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3 years ago

What is the equation of the line that is parallel to the given line and passes through the point (–2, 2)?

Mathematics

1 answer:

uysha [10]3 years ago

5 0

Answer:

y = 1/5 x + 12/5

Step-by-step explanation:

We are looking for the equation of a line.

We satart with the slope-intercept form of the equation of a line.

y = mx + b

We need to find m, the slope, and b, the y-intercept.

The given line is y = 1/5 x - 3.

Its slope is 1/5.

Parallel lines have equal slopes, so the slope we need is also 1/5.

Now we have for our line

y = 1/5 x + b

Now we find b.

We are given a point on the line, (-2, 2). Using the slope we already know, plug in (-2, 2) in the equation and solve for b.

2 = 1/5 * (-2) + b

10 = -2 + 5b

12 = 5b

b = 12/5

Now that we know the y-intercept, we can write the equation of the line.

y = 1/5 x + 12/5

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If x²-1 is a factor of the polynomial, both x-1 and x+1 are factors of it.

According to the remainder theorem, if a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

If x-1 and x+1 are factors of the polynomial p(x)=ax⁴+bx³+cx²+dx+e, then p(1)=0 and p(-1)=0.

$p(1)=a \times 1^4+b \times 1^3 + c \times 1^2+ d \times 1+e \\
p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\
p(1)=a+b+c+d+e \\
p(-1)=a-b+c-d+e \\ \\
p(1)=0 \\
p(-1)=0 \\ \\ \hbox{add both equations:} \\
a+b+c+d+e=0 \\
\underline{a-b+c-d+e=0} \\
2a+2c+2e=0 \\
2(a+c+e)=0 \\
a+c+e=0 \\ \\
\hbox{substitute 0 for a+c+e in the first equation:} \\
a+b+c+d+e=0 \\
(a+c+e)+b+d=0 \\
0+b+d=0 \\
b+d=0 \\ \\
\boxed{a+c+e=b+d=0} \\
\hbox{proved } \checkmark$

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