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It is known that the intercept of lines L1 and L2 on x-axis is equal, and their inclination angles are complementary, and the line L1 passes through point P (- 3,3). If the distance between point Q (2,2) and L2 is 1, the equation of L2 is solved
Let the equation of line L2 be y = K (x-a), then the equation of line L1 is y = - K (x-a). ∵ the distance from point Q (2,2) to L2 is 1, ∵ K (2 − a) − 2| 1 + K2 = 1. (1) because line L1 passes through point P (- 3,3), then 3 = - K (- 3-A); (2) from (2) to get Ka = 3-3k, substitute (1), get
Find out the intersection of line L1: y = x + 1 and line L2: y = 3x-3, and the intercept is equal on the two coordinate axes
Let y = ax + B
y=x+1
y=3x-3
Two equations constitute a system of equations to find the intersection coordinates m (2,3)
And because the intercept of a line on two axes is equal
So the intersection coordinates of X and y are a (0, a), B (B, 0)
Substituting the line y = ax + B
The result is: x = - B / A, y = B
Cross point coordinates m (2,3)
So B = - 2 / 3
So the linear equation y = - 2 / 3x + 3
That's all
When the line L1: y = √ 3x + √ 3 is rotated 90 ° anticlockwise around its intersection with the X axis, the intercept of L2 on the Y axis is
When the line L1: y = √ 3x + √ 3 is rotated 90 ° anticlockwise around its intersection with the X axis, the intercept of L2 on the Y axis is obtained
The x-axis intersection is (- 1,0)
The original slope is √ 3, and after 90 ° rotation is - √ 3 / 3
So the equation is
y=-√3/3x-√3 /3
The intercept on the y-axis is - √ 3 / 3
The equation of a line perpendicular to the line 3x + 2Y + 1 = 0 and whose intercept on the X axis is - 4 is?
A.2x-3y+8=0 B.2x-3y-8=0
C.2x-3y-12=0 D.2x+3y+12=0
A is correct. Y = 0, x = - 4 (the intercept on the X axis is - 4), directly excluding B / C / d. multiple choice questions need to be answered like this, so that I have more time to do other questions
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