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What is the y-intercept of the equation of the line that is perpendicular to the e y = 3/5x + 10 and passes through the point (15, –5)?

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The perpendicular equation is y = -5/3x + 20.

You can find this by first realizing that perpendicular lines have opposite and reciprocal slopes. So since it starts at 3/5 we flip it and make it a negative and the new slope is -5/3. Now we can use that and the point to get the y intercept using slope intercept form.

y = mx + b

-5 = (-5/3)(15) + b

-5 = 25 + b

20 = b

And now we can use our new slope and new intercept to model the equation.

y = -5/3x + 20

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