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In the xy-plane, the points P, Q and S have coordinates

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Carcass
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In the xy-plane, the points P, Q and S have coordinates [#permalink] New post   03 Dec 2019, 02:27
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In the xy-plane, the points P, Q and S have coordinates (14,10), (1,0), and (6,0), respectively. What is the area of triangular region PQS?


A. 25

B. 30

C. 35

D. 42

E. 50


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A
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Re: In the xy-plane, the points P, Q and S have coordinates [#permalink] New post   20 Apr 2020, 15:51
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Area= |14(0-0)+1(0-10)+6(10-0)/2|
=|50/2|
=25

IMO-A
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nurakib
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Re: In the xy-plane, the points P, Q and S have coordinates [#permalink] New post   27 Feb 2021, 03:27
Carcass wrote:
Area= |14(0-0)+1(0-10)+6(10-0)/2|
=|50/2|
=25

IMO-A



What is the formula for finding Area? Please share some resources if possible.
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Re: In the xy-plane, the points P, Q and S have coordinates [#permalink] New post   27 Feb 2021, 03:51
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See here https://gre.myprepclub.com/forum/gre-math- ... -2609.html

Geometry chapter
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Re: In the xy-plane, the points P, Q and S have coordinates [#permalink] New post   27 Feb 2021, 05:31
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Carcass wrote:
In the xy-plane, the points P, Q and S have coordinates (14,10), (1,0), and (6,0), respectively. What is the area of triangular region PQS?


A. 25

B. 30

C. 35

D. 42

E. 50

Kudos for the right answer and explanation


We can do it without the formula too!

Just plot the points on a rough co-ordinate system. You will notice that the base is 5 and height is 10
Base = (6 - 1) = 5
Height = 10 [vertical distance of point P from x-axis]

Area = \(\frac{1}{2}bh = \frac{1}{2}(5)(10) = 25\)

Hence, option A
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Re: In the xy-plane, the points P, Q and S have coordinates [#permalink] New post   11 Jul 2023, 09:10
what's the source?
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Re: In the xy-plane, the points P, Q and S have coordinates [#permalink] New post   11 Jul 2023, 10:42
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Simply drop a perpendicular to the X axis from the point (14,10), the height of this perpendicular is the height of the triangle, 10 in this case. The base is the distance between the points (1,0) and (6,0) which is 5.

So Area = 1/2*5*10 = 25.

In cases where one of the sides of the triangle does not lie solely along the X or Y axes, I would simply extend the vertices of the relevant side to form a right angled triangle and employ the Pythagoras theorem to find the length of that side.
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