Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

heres another one, write an equation for the line that is perpendicular to the given line and that contains the given point. 3x + 2y = -10; (-9,-2)

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Perpendicular is a different story. If the orginal line have the slope of 3. Then a new perpendicular line must have a slope of the negative reciprocal. This means that the slope must be -(1/3).
What is the slope of the line 3x + 2y = -10? If you put it into slope-intercept form, you'll see that y = (-3/2) x - 5. So the slope is -3/2. The slope of a perpendicular line to 3x + 2y = -10 must then have slope 2/3 (the negative reciprocal of -3/2). So then we want this perpendicular line to pass through (-9, -2). So we can use “point-slope” form of a line: y – y1 = m*(x – x1). So y – (-2) = (2/3) * (x – (-9)). Simplify to get the equation of the line which is perpendicular to your first line and going through the point (-9, -2).

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question