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If the coordinates of point B are (-3, 4) and the coordinate

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If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 02 Nov 2018, 11:42
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57% (01:02) correct 42% (00:49) wrong based on 26 sessions
If the coordinates of point B are (-3, 4) and the coordinates of point C are (-7, 7), what is the area of the parallelogram?

A. 1

B. 2 \(\sqrt{7}\)

C. 7

D. 8

E. 7 \(\sqrt{2}\)
[Reveal] Spoiler: OA

Last edited by Sawant91 on 02 Nov 2018, 11:55, edited 1 time in total.
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 02 Nov 2018, 11:45
I solved as below:

Area of parallelogram = base * height

According to me base is 7-4=3
height = -3+7=4

put this value in formula and i got an answer is 12 which is wrong.

Please help me to solve this question
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 02 Nov 2018, 11:53
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Sawant91 wrote:
I solved as below:

Area of parallelogram = base * height

According to me base is 7-4=3
height = -3+7=4

put this value in formula and i got an answer is 12 which is wrong.

Please help me to solve this question


Hi, there was an image associated with this. Could you share the question number and the test?
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 02 Nov 2018, 11:57
Quantitative Section Test1 -> Question number 13

Image uploaded.
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 02 Nov 2018, 16:13
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Sawant91 wrote:
Quantitative Section Test1 -> Question number 13

Image uploaded.



You have the base and height wrong. In a parallelogram shown in the figure below the base and height are marked. In this case, it would be a very long calculation to calculate the base and height from the given coordinates.

Attachment:
InkedCapture_LI.jpg
InkedCapture_LI.jpg [ 815.79 KiB | Viewed 3637 times ]


The problem is solved by calculating the area of the triangle ABD.

We can find area of triangle with Shoelace formula:

Area=\(\frac{1}{2}(x_1(y_2−y_3)+x_2(y_3−y_1)+x_3(y_1−y_2))\)

Where \(x_n, y_n\) are coordinates of the vertices of a triangle. Now one vertex is (0,0) so the formula reduces to :

Area=\(\frac{1}{2}(x_1(y_2)+x_2(−y_1))\) \(x_3=0\) and \(y_3=0\).

And the area of the triangle is indeed \(\frac{1}{2}\) area of the parallelogram ABCD.

Hence the answer is 7.
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 03 Nov 2018, 05:44
Thanks for your response.

I tried to solve by matrix method and got the answer is 7/2 for triangle BCD. I assumed the area of parallelogram is twice the triangle as there are two triangle present.

Is my understanding correct?
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 04 Nov 2018, 05:27
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Sawant91 wrote:
Thanks for your response.

I tried to solve by matrix method and got the answer is 7/2 for triangle BCD. I assumed the area of parallelogram is twice the triangle as there are two triangle present.

Is my understanding correct?


Yes, that is correct. Well done, it is a tough question.
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 16 Nov 2018, 18:40
Is there any other easy approach?
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 18 Nov 2018, 14:02
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AE wrote:
Is there any other easy approach?


Unfortunately, I think this might be the easiest method available.
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 18 Nov 2018, 20:56
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AE wrote:
Is there any other easy approach?


You may look at it this way..
1) measure if diagonals
a) C is (-7,7) so we are looking for hypotenuse with sides 7 each
So \(\sqrt{7^2+7^2}=7√2\)
b) diagonal BD
Since B is (-3,4), D is mirror image from line AC which is x=y
So D is (-4,3)
Diagonal BD is hypotenuse with sides (-3-(-4) and (4-3) so 1 each..
Diagonal = \(\sqrt{1^2+1^2}=√2\)


Now area will be half of product of diagonals = 7√2*√2*1/2=7*2/2=7
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 22 Sep 2019, 05:10
sandy wrote:
Sawant91 wrote:
Quantitative Section Test1 -> Question number 13

Image uploaded.



You have the base and height wrong. In a parallelogram shown in the figure below the base and height are marked. In this case, it would be a very long calculation to calculate the base and height from the given coordinates.

Attachment:
InkedCapture_LI.jpg


The problem is solved by calculating the area of the triangle ABD.

We can find area of triangle with Shoelace formula:

Area=\(\frac{1}{2}(x_1(y_2−y_3)+x_2(y_3−y_1)+x_3(y_1−y_2))\)

Where \(x_n, y_n\) are coordinates of the vertices of a triangle. Now one vertex is (0,0) so the formula reduces to :

Area=\(\frac{1}{2}(x_1(y_2)+x_2(−y_1))\) \(x_3=0\) and \(y_3=0\).

