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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]  02 Nov 2018, 11:42
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Question Stats:

54% (01:12) correct 45% (00:49) wrong based on 24 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]  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.

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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink]  02 Nov 2018, 11:53
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Expert's post
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.

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]  02 Nov 2018, 11:57
Quantitative Section Test1 -> Question number 13

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

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 [ 815.79 KiB | Viewed 3630 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.

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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink]  03 Nov 2018, 05:44

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]  04 Nov 2018, 05:27
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Sawant91 wrote:

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]  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]  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]  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]  22 Sep 2019, 05:10
sandy wrote:
Sawant91 wrote:
Quantitative Section Test1 -> Question number 13

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.

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]  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

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 [ 25.69 KiB | Viewed 2239 times ]

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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink]  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

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]  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:
(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]  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:
(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
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink]  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:
(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]  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:
(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]  29 Jul 2020, 11:57
sandy wrote:
Sawant91 wrote:

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]  29 Jul 2020, 12:09
Expert's post
ant99 wrote:
sandy wrote:
Sawant91 wrote:

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?

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]  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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