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It is known that the coordinates of each vertex of triangle ABC are a (- 1,3) B (- 2, - 1) C (2,0) respectively. Find the area of triangle?
Area of triangle = 4 * 4-1 / 2 (4 * 1 + 3 * 3 + 1 * 4) = 16-1 / 2 * 17 = 16-8.5 = 7.5
As shown in the figure, the coordinates of the three vertices of triangle ABC are a (- 3,0), B (2.5,1) and C (0, - 2) respectively. Calculate the area of triangle ABC
Let l be a straight line BC, then it is not difficult to find that the equation of L is y = 6 / 5x-2
L intersects with X axis and D, then D is (5 / 3,0) ad = 14 / 3
SABC=SADB+SADC
SADB=1/2*AD*1=7/3
SADC=1/2*AD*2=14/3
SABC=7/3+14/3=7
Given that the three vertex coordinates of triangle ABC are a (- 7,0) B (1,0) C (- 5,4), then the area of triangle ABC is equal to? Such as title`
You should be able to calculate the length of AB BC AC? The length is calculated by the distance formula between two points, then the angle between AB and BC is calculated by cosine theorem, and then the area formula s = 1 / 2Ab times BC
In the known triangle ABC, the coordinates of the three vertices are a (- 5, - 1), B (4,1), C (0,4), and the area of the triangle is calculated If the quadrilateral ABCD is a parallelogram, the coordinates of point d
Area 35 / 2, D (- 9,2) or (9,6)
Side length BC = 5
The linear BC equation is 4Y + 3x-16 = 0
So the distance from a to straight line BC is 7
So the area is 5 * 7 / 2 = 35 / 2
Let D (x, y)
Vector AB = (9,2), vector DC = (- x, 4-y)
From (9,2) = (- x, 4-y) or (9,2) = - (- x, 4-y), x = - 9, y = 2 or x = 9, y = 6
So D (- 9,2) or (9,6)
Given the two vertex coordinates a (- 4,0) B (2,0) of equilateral triangle ABC, the coordinate of point C is? The area of triangle ABC is?
In 1 equilateral triangle ABC, a (- 4,0) B (2,0) let C be (x, y). Obviously, a B can see that C is - 1 on the X axis on the X axis. There are two cases on the Y axis, that is, 9 + y square = 36, y = 2 √ 3 or - 2 √ 3, so C is (- 1, ± 3 √ 3) 2 s △ ABC = 1 / 2 bottom * height = 1 / 2 * 6 * 3 √ 3 = 9 √ 3. The other
In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, where C = 2 and C = π 3. If the area of △ ABC is equal to 3. Find a, B
From the cosine theorem and known conditions, A2 + B2 AB = 4
And the area of ≓ △ ABC is equal to
3．
∴1
2absinC=
3, ab = 4
Simultaneous equations
a2+b2−ab＝4
ab＝4 ，
The solution is a = 2, B = 2
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