And the area of the triangle is indeed \(\frac{1}{2}\) area of the parallelogram ABCD.

Hence the answer is 7.


Hi Sandy,
How we want to find the area of ABD and we are using the coordinate of C?
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 22 Sep 2019, 09:23
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Asmakan wrote:
Hi Sandy,
How we want to find the area of ABD and we are using the coordinate of C?


Hi,
Unfortunately Sandy, is not with us :(

However, to answer your ques. - Simple trick don't complicate with formulas 8-)

Plz see the attach diag.

Let us extended the line from co-ordinate C to point M (-7,0) and to Point O (0,7)
Extend the line from the co-ordinate B (-3,4) to Point O (0,7)

Extend the line from the co-ordinate D (4,-3) to Point M (-7,0) ( co-ordinate of point D = mirror image of the co-ordinate of Point B and hence the sign changes and value gets interchanged)

Now you can see 4 \(\triangle\) s

Since we need the Area of the parallelogram ABCD and that can be found = Area of the square AMCO - Sum of the Areas of all the 4 \triangle s

i.e. Area of the parallelogram ABCD = Area of the square AMCO - { Area of \(\triangle\) AMD + Area of \(\triangle\) DMC + Area of \(\triangle\) COB + Area of \(\triangle\) OAB }

First :

Area of the square \(AMCO = 7 * 7 = 49\) (distance between each point =7)

Now,

Area of \(\triangle\) AMD = \(\frac{1}{2} * 7 * 3\) = \(\frac{21}{2}\)( base =7 and height = distance from base to point D = 3)

Area of \(\triangle\) DMC = \(\frac{1}{2}* 7 * 3\) = \(\frac{21}{2}\)( base =7 and height = distance from base to point D = 3)

Area of \(\triangle\)COB = \(\frac{1}{2} * 7 * 3\) =\(\frac{21}{2}\)( base =7 and height = distance from base to point B = 3)

Area of \(\triangle\) OAB = \(\frac{1}{2} * 7 * 3\) =\(\frac{21}{2}\)( base =7 and height = distance from base to point B = 3)


Sum of all 4 \(\triangle\) s = \({4* \frac{21}{2}} = 42\)

Therefore Area of the parallelogram ABCD = 49 - 42 =7
Attachments

InkedCapture_LI.jpg
InkedCapture_LI.jpg [ 25.69 KiB | Viewed 2246 times ]


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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 23 Dec 2019, 12:36
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pranab01 wrote:
Asmakan wrote:
Hi Sandy,
How we want to find the area of ABD and we are using the coordinate of C?


Hi,
Unfortunately Sandy, is not with us :(

However, to answer your ques. - Simple trick don't complicate with formulas 8-)

Plz see the attach diag.

Let us extended the line from co-ordinate C to point M (-7,0) and to Point O (0,7)
Extend the line from the co-ordinate B (-3,4) to Point O (0,7)

Extend the line from the co-ordinate D (4,-3) to Point M (-7,0) ( co-ordinate of point D = mirror image of the co-ordinate of Point B and hence the sign changes and value gets interchanged)

it is the best approach. Thanks!

Now you can see 4 \(\triangle\) s

Since we need the Area of the parallelogram ABCD and that can be found = Area of the square AMCO - Sum of the Areas of all the 4 \triangle s

i.e. Area of the parallelogram ABCD = Area of the square AMCO - { Area of \(\triangle\) AMD + Area of \(\triangle\) DMC + Area of \(\triangle\) COB + Area of \(\triangle\) OAB }

First :

Area of the square \(AMCO = 7 * 7 = 49\) (distance between each point =7)

Now,

Area of \(\triangle\) AMD = \(\frac{1}{2} * 7 * 3\) = \(\frac{21}{2}\)( base =7 and height = distance from base to point D = 3)

Area of \(\triangle\) DMC = \(\frac{1}{2}* 7 * 3\) = \(\frac{21}{2}\)( base =7 and height = distance from base to point D = 3)

Area of \(\triangle\)COB = \(\frac{1}{2} * 7 * 3\) =\(\frac{21}{2}\)( base =7 and height = distance from base to point B = 3)

Area of \(\triangle\) OAB = \(\frac{1}{2} * 7 * 3\) =\(\frac{21}{2}\)( base =7 and height = distance from base to point B = 3)


