It is known that A 2,0 ,B 4,0 ,C 2,4 , 1 find the area of triangle ABC

The first question is based on ab, the ordinate of c is high, and the area is (4+2) 4 1 2=12

In the second question, the coordinates are p(x,0), the area of the triangle apc is ap, that is, the ap is the base, the ordinate of c is high, and the s triangle apc = |x-(-2)|×4×1/2=|2x+4|, S triangle PBC is the same, with PB as the base, the ordinate of point C is high, and the S triangle PBC = |x-4|×4×1/2=|2x-8|, that is, to find |2x+4|=1/2|2x-8|, i.e. |2x+4|=|x-4|, divided into 2x+4=x-4 and 2x+4=4-x to remove the absolute value, and finally solve x=-8 or x=0, and bring back the original test is true, so p is (0,0) or (-8,0).

It's a lot of effort to write, q q, if the landlord can understand it, give a good one, if the landlord doesn't understand, you can draw a diagram, if you have a diagram, you will understand a lot, all triangles can be used as the x-axis as the bottom edge to find the area

Solution: Let the coordinates of p be (x,0).

x-(-2)|×4×1/2=1/2|x-4|×4×1/22x+4|=1/2|2x-8|

2x+4|=|x-4|

x=-8 or 0

Answer: p is (0,0) or (-8,0).

Make the point d(-1,4), the point e(4,4), the point f(4,-2), so that the triangle is not formed and quietly merges the quadrilateral DAFE, then the triangle is extinct and hail ADC=dc*DA 2=3*6 2=9, the triangle CEB=2*4 2=4, the triangle AFB=2*5 2=5, then the triangle ABC=5*6-9-4-5=12,6,bc equation is: y (x-4)=4 (-2)=-2,2x+y-8=0,bc|= 4+16)=2 5,A to bc distance h=12 5,s abc=|bc|*h/2=(2√5*12/√5)/212.,0,

Since ab is all on the x-axis, it is known that ab=6 is easy to obtain high c, ordinate 4, then the area is 12 (2), since the height is the same area ratio is equal to the bottom ratio, then ap is 2 long, then p(1,0).

(1) Its area is (2 4) * 4 * half = 12

2) When P is right in B, the SPBC is smaller than the SAPCHouse.

When p is left in a, then sapc=sabc=12then ap=ab= is (-8,0)

When P is between AB, then SAPC = one-third * SABC = 4Push ap=2, i.e. p(0,0)

In summary, p(0,0) or (-8,0).

After passing A and C as the perpendicular lines of the Y axis, and the perpendicular feet are A1 and C1 respectively, then the area of the triangle ABC is equal to the area of the trapezoidal ACC1A1 minus the area of the triangle ABA1, and then subtracting the area of the triangle CBC1, so there is: the area sought = (2+4)x(3+1) 2-3x4 2-1x2 2=12-6-1=5

Method 1: Calculate the length of the three sides is 5, 2 5, 5 is RT, and the area is 5 Method 2: Use the outer product formula.

ab=（4，-3），bc=(-2，-1）

ab×bc=ijk4-30

0=-10k

So 2s = 10s = 5

Solution: The distance from the bottom A to the X axis on the bottom edge of ABC, so BC=4-(-2)=6

h=|-5|=5

s△abc=5*6/3=15

If you still have questions, feel free to ask!

You put the first.

3. The fourth quadrant is seen as a rectangle, and the area of the rectangle - the area of 3 small right triangles = the area of the triangle abc.

You try, you can do it.

Known: a(-4,-5) b(-2,0) c(4,0), find the area of the triangle abc.

BC=6H=A The distance from BC to BC is 5

Area of the triangle ABC = H*BC2=15