Asked by james
Find the area of the triangle whose vertices (on cartesian graphs) are (-1,5) , (-2,-3) & (10,1)
Answers
Answered by
Reiny
I don't know at what of math this is from.
There are several ways to do this, I will show you two ways:
1. pick one line segment to be the base of your triangle,
I picked the line from (-1,5) to (-2,-3)
slope = (-3-5)/(-2+1) = -8/-1 = 8
equation:
y-5 = 8(x+1)
8x + 8 = y-5
8x - y + 13 = 0
distance form an external point (a,b) to a line
Ax + By + C = 0 is |aA + bB + C|/√(A^2 + B^2)
so from (10,1) to our line
= |10(8) + 1(-1) + 13|/√(64+1) = 92/√65
length of base = √( (-3-5)^2 + (-2+1)^2) = √65
area of triangle = (1/2) base x height
= (1/2)(√65)(92/√65) = 46
2. The simplest way involves the cross-product of vectors
line up your 3 points in a column, repeating the point you started with at the end
-1 5
-2 -3
10 1
-1 5
Area = (1/2) | sum of downproducts - sum of upproducts |
= (1/2) |(3 - 2 + 50) - (-10 -30 -1) |
=(1/2) | 51 + 41|
= (1/2)(92) = 46
BTW, this works for all convex polygons if you know the points, make sure you list them in sequence, repeating the one you started with
There are several ways to do this, I will show you two ways:
1. pick one line segment to be the base of your triangle,
I picked the line from (-1,5) to (-2,-3)
slope = (-3-5)/(-2+1) = -8/-1 = 8
equation:
y-5 = 8(x+1)
8x + 8 = y-5
8x - y + 13 = 0
distance form an external point (a,b) to a line
Ax + By + C = 0 is |aA + bB + C|/√(A^2 + B^2)
so from (10,1) to our line
= |10(8) + 1(-1) + 13|/√(64+1) = 92/√65
length of base = √( (-3-5)^2 + (-2+1)^2) = √65
area of triangle = (1/2) base x height
= (1/2)(√65)(92/√65) = 46
2. The simplest way involves the cross-product of vectors
line up your 3 points in a column, repeating the point you started with at the end
-1 5
-2 -3
10 1
-1 5
Area = (1/2) | sum of downproducts - sum of upproducts |
= (1/2) |(3 - 2 + 50) - (-10 -30 -1) |
=(1/2) | 51 + 41|
= (1/2)(92) = 46
BTW, this works for all convex polygons if you know the points, make sure you list them in sequence, repeating the one you started with
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