Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

Please help me with my homework i really dont know how to do these its algerbra please explain how to do these because i have slow mind so please explain each problem thanks 1. Estimate the slope of a line that is perpendicular to the line through (2.07,8.95) and (-1.9,25.07). 2. Explain in words how tp write an equation in slope-intercept form that describes a line parallel to y-3=-6(x-3).

OpenStudy (anonymous):

slope = m, \[m=\frac{ y2-y1 }{ x2-x1 }\]

OpenStudy (mathstudent55):

For part 1. (a) First find the slope of the line through the points you're given, and call it m. Then (b) the slope of the perpendicular line is -1/m

OpenStudy (mathstudent55):

Use the formula that TaRJeel wrote to get the slope, m. Once you know m, do -1/m to get slope of perpendicular line.

OpenStudy (anonymous):

its says to estimate

OpenStudy (mathstudent55):

For part 2. First you need to get the given equation in slope-intercept form. To do that, you simply solve for y and write it as y = mx + b. After doing that, any line that has the same value of m and a different b is a parallel line.

OpenStudy (anonymous):

explain that please cause i dont understand a thing you're saying

OpenStudy (mathstudent55):

You still need to use the formula, but use approximations for the numbers: use (2,9) and (-2,25). Those two points are close to the ones you were given.

OpenStudy (mathstudent55):

For part 1. To estimate the slope, you have approx (2,9) and (-2,25). Using the slope formula, that gives: m = (25 - 9)/(-2 - 2) = 16/(-4) = -4

OpenStudy (anonymous):

ok let me do that hold up give me like 3 min

OpenStudy (mathstudent55):

That means the line you have has a slope of approximately -4.

OpenStudy (mathstudent55):

ok

OpenStudy (anonymous):

ok i did that a got the same answer...so whats next

OpenStudy (mathstudent55):

m = -4 is the approximate slope of the line you were given. The slope of a line perpendicular to a line of slope m is -1/m. In other words, if one line has slope m, and you want the slope of a perpendicular to it, the slope of the perpendicular is -1/m

OpenStudy (mathstudent55):

-1/m means divide -1 by m and change the sign

OpenStudy (anonymous):

im still confused im so sorry

OpenStudy (mathstudent55):

You understood the slope of -4 right?

OpenStudy (anonymous):

yes that was what i understand out of this

OpenStudy (mathstudent55):

Ok the next step of part 1 is to find the slope of the perpendicular line. What you need to know is that the slopes of 2 perpendicular lines multiply to give you -1. That means that if you divide -1 by the slope of a line you get the slope of its perpendicular.

OpenStudy (mathstudent55):

Since line you were given has slope -4, the slope of the perpendicular is -1/-4 (negative 1 divided by negative 4) and is -1/-4 = 1/4. So the slope of the perpendicular is 1/4

OpenStudy (anonymous):

ok i think i get it

OpenStudy (mathstudent55):

That takes care of part 1

OpenStudy (anonymous):

ok then what do i do next

OpenStudy (mathstudent55):

Now for part 2. An equation in slope-intercept form is an equation in the form: y = mx + b. Take the equation of part 2, and solve for y and write in the form y = mx + b: y - 3 = -6(x - 3) You want y alone on left side

OpenStudy (mathstudent55):

First you must distribute the -6 on the right side: y - 3 = -6x + 18 Now you add 3 to both sides: y = -6x + 21 Now you have an equation with only y on the left side and the right side is in the form mx + b, where m = -6 and b = 21

OpenStudy (mathstudent55):

The last step in part 2 is: A line parallel to the given line has equation y = -6x + b, where b is a number not 21

OpenStudy (anonymous):

hold on please

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!