)

Simplifying:

m_perpendicular = -a/2

Now that we have the slope of our perpendicular line, we can use it along with the given point (3, a) to find its equation using the point-slope form:

y - y1 = m(x - x1)

Substituting in our values:

y - a = (-a/2)(x - 3)

Expanding and simplifying:

y - a = (-a/2)x + (3a/2)

y = (-a/2)x + (3a/2) + a

y = (-a/2)x + (3a/2) + (2a/2)

y = (-a/2)x + (5a/2)

Therefore, the equation of the line that passes through the point (3, a) and is perpendicular to the line ay = 2x + 4 is y = (-a/2)x + (5a/2).

Explanation:

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