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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink]
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]
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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 9661 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]
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]
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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]
Is there any other easy approach?
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink]
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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]
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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]
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]
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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
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink]
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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]
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]
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
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Re: If the coordinates of point B are (-3, 4) and the coordinate [#permalink]
1
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]
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]
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]
Expert Reply
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]
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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