Could somebody help me understand this SAT Prep question
the explanation is below, what is not clear there
I don't get it.
Is not clear enough
ok, just start doing the things the paragraph says, plot R and plot S maybe if you want first |dw:1452204504529:dw|
can't tell where point T is yet, they also tell you that the sum of the 3 slopes for the sides of the triangle add up to 1, first get the slope of the thing ya know RS, 3/3 or 1 just that side has the entire slope sum..how can that be, if you have to add 2 more sides to that?
recall using the same idea for calculating the slope, a horizontal line has a slope = 0, and a vertical line has a undefined slope (rise/0)-und.
so for the two remaining sides to not have slope values, since RS already is the entire total slope of 1, the two unknown sides must be horizontal and vertical, and perpendicular to each other
So since the slope is = 1 is 45°
I get horizontal so its sum is equal to zero but why vertical ?
And perpendicular to each other ?
that is true a line slope 1 forms a 45 angle with the x axis, or the horizontal side, but not really needed to solve the prob
vertical has an undefined slope, rise/run is some rise an d zero run, that is undefinedm dividing by 0
horizontal slope - 0 / # = 0 vertical slope - # / 0 = undefined
I get that but I don't understand why you say the other two have to be horizontal, vertical and perpendicar go each other ? Is because it is a triangle ?
*to each other
the sum of the 3 slopes for the sides of this triangle total to 1 the given side has slope 1, the remaining 2 sides must not have any slope value
oh the answer is A - NONE of the angles are right, forget that, i thought adding a undefined slope would break something, forget all that..
And because they need to intersect it is impossible
so the remaining two sides should have 2 slopes that total to 0, right, one + and one -
And is it possible to get such triangle ?
|dw:1452205612762:dw| random point T(x,y)q
I get that but I don't get how to get the other two line segments
Total of 3 sides slopes = 1 slope RS = 1 so slope of RT and slope of RS will have the same value, but opposite sign
sorry RT and TS must have same slopes with opposite sines +/-, since they have to total to 0
Ohh OK. I see however how would that look like ?
Would that actually form a triangle ?
yes, you can make a triangle with those conditions, but not a right triangle i am pretty sure, slope RT = (y - 3)/(x - 2) Slope ST = (y - 6) / (x - 5)
could you please please draw it? is just hard to visualize it :(
i mean the possible triangle based on those slopes
those must total to 0 , so slope RS = - slope ST or - slope RS = slope ST the possible point can be above RS or below, forming a parallelogram
|dw:1452206325764:dw|
start at R , go to T with a neg slope, then from T to S with the same positive slope, there are probably infinite answers, you can get the coordinate of T(x,y) as a equation y in terms of x, and that will have the possible solution values i think, in either case, there will be no 90 angles in the triangle,
none of those slopes can be perpendicular
in general slopes 1 , +m, and -m total to 1 like it says +m and -m are not perpendicular, recall perpendicular slopes are m and -1/m
to be perpenticular it was need an undefined slope and a horizontal one
needed*
to be perpendicular to RS with slope 1 , you need a slope of -1/1 = -1 then the third slope would also have to be +1 to total all 3 to 1, that will not make a triangle, you have 2 sides that are parallel with sloipe 1
that shows you cant have a right angle triangle with those givens slopes in the triangle sides are --- 1 , +m , -m cant do it and have a right angle
you can do what i think i was trying, solve th eslopes as a equation of y and x y(x) may say what points actually will work, the possible intervals
I thought to have a right traingle you needed a horizontal line. A slope of 1 and an undefined slope
yeah i said that at first also as a way to make a right triangle here, but adding undefined slope doesnt give 1 for the total of the 3 slopes, no vertical line possible
sooo to see that i am understanding
1 + 0 + undefined = undefined, not 1
The part that i am confused on is that you say that you can make a right triangle by having slopes 1, -m and m I know that is the only way to have a slope of 1 but those that actually makes a right triangle?
does that *
no i am saying that it is impossible to make a right triangle, just from that fact. +m and -m, not right angle a right angle to RS will create a side with slope -1/1 or -1 this is not possible with the rules
-1/1 what do you mean?
in the case you draw a perpendicular line to RS, and try that way, no way to make a triangle , the slopes will be 1 , -1 , 1 2 sides are parallel slope 1, not triangle
RS has slope 1, if you want a right angle at R or S, the slope or RT or ST will become -1/1 = -1 a perpendicular slope is the negative reciprical of the given slope
ohh ok I see
ohh i see what you are saying now. It is impossible because based on the given conditions, if you want to slopes to add up to 3 to line segments need to be parallel
two line segments*
just for fun, since RT and ST have opposite slopes, plugging in the values and solving for y in slope RT = - slopeST i got \[y = \frac{ 9x - 27 }{ 2x -7 }\] those values are the possible (x,y) points that work to make a triangle
How you did that?
just let T be an arbitrary point T(x,y) the slopes RT and ST can be formed with that, then equate them and make one the negative of the other, solve for y
notice there is a vertical asmytpe value when x=7/2, the denominator in that becomes 0 i bet when x=7/2, that is where you can get the standard horizontal vertical right triangle with RS
no right triangle possible, and from y(x) there, any other value you want for x, and the resulting y, will work for the point T to make a triangleRST
lol, i am done, for the actual prob realize side slopes 1, +m,-m can not make a right angle at any of those vertex points
Thanks i got do confused after that lol
Thanks a lot for taking The time to explain to me :)
no prob, you can make any triangle you wish, any point i think will work, except (7/2 , 0) and the mirror point over RS the location for T can be anything but that, and have slope values arbitrarily close to vertical or horizontal,
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