Sum of all 4 \(\triangle\) s = \({4* \frac{21}{2}} = 42\)

Therefore Area of the parallelogram ABCD = 49 - 42 =7


It is the best approach! Thanks!
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 22 Jul 2020, 00:47
Formula of Area of Parallelogram = 2 * Area of Triangle
Formula of Area of Triangle when 2 coordinate points are given:
|ad-bc| / 2
(a,b) = (-3,4)
(c,d) = (-7,7)

Area of Triangle: (ad - bc) /2 = (-3*7) - (4*-7) / 2 = (-21 + 28) / 2 = 7/2
Area of Parallelogram = 2 * 7/2 = 7 Option C


Please correct me if I am wrong
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 22 Jul 2020, 18:48
Farina wrote:
Formula of Area of Parallelogram = 2 * Area of Triangle
Formula of Area of Triangle when 2 coordinate points are given:
|ad-bc| / 2
(a,b) = (-3,4)
(c,d) = (-7,7)

Area of Triangle: (ad - bc) /2 = (-3*7) - (4*-7) / 2 = (-21 + 28) / 2 = 7/2
Area of Parallelogram = 2 * 7/2 = 7 Option C


Please correct me if I am wrong


Hi
This approach is not correct,

when you mention "ad" - it doesn't mean = a * d, infact it is the distance from "a" to "d"

Here is an example from khan academy



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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 22 Jul 2020, 23:40
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pranab223 wrote:
Farina wrote:
Formula of Area of Parallelogram = 2 * Area of Triangle
Formula of Area of Triangle when 2 coordinate points are given:
|ad-bc| / 2
(a,b) = (-3,4)
(c,d) = (-7,7)

Area of Triangle: (ad - bc) /2 = (-3*7) - (4*-7) / 2 = (-21 + 28) / 2 = 7/2
Area of Parallelogram = 2 * 7/2 = 7 Option C


Please correct me if I am wrong


Hi
This approach is not correct,

when you mention "ad" - it doesn't mean = a * d, infact it is the distance from "a" to "d"

Here is an example from khan academy



Regards


Thanks for correcting me. I saw this explanation which I have written on GREClub test explanations.
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 23 Jul 2020, 06:02
Farina wrote:
pranab223 wrote:
Farina wrote:
Formula of Area of Parallelogram = 2 * Area of Triangle
Formula of Area of Triangle when 2 coordinate points are given:
|ad-bc| / 2
(a,b) = (-3,4)
(c,d) = (-7,7)

Area of Triangle: (ad - bc) /2 = (-3*7) - (4*-7) / 2 = (-21 + 28) / 2 = 7/2
Area of Parallelogram = 2 * 7/2 = 7 Option C


Please correct me if I am wrong


Hi
This approach is not correct,

when you mention "ad" - it doesn't mean = a * d, infact it is the distance from "a" to "d"

Here is an example from khan academy



Regards


Thanks for correcting me. I saw this explanation which I have written on GREClub test explanations.


No worries,

Would you mind providing the question ID and question set of the GRECLUB test.

Regards
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 29 Jul 2020, 11:57
sandy wrote:
Sawant91 wrote:
Thanks for your response.

I tried to solve by matrix method and got the answer is 7/2 for triangle BCD. I assumed the area of parallelogram is twice the triangle as there are two triangle present.

Is my understanding correct?


Yes, that is correct. Well done, it is a tough question.


Hi, I tried solving this by finding the distance between the two points. For example, I first found the distance between point B and origin. Then point B and C. I was obviously wrong in this approach, but could you tell me how I was wrong here?
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 29 Jul 2020, 12:09
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ant99 wrote:
sandy wrote:
Sawant91 wrote:
Thanks for your response.

I tried to solve by matrix method and got the answer is 7/2 for triangle BCD. I assumed the area of parallelogram is twice the triangle as there are two triangle present.

Is my understanding correct?


Yes, that is correct. Well done, it is a tough question.


Hi, I tried solving this by finding the distance between the two points. For example, I first found the distance between point B and origin. Then point B and C. I was obviously wrong in this approach, but could you tell me how I was wrong here?


Please Sir,

read the explanations above. They go in depth
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink] New post 29 Jul 2020, 17:48
pranab223 wrote:

No worries,

Would you mind providing the question ID and question set of the GRECLUB test.

Regards

Quant Section 1: ID: Q02-51
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Re: If the coordinates of point B are (-3, 4) and the coordinate   [#permalink] 29 Jul 2020, 17:48
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