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It is known that the intercept of the line y = KX + B on the y-axis is 4, and the line passing through the point C (3,2) intersects with the x-axis and y-axis at two points a and B. in addition, there is a point D on the x-axis (3 / 2, 0) finding the area of a quadrilateral bcdo fast
If the intercept of the line y = KX + B on the Y axis is 4, then: B = 4,
Passing through point C (3,2), substituting the coordinates of the point, we get:
3k+4=2,
k=3/2.
So the linear equation is y = 2 / 3 * x + 4,
Let x = 0, then: y = 4;
Let y = 0, then: x = - 6
So point a (- 6,0), point B (0,4), and,
D (2 / 3,0), C (3,2), D (2 / 3,0),
therefore
Area of quadrilateral bcdo = triangle BCO area + triangle CDO area
=1/2*|BO|*|xC|+1/2*|DO|*|yC|
=1/2*4*3+1/2*2/3*2
=6+2/3=20/3.
PS: Xc, YC are abscissa and ordinate of point C respectively
The line y = KX + B is parallel to y = - 1 / 2x, and the intercept of Y axis is 3,
The line y = KX + B is parallel to y = - 1 / 2x
∴k=-1/2
The intercept of y-axis is 3
∴b=3
∴y=-1/2x+3
Let the symmetry axis of the image of the quadratic function y = ax ^ 2 + BX + C be 2x-3 = 0, the reciprocal sum of the intercepts on the X axis be 2, and pass through the point (3, - 3)
(1) Try to find the values of a, B and C;
(2) When x is in what range, y ＞ 1 or Y ＜ - 3
(3) When what is the value of X, y has the maximum value? And find the maximum value
1/x1+1/x2
=(x1+x2)/x1x2
=2
x1+x2=-b/a
x1x2=c/a
So - B / C = 2
The axis of symmetry is x = 3 / 2 = - B / 2a, B = - 3a
-3=9a-9a+3a/2
a=-2 b=6 c=-3
y=-2x²+6x-3
(2) -2x²+6x-3>1 x²-3x+2
Given that the intercept of AX + by + C = 0 on the x-axis is greater than that on the y-axis, then a, B and C should satisfy the condition ()
A. A＞BB. A＜BC. CA+CB＞0D. CA−CB＜0
∵ the intercept of AX + by + C = 0 on the x-axis is greater than that on the y-axis, that is − Ca ＞ − CB, ∵ Ca − CB ＜ 0
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