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eduard
3 years ago
7

identify an equation in point-slope form for the line perpendicular to y=-4x-1 that passes through (-2,7)

Mathematics
1 answer:
german3 years ago
7 0

I find it simplest to convert to standard form, find the perpendicular, convert back.

y = -4x - 1

For standard form we move the x term to the left side:

4x + y = -1

The perpendiculars are given by swapping the x and y coefficients, negating one. The right side is directly determined by the point (-2,7):

x - 4y = -2 - 4(7) = -30

Solve for y for point slope form:

4y = x + 30

y = \frac 1 4 x + \frac{15}{2}

That's the answer.


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<h3>Answer: 1</h3>

where x is nonzero

=======================================================

Explanation:

We'll use two rules here

  • (a^b)^c = a^(b*c) ... multiply exponents
  • a^b*a^c = a^(b+c) ... add exponents

------------------------------

The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.

Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)

-------------------------------

After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2),  (b^2-c^2) and (c^2-a^2)

Add up those exponents (using rule 2 above) and we get

(a^2-b^2)+(b^2-c^2)+(c^2-a^2)

a^2-b^2+b^2-c^2+c^2-a^2

(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)

0a^2 + 0b^2 + 0c^2

0+0+0

0

All three exponents add to 0. As long as x is nonzero, then x^0 = 1

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