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3 years ago

Write an equation in point slope form for the line that passes through (-5,6) and (-3,-9).

1 answer:

8 0

$\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{6})\qquad
(\stackrel{x_2}{-3}~,~\stackrel{y_2}{-9})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-9-6}{-3-(-5)}\implies \cfrac{-9-6}{-3+5}\implies -\cfrac{15}{2}$

$\bf \begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-6=-\cfrac{15}{2}[x-(-5)]\implies y-6=-\cfrac{15}{2}(x+5)
\\\\\\
y-6=-\cfrac{15}{2}x-\cfrac{75}{2}\implies y=-\cfrac{15}{2}x-\cfrac{75}{2}-6\implies y=-\cfrac{15}{2}x-\cfrac{87}{2}$

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4 years ago

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katrin2010 [14]

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3×=13.50
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4 years ago

Given the points (–3,k) and (2,0), for which values of k would the distance between the points be √34 ?

myrzilka [38]

Answer:

The values of k are 3 and -3

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$(-3,k)\\(2,0)$

$d=\sqrt{34}$

substitute

$\sqrt{34}=\sqrt{(0-k)^{2}+(2+3)^{2}}$

$\sqrt{34}=\sqrt{(-k)^{2}+(5)^{2}}$

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squared both sides

$34=k^{2}+25$

$k^{2}=34-25$

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4 years ago

Read 2 more answers

What is a good science fair projects for 5th graders

tino4ka555 [31]

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4 years ago

16x-10y=10<br>-8x-6y=6<br><br>can someone help me understand this question?

kkurt [141]

This is a system of equations, and I am assuming that you are being asked to solve it.

This is finding the point where these two lines intersect.

The key to solving these algebraically is finding one part in terms of another part and substituting.

We can solve this system by Gaussian elimination, first by multiplying the second equation in its entirety by 2, so that the x-terms will cancel.

We have
$16x-10y=10$
and
$2(-8x-6y=6)$ .

The last equation is rewritten as $-16x-12y=12$ .

Now, we cancel the x-terms.

$16x-10y=10$
$-16x-12y=12$

$-22y=22 \\ y = -1$

We have our value for y, so we just plug this back into any of the original equations to get x.

$16x-10(-1)=10\\16x=0\\x=0$

Thus, the solution to the system of equations is the point $(0,-1)$ .

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4 years ago