Is anyone good at math? because I need some help with this one math problem please.?

Question

The math problem is :

Lines p,q, and r all pass through point (-3,4). Line p has slope 4 and is perpendicular to line q. Line r passes through quadrants 1 and 2 only. Write and equation for each line.

Answer 1

i miss doing those...

ok, lets do this step by step if u dont understand:



First, let's look at Line P. The problem already gives u the slope and a point the line goes thru. So, the easiest thing to do is write an equation for it in point-slope form, which is y – y1 = m (x – x1). The 1's r actually subscripts (they're just there to help u remember), and x1 & y1 r points on the line. M is the slope. So, you'd plug in 4 for y1 and -3 for x1, and 4 for m. The equation for P is:



y - 4 = 4 (x - (-3))

y - 4 = 4 (x + 3) <---- equation for P



If u need it in slope-intercept form (y=mx+b) just solve for y.





The problem states that line Q is perpendicular to line P, so the slope of line Q has to be the negative reciprocal of the slope of line P. In other words, since slope of P is 4, slope of Q must be -1/4. Then u just plug everything into the equation for point-slope form:



y - 4 = -1/4 (x - (-3))

y - 4 = -1/4 (x+3) <---- equation for Q



If u need it in slope-intercept form (y=mx+b) just solve for y.





For line R, the problem says that it only passes thru quadrants 1 & 2. So, it can't have a slope other than 0 b/c if it did, it would gradually increase or decrease until it crosses the x-axis into one of the other quadrants. Again using point slope, the equation would be:



y - 4 = 0 (x+3)

y - 4 = 0

y = 4 <---- equation for R





=)

Answer 2

The equation for line p:



Knowns: slope and one point (-3.4)



y = 4x + b (b is the y-intercept)



4 = 4 * (-3) + b



4 = b - 12



[correction: sorry, got the addition wrong]

b = 16



For line p: y = 4x + 16



-------------------------



Repeat the process for line q using -(1/4) for the slope value since it's perpendicular to line p.



-------------------------



Sorry, I've forgotten the quadrant numbering scheme. I'm assuming that quadrants 1 and 2 are the portions of the graph on one one side or the other of an axis.



If quadrants 1 and 2 encompass all positive y values, the line must be horizontal: y = 4.



If quadrants 1 and 2 encompass all negative x values, the line must be vertical: x = -3.

Answer 3

Eqn for p:

(y - 4)/(x + 3) = 4

y - 4 = 4(x + 3) = 4x + 12

y = 4x + 16



Eqn for q:

(y - 4)/(x + 3) = - 1/4, the perpendicular line

4 (y - 4) = - (x + 3)

4y - 16 = -x - 3

4y + x = 13



Eqn for r:

For it to stay in quadrants 1 and 2, it must be a horizontal line

or a line of 0 slope

(y - 4) / (x + 3) = 0

y - 4 = 0

Answer 4

Line P

--point: -3, 4

--slope: 4

--equation: y - y1 = m ( x-x1)

so.. y-4 = 4 (x - -3)

so.. y-4 = 4 (x + 3)

so.. y = 4x + 12 + 4

so.. y = 4x + 16



Line Q

--point: -3, 4

--slope: -1/4

--equation: y - y1 = m ( x-x1)

so.. y-4 = -1/4 (x - -3)

so.. y-4 = -1/4 (x + 3)

so.. y = -1/4x - 3/4 + 4

so.. y = -1/4x + 13/4



Too little info for Line R

Answer 5

P:

m = 4 and (x1,y1) = (-3, 4)

now substitute in point-slope: y-y1 = m(x-x1)



Q:

m = -1/4

point-slope



R: must pass through the origin (0, 0)

m = (4-0)/(-3-0) = - 4/3

point-slope

Answer 6

ooh i miss these..



(4 - y)/(-3 - x) = 4 solve that its your equation for p



(4 - y)/(-3 - x) = -1/4 for q



i dont get the r bit because i havent done these in years.. and i probably shouldnt be helping you.. lol