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The intersection of the straight line y = 2x + 3 and the parabola y = x square is a and B. find the OAB area of the triangle
y=2x+3
y=x^2
=>x=3,-1
A(3,9) B(-1,1)
Let the projections of a and B on the coordinate axis X be a ', and B'
A'(3,0),B'(-1,0)
Triangle OAB area = trapezoid aa'b'b area triangle oaa'area triangle obb'area
=(3+9)*[3-(-1)]/2-9*3/2-3*1*2
=9
Find the area of triangle surrounded by line l1:2x-y + 1 = 0, line l2:2x + y-4 = 0 and x-axis
25/8
Line L1: y = 2x + 1 line L2: y = - 2x + 3 1: calculate the area of the triangle enclosed by L1L2 and Y axis 2: calculate the area of the triangle enclosed by L1L2 and X axis
The intersection of L1 and Y axis is a (0,1) and the intersection of X axis is C (- 1 / 2,0)
The intersection of L2 and Y car is B (0,3) and the intersection of X axis is d (3 / 2,0)
To solve the equations y = 2x + 1 y = - 2x + 3
We get x = 1 / 2, y = 2
So the intersection of two lines is p (1 / 2,2)
Area of triangle ABP = 1 / 2x (3-1) X1 / 2 = 1 / 2
The area of triangular CDP = 1 / 2x (3 / 2 + 1 / 2) x2 = 2
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