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Novosadov [1.4K]
3 years ago
15

What is the equation, in slope-intercept form, of the line that is perpendicular to the line

Mathematics
1 answer:
NeTakaya3 years ago
4 0

Answer:

y = -x - 4

Step-by-step explanation:

First find the slope

y - 4 = (x - 6)

y - 4 = x - 6

y = x - 6 + 4

y = x - 2 (y = mx + C)

Slope m = 1

It is said that the line is perpendicular to a point ( -2, -2)

If two lines are perpendicular, their slope will be negative reciprocal

The negative reciprocal of slope = 1 is -1

Using a slope intercept form as requested by the question

y = mx + C

Inserting the values given

(-2, -2)

We are using point slope form

y - y_1 = m ( x - x_1)

x_1 = -2

y_1 = -2

m = -1

Insert the values

y - ( -2)) = -1( x - (-2))

y + 2 = -1 ( x + 2)---- point slope form

But we are requested to give the answer in slope intercept form

y = mx + C

We have to open the bracket

y + 2 = -1(x + 2)

y + 2 = -x - 2

y = -x - 2 - 2

y = -x - 4 ( slope - intercept form)

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Y=-1/3x^2-4x-5 in vertex form
grigory [225]

Answer:

Y=-\frac{1}{3}(x+6)^2+7   [Vertex form]

Step-by-step explanation:

Given function:

Y=-\frac{1}{3}x^2-4x-5

We need to find the vertex form which is.,

y=a(x-h)^2+k

where (h,k) represents the co-ordinates of vertex.

We apply completing square method to do so.

We have  

Y=-\frac{1}{3}x^2-4x-5

First of all we make sure that the leading co-efficient is =1.

In order to make the leading co-efficient is =1, we multiply each term with -3.

-3\times Y=-3\times\frac{1}{3}x^2-(-3)\times4x-(-3)\times 5

-3Y=x^2+12x+15

Isolating x^2 and x terms on one side.

Subtracting both sides by 15.

-3Y-15=x^2+12x-15-15

-3Y-15=x^2+12x

In order to make the right side a perfect square trinomial, we will take half of the co-efficient of x term, square it and add it both sides side.  

square of half of the co-efficient of x term = (\frac{1}{2}\times 12)^2=(6)^2=36

Adding 36 to both sides.

-3Y-15+36=x^2+12x+36

-3Y+21=x^2+12x+36

Since x^2+12x+36 is a perfect square of (x+6), so, we can write as:

-3Y+21=(x+6)^2

Subtracting 21 to both sides:

-3Y+21-21=(x+6)^2-21

-3Y=(x+6)^2-21

Dividing both sides by -3.

\frac{-3Y}{-3}=\frac{(x+6)^2}{-3}-\frac{21}{-3}

Y=-\frac{1}{3}(x+6)^2+7   [Vertex form]

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Answer:

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Step-by-step explanation:

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