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3 years ago

The midpoint of AB is M(4, 1). If the coordinates of A are (2, 8), what are the coordinates of B?

Mathematics

1 answer:

Lisa [10]3 years ago

8 0

Answer:

B (6 , 0)

Step-by-step explanation:

(2,8) Add 2 to x-axis and subtract 1 from y-axis

(4,1)

(6,0)

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Square's are equal on all sides so if one leg of the square is eight then the others are also and in order to find the area of a square you multiply the height by the width so 8^2 or eight squared is the expression. Your answer is the area is 64 ft^2 and the expression is : 8^2=64

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Step-by-step explanation:

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Let f be a linear function such that f(2) =5 and f(6) = -1. find an equation for f(x).

astra-53 [7]

Using f(x) = y, we know that a graph of the function contains the (x,y) points (2,5) and (6,-1). first find the slope of that line,
m = (y2 - y1)/(x2 - x1) ⇒ -6/4⇒-3/2

then using either point (I'll use the first one) solve for b in y = mx + b.
5 = (-3/2)(2) + b⇒ 5 = -3 + b⇒ 8 = b.

So y = (-3/2)x + 8 ⇒ f(x) = (-3/2)x + 8.

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3 years ago

What is the equation of the quadratic graph with a focus of (4, −3) and a directrix of y = −6?

Artyom0805 [142]

Answer:

$y=\frac{1}{6} (x-4)-\frac{9}{2}$

Step-by-step explanation:

∵ The quadratic equation form is :

y = [1/2(b - k)] (x - a)² + (b + k)/2

Where (a , b) is the focus and directrix y = k

∵ The focus is (4 , -3) and directix is y = -6

∵ $y=\frac{1}{2(-3-(-6))} (x-4)^{2}+\frac{(-3)+(-6)}{2}$

∴ $y=\frac{1}{6} (x - 4)^{2}+\frac{-9}{2}$

∴ $y=\frac{1}{6} (x-4)-\frac{9}{2}$

another way:

Assume that (x , y) is the general point on the parabola

∵ The distance between the directrix and (x , y) = the distance between the focus and (x , y)

By using the distance rule:

∵ (y - -6)² = (x - 4)² + (y - -3)² ⇒ (y + 6)² = (x - 4)² + (y + 3)

∴ y² + 12y + 36 = (x - 4)² + y² + 6y + 9

∴ 12y - 6y = (x - 4)² + 9 - 36

∴ 6y = (x - 4)² - 27 ⇒ ÷ 6

∴ y = 1/6 (x - 4)² - 9/2

4 0

3 years ago