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Find the equation of a line through (3, 1) and perpendicular to 10x + 3y = 5

User Andrewm
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1 Answer

23 votes
23 votes

Answer:

y = (3/10)x + 1/10

Explanation:

The general structure of an equation in slope-intercept form is:

y = mx + b

In this form, "m" represents the slope and "b" represents the y-intercept. First, we need to rearrange the given equation to determine the slope of the original line.

10x + 3y = 5 <----- Original equation

3y = -10x + 5 <----- Subtract 10x from both sides

y = (-10/3)x + 5 <----- Divide both sides by 3

Now, we can tell that the slope of the original line is m = -10/3. The perpendicular line should have an opposite-signed, reciprocal slope to that of the original. Therefore, the new slope is m = 3/10.

We can find the value of "b" by plugging the new slope and values from the Point (3, 1) into slope-intercept form.

m = 3/10

x = 3

y = 1

y = mx + b <----- Slope-intercept form

1 = (3/10)(3) + b <----- Insert values

1 = 9/10 + b <----- Multiply 3/10 and 3

10/10 = 9/10 + b <----- Create common denominators

1/10 = b <----- Subtract 9/10 from both sides

Now that we know the values for "m" and "b", we can construct the equation of the perpendicular line.

y = (3/10)x + 1/10

User DeA
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2.9k points