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4 years ago

Write an equation in standard form for a line that goes through the points(5,1) and (7,-3).

Mathematics

1 answer:

PolarNik [594]4 years ago

7 0

Answer:

Equation of line =

y = -2x + 11

Step-by-step explanation:

Using the formula for equation of a line with two points

(y_2 - y_1) / (x_2 - x_1) = (y - y_1) / (x - x_1)

Using the points we are provided with

( 5,1)(7,-3)

x_1 = 5

y_1 = 1

x_2 = 7

y_2 = -3

Inserting those values

( -3 - 1) / (7 - 5) = (y - 1) / (x - 5)

-4/2 = (y - 1) / (x - 5)

Cross multiply

-4(x - 5) = 2 ( y - 1)

Open the brackets

-4x + 20 = 2y - 2

-4x +20 +2 = 2y

-4x + 22 = 2y

Divide through by 2 to get the value of y following the rules

y = mx + C

-4x + 22 = 2y

-4x + 22 / 2 = 2y/2

-4x + 22 / 2 = y

We can still separate

-4x / 2 + 22/ 2 = y

-2x + 11 = y

y = -2x + 11

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Which of the following is false for the hyperbola (x+5)^2/4^2 - (y-2)^2/3^2=1

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Answer:

C)One of the vertices is (-10,2)

Step-by-step explanation:

Given the hyperbola: $\dfrac{(x+5)^2}{4^2} -\dfrac{ (y-2)^2}{3^2} =1$

The standard equation for a hyperbola with a horizontal transverse axis is:

$\dfrac{(x-h)^2}{a^2} -\dfrac{ (y-k)^2}{b^2} =1$ where the center is at (h, k).

For our given hyperbola, the center (h,k)=(-5,2)

Now:

$c^2=a^2+b^2\\c^2=4^2+3^2\\c^2=25\\c=5$

Since the center is at (-5,2), its foci (-c,0) and (c,0) are:

(-5-5,2) and (-5+5,2)= (-10,2) and (0,2)

Similarly, since the center is at (-5,2), its vertices, (-a,0) and (a,0) are:

(-5-4,2) and (-5+4,2)= (−9,2), (−1,2).

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