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The vertices of triangle Q are located

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New post 20 Jul 2018, 08:47
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E

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68% (02:00) correct 32% (01:25) wrong based on 71 sessions

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The vertices of triangle Q are located at the origin and at the x- and y- intercepts of line q, respectively. If live q is described by the equation y=Ax+B, then which of the following values of A and B will result in the greatest area of triangle Q?

a. A=1, B=5
b. A=2, B=3
c. A=0.2, B=1
d. A=-5, B=-1
e. A=.01, B=-10
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Re: The vertices of triangle Q are located  [#permalink]

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New post 20 Jul 2018, 09:54
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Option E

y=Ax+B


- x- and y- intercepts:

x=0 > y=B

y = 0 > x = -B/A

- area of triangle Q = (1/2)*(|x- * y- intercepts|) = |- (B^2)/2A|


Therefore, Option E.
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New post 20 Jul 2018, 10:48
HI FillFM how did you get y = 0 > x = -B/A?

Thanks!
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Re: The vertices of triangle Q are located  [#permalink]

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New post 20 Jul 2018, 10:58
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franteraoka wrote:
HI FillFM how did you get y = 0 > x = -B/A?

Thanks!


x intercept, so y = 0.

y = Ax + B. So, if y = 0, Ax + B = 0 ...... Ax = -B .... and x = -B/A
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New post 20 Jul 2018, 11:00
FillFM makes sense! Thank you so much!
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New post 20 Jul 2018, 11:32
@fillfim sorry one more question, why is it |x- * y- intercepts| and not just x-*y-intercepts? Thanks again!
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Re: The vertices of triangle Q are located  [#permalink]

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New post 20 Jul 2018, 11:43
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franteraoka wrote:
@fillfim sorry one more question, why is it |x- * y- intercepts| and not just x-*y-intercepts? Thanks again!


sorry franteraoka !

is the modulus of (x intercept)*(y intercept)

= |-(B/A)*B |
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New post 20 Jul 2018, 14:03
FillFM makes sense, thank you very much!
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Re: The vertices of triangle Q are located  [#permalink]

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New post 23 Jul 2018, 20:10
franteraoka wrote:
The vertices of triangle Q are located at the origin and at the x- and y- intercepts of line q, respectively. If live q is described by the equation y=Ax+B, then which of the following values of A and B will result in the greatest area of triangle Q?

a. A=1, B=5
b. A=2, B=3
c. A=0.2, B=1
d. A=-5, B=-1
e. A=.01, B=-10



Just look at the orders of magnitude between A and B to get the E.

If the vertices at are X-intercept, Y-intercept and origin, that means the height of the triangle is 10 (0 --> -10) and when y=0, X will = 100 (0=.1x - 10)
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Re: The vertices of triangle Q are located  [#permalink]

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New post 26 Jul 2018, 05:20
FillFM
Can you please explain it by drawing a diagram?
I'm not able to visualise it
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Re: The vertices of triangle Q are located  [#permalink]

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New post 26 Jul 2018, 05:40
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rraman wrote:
FillFM
Can you please explain it by drawing a diagram?
I'm not able to visualise it


rraman you just need to substitute x=0 (in y = Ax+B) to find the value of y-intercept.
and then substitute y=0 (in y = Ax+B) to find the value of x-intercept.

and then put this values in the formula of the area of the triangle.

Hope it helps.
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Re: The vertices of triangle Q are located  [#permalink]

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New post 03 Aug 2018, 01:09
franteraoka wrote:
The vertices of triangle Q are located at the origin and at the x- and y- intercepts of line q, respectively. If live q is described by the equation y=Ax+B, then which of the following values of A and B will result in the greatest area of triangle Q?

a. A=1, B=5
b. A=2, B=3
c. A=0.2, B=1
d. A=-5, B=-1
e. A=.01, B=-10



I can understand the solution but why can't we just use the theory that isosceles right triangle has the biggest area? :roll:
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New post 03 Aug 2018, 10:09
rraman wrote:
FillFM
Can you please explain it by drawing a diagram?
I'm not able to visualise it


OA:E
Using Intercept form of line,The equation of a straight line whose x and y intercepts are a and b, respectively can be represented as
\(\frac{x}{a}+\frac{y}{b}=1\)
Attachment:
ab.PNG
ab.PNG [ 30.46 KiB | Viewed 175 times ]

Then area of triangle thus formed is given \(\frac{1}{2}|a|*|b|\),as third vertice is at origin.
Mode of x and y intercept is required as length of sides of triangle is always positive.
In the question stem , equation of line is given as \(y=Ax+B\)
In intercept form, it can be written as

\(\frac{x}{\frac{-B}{A}}+\frac{y}{B}=1\)

\(a. A=1, B=5\)
\(\frac{x}{\frac{-5}{1}}+\frac{y}{5}=1\)
Area \(=\frac{1}{2}*|\frac{-5}{1}|*|5| = \frac{25}{2}\)

\(b. A=2, B=3\)
\(\frac{x}{\frac{-3}{2}}+\frac{y}{3}=1\)
Area \(=\frac{1}{2}*|\frac{-3}{2}|*|3| = \frac{9}{4}\)

\(c. A=0.2, B=1\)
\(\frac{x}{\frac{-1}{0.2}}+\frac{y}{1}=1\)
Area \(=\frac{1}{2}*|\frac{-1}{0.2}|*|1| = \frac{5}{2}\)

\(d. A=-5, B=-1\)
\(\frac{x}{\frac{+1}{-5}}+\frac{y}{-1}=1\)
Area \(=\frac{1}{2}*|\frac{-1}{5}|*|-1| = \frac{1}{10}\)

\(e. A=.01, B=-10\)
\(\frac{x}{\frac{+10}{0.01}}+\frac{y}{-10}=1\)
Area \(=\frac{1}{2}*|1000|*|-10| = 5000\)
Clearly E gives largest area.
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