The problem of linear equation Given that the line L passes through the point P (- 5, - 4), and the area of the triangle formed by L and two coordinate axes is 5, the equation of line L is obtained

The problem of linear equation Given that the line L passes through the point P (- 5, - 4), and the area of the triangle formed by L and two coordinate axes is 5, the equation of line L is obtained

Let the linear equation ly + 4 = K (x + 5) = KX + 5K
Y = KX + 5k-4, x = 0, y = 5k-4
y=0 x=(4-5k)/k
S=1/2*|5k-4|*|4/k-5|=5
|40-35k-16/k|=10
(1) 40-35k-16 / k = 10 discriminant test

If the light l emitted from point a (- 3, 3) hits the x-axis and is reflected by the x-axis, and the reflected light is tangent to the circle C: x2 + y2-4x-4y + 7 = 0, then the linear equation of light L is______ ．

The standard equation of a known circle is (X-2) 2 + (Y-2) 2 = 1, and the equation of its symmetric circle about X axis is (X-2) 2 + (y + 2) 2 = 1. Let the equation of the line where the light L is located be Y-3 = K (x + 3) (where the slope k is to be determined)

Positional relationship between line and circle (17 11:15:15)
Given two circles: x2 + y2-10x-10y = 0, X2 + Y2 + 6x-2y-40 = 0, find
1. The equation of the line where their common line lies
2. Common chord length

To find the common chord, we only need to connect the two equations with an equal sign
x^2+y^2-10x-10y=x^2+y^2+6x-2y-40
The linear equation L: 2x + Y-5 = 0 is obtained by eliminating the power of two sides
2. The common chord length is solved by Pythagorean theorem in triangles
The center of the circle x ^ 2 + y ^ 2-10x-10y = 0 is O (5,5), the distance to the straight line d = 2 * (5 ^ 0.5), and the perpendicular foot is a
The distance from the center O of the circle to the intersection B of the line and the circle is the radius of the circle r = 5 * (2 ^ 0.5)
Half of chord length = (R ^ 2-D ^ 2) ^ 0.5 = 30 ^ 0.5
The chord length is 2 * 30 ^ 0.5