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Mathematics 14 Online
OpenStudy (anonymous):

Will somebody help explain some math problems to me? I will give a medal and become a fan.

hartnn (hartnn):

start posting the questions you need help with

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Here is my first question thank you 7. What is the slope of the line that passes through the pair of points (2, 5) and (8, 3)? (1 point) 1/3 -1/3 3 –3

hartnn (hartnn):

The slope of the line through points (x1,y1) and (x2,y2) is given by : \(\huge m=\frac{y_1-y_2}{x_1-x_2}\) now,just put the values and find m.

OpenStudy (anonymous):

ok thank you i will solve it right now

OpenStudy (anonymous):

Is it negative 1/3?

hartnn (hartnn):

YES! good work :)

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

here is the next question 11. Write an equation in point-slope form for the line through the given point with the given slope. (5, 2); m = 3 (1 point) y + 2 = 3(x – 5) y + 2 = 3x – 5 y – 2 = 3(x + 5) y – 2 = 3(x – 5)

hartnn (hartnn):

slope point form : \(\large y-y_1 = m (x-x_1)\) where point is x1,y1 = 5, 2 here

OpenStudy (anonymous):

so would it be a? this one is a little harder

hartnn (hartnn):

'a' has y+2 but according to the formula, left side = y-y1 with y1 = 2 so left side = y-2

OpenStudy (anonymous):

so d?

hartnn (hartnn):

\(\Large \checkmark \)

OpenStudy (anonymous):

thank you again. do you mind helping me on a couple more i am doing a unit test? :)

OpenStudy (anonymous):

18. Write an equation for the line that is parallel to the given line and passes through the given point. y = 3x + 7; (2, 10) (1 point) y = 3x + 11 y = 3x + 4 y = –3x + 4 y = ( )x + 4

OpenStudy (anonymous):

i just realized i for got to separate them 18. Write an equation for the line that is parallel to the given line and passes through the given point. y = 3x + 7; (2, 10) (1 point) y = 3x + 11 y = 3x + 4 y = –3x + 4 y = (-1/3 )x + 4

hartnn (hartnn):

parallel lines have same slope! so, first do u know how to find slope of y = 3x +7 ?

OpenStudy (anonymous):

Not really :(

hartnn (hartnn):

compare y= 3x+7 with y= mx+c then m= slope

OpenStudy (jdoe0001):

\(\large \bf y= \color{red}{3}x+7\\ y= \color{red}{m}x+c \)

OpenStudy (anonymous):

so would the answer be b?

hartnn (hartnn):

YES! good :)

OpenStudy (anonymous):

20. Tell whether the lines for each pair of equations are parallel, perpendicular, or neither. y = ( )x + 8 –2x + 8y = 4 (1 point) parallel perpendicular neither

hartnn (hartnn):

whats in the bracket ?? "()"

OpenStudy (anonymous):

oh -1/4

hartnn (hartnn):

ok,so comparing y = (-1/4)x + 8 with y= mx+c , we get slope of 1st line as m = -1/4

hartnn (hartnn):

to get the slope of 2nd line, we will have to isolate 'y' from –2x + 8y = 4 can you try ? start by adding 2x on both sides

OpenStudy (anonymous):

Would it be perpendicular ?

hartnn (hartnn):

nopes. –2x + 8y = 4 add 2x on both sides, what u get ?

hartnn (hartnn):

-2x +2x +8y = 4 + 2x so, 0 +8y = 2x+4 8y = 2x+4 now divide by 8 on both sides

OpenStudy (anonymous):

y=1/4x+1/2

hartnn (hartnn):

correct, so comparing that with y=mx+c we get slope = m' = +1/4 so they are neither parallel, nor perpendicular :)

OpenStudy (anonymous):

Ok thank you I will see if i need more help 1 sec

hartnn (hartnn):

sure :)

OpenStudy (anonymous):

21. Write the equation of a line that is perpendicular to the given line and that passes through the given point. y – 2 =7/3 (x + 5); (–4, 9) (1 point) y – 9 = (-3)/7 (x – 4) y – 9 =3/7 (x – 4) y – 9 = (-3)/7 (x + 4) y – 9 =3/7 (x + 4) the (-3)/7 means that it is negative 3 over 7

hartnn (hartnn):

y – 2 =7/3 (x + 5) means the slope of this line is m' = 7/3 the slope of perpendicular line will be m =-1/m' = -3/7 and you have the point x1,y1 =-4,9 use the same formula again, \(y-y_1 =m (x-x_1)\)

OpenStudy (anonymous):

would it be c?

hartnn (hartnn):

\(\huge \checkmark \quad \ddot \smile \)

OpenStudy (anonymous):

oh good i got another one right yay :)

hartnn (hartnn):

yaay, congrats :)

OpenStudy (anonymous):

I want to thank everybody for the great help Thank you :)

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