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3 years ago

Someone help please i am so bad at point slope stuff

1 answer:

3 0

1) Write the point-slope form of the equation of the horizontal line that passes through the point (2, 1). y = 1/2x

2)Write the point-slope form of the equation of the line that passes through the points (6, -9) and (7, 1).
m = (-9 - 1) / (6 - 7) = -10/-1 = 10
y + 9 = 10 (x - 6)
y = 10x - 69

3) A line passes through the point (-6, 6) and (-6, 2). In two or more complete sentences, explain why it is not possible to write the equation of the given line in the traditional version of the point-slope form of a line.

4)Write the point-slope form of the equation of the line that passes through the points (-3, 5) and (-1, 4).
m = (5 - 4) / (-3 - -1) = 1/-2
y - 5 = (-1/2) (x +3)
y = (-1/2)x + 7/2

5) Write the point-slope form of the equation of the line that passes through the points (6, 6) and (-6, 1).
m = (6-1)/(6 - -6) = 5 / 12
y - 6 = (5/12) (x-6)
y = (5/12)x + 17 / 2

6) Write the point-slope form of the equation of the line that passes through the points (-8, 2) and (1, -4).
m = (2 - -4) / (-8 -1) = 6 / -7
y - 2 = (-6/7) (x + 8)
y = (-6/7)x - 50 / 7

7) Write the point-slope form of the equation of the line that passes through the points (5, -9) and (-6, 1).
m = (-9 - 1) / (5 - -6) = -10 / 11
y + 9 = (-10 / 11) (x - 5)
y = (-10 / 11)x -49/11

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The price of a gallon of heating oil rose from 1.60 a gallon to 1.92 . by what percent did the price increase?

AURORKA [14]

Answer:

20%

Step-by-step explanation:

1.92 - 1.60 = 0.32

percentage increase = (0.32/1.60) * 100 =

0.2 * 100 = 20%

3 0

3 years ago

Read 2 more answers

Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6.

jok3333 [9.3K]

Answer:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Step-by-step explanation:

We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.

First, we will need the distance formula, given by:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

$\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}$

And, the distance from P to (0, -2) will be:

$\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}$

So sum of the two distances must be 6. Therefore:

$d_1+d_2=6$

Now, by substitution:

$(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6$

Simplify. We can subtract the second term from the left:

$\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}$

Square both sides:

$(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)$

We can cancel the x² terms and continue squaring:

$y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4$

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

$-8y=36-12\sqrt{x^2+(y+2)^2}$

We can divide both sides by -4:

$2y=-9+3\sqrt{x^2+(y+2)^2}$

Adding 9 to both sides yields:

$2y+9=3\sqrt{x^2+(y+2)^2}$

And, we will square both sides one final time.

$4y^2+36y+81=9(x^2+(y^2+4y+4))$

Distribute:

$4y^2+36y+81=9x^2+9y^2+36y+36$

The 36y will cancel. So:

$4y^2+81=9x^2+9y^2+36$

Subtracting 4y² and 36 from both sides yields:

$9x^2+5y^2=45$

And dividing both sides by 45 produces:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.

5 0

3 years ago

Read 2 more answers

Alguem pra me ajudar na minha prova de matematica daqui 10 min? 8 ano

o-na [289]

Noenoenoenoenoenoenoenoenoe

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2 years ago

Solve the linear system using multiplication Elimination. 4х + Зу = 8 X - 2y = 13

motikmotik

X=5, y=-4
(5,-4)
do you need the work?

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