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What is an equation of the line that passes through the point (6,5) and is perpendicular to the line 3x 5y=40

User Baloo
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Answer:

To find an equation of a line that is perpendicular to another line, we need to determine the negative reciprocal of the slope of the given line.

Let's begin by putting the given line, 3x - 5y = 40, into slope-intercept form (y = mx + b), where m represents the slope:

3x - 5y = 40 (subtract 3x from both sides)

-5y = -3x + 40 (divide by -5)

y = (3/5)x - 8

The slope of this line is 3/5. The negative reciprocal of 3/5 is -5/3.

Now, we can use the point-slope form of a line to find the equation. Using the given point (6, 5) and the negative reciprocal slope (-5/3):

y - y1 = m(x - x1)

y - 5 = (-5/3)(x - 6)

Simplifying further:

y - 5 = (-5/3)x + 10/3

To convert the equation into a more standard form, we can multiply both sides by 3 to eliminate the fraction:

3(y - 5) = -5x + 10

3y - 15 = -5x + 10

Finally, rearranging the equation:

5x + 3y = 25

Therefore, an equation of the line that passes through the point (6, 5) and is perpendicular to the line 3x - 5y = 40 is 5x + 3y = 25.

User Elydasian
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