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The linear equation of the line x-2y + 1 = 0 with respect to the symmetry of the line Y-X = 1 More detailed steps, thank you
The intersection of the two equations is a (- 1,0)
Take a point B (1,1) on x-2y + 1 = 0, and then find its symmetric point B '(a, b) with respect to Y-X = 1, then ab' is the straight line
Y-X = 1, y = x + 1, its slope is 1, the slope of BB 'is - 1, the equation is Y - 1 = - (x - 1), y = 2-x
The intersection of BB 'and Y-X = 1 is C (1 / 2, 3 / 2)
C is the midpoint of BB ': 1 / 2 = (a + 1) / 2, a = 0
3/2 = (b+1)/2, b = 2
B'(0, 2)
The equation of ab 'is: X / (- 1) + Y / 2 = 1
y = 2x + 2
The linear equation of x-2y-1 = 0 symmetry about x + Y-1 = 0
RT
First, find the coordinates (1,0) of the intersection of two lines
Take any point on x-2y-1 = 0, if (3,1)
Find the symmetric point of (3,1) with respect to x + Y-1 = 0
Let the coordinates of the point be (a, b)
According to (3,1) (a, b), the slope of two points is 1,
The midpoint is on X + Y-1 = 0
Find a = 0, B = - 2
From (0, - 2), (1.0)
The linear equation is 2x-y-2 = 0
The equation of a line passing through (- 1,3) and parallel to the line x-2y + 3 = 0 is ()
A. x-2y=0B. 2x+y-1=0C. x-2y+7=0D. 2x+y-5=0
From the parallel relation, we can assume that the linear equation is x-2y + C = 0, substitute the point (- 1,3) to get - 1-2 × 3 + C = 0, and solve C = 7. The equation of the straight line is x-2y + 7 = 0, so we choose: C
Given that the line L passes through the point P (2,1), and the angle between the line L and the line 5x + 2Y + 3 = 0 is equal to 45 degrees, the equation of the line L is obtained
I first use the slope of L as K, and then convert the K value of the straight line 5x + 2Y + 3 = 0. The title tells me to substitute 45 degrees with the included angle formula. It's stuck in the absolute value and can't be decomposed. Please. I'll wait online
Let the linear equation be Y-1 = K (X-2), that is, y = kx-2k + 1
5x+2y+3=0,--->y=-2.5x-1.5
Let a = 45
tana=tan45=|(k2-k1)/(1+k1k2)|=|(k-(-2.5))/(1-2.5k)|=1
|(k+2.5)/(1-2.5k)|=1
(k+2.5)/(1-2.5k)=±1
The solution is k = - 3 / 7 or K = 7 / 3
The linear equation is y = - 3 / 7X + 13 / 7 or y = 7 / 3x-11 / 3
Through the point (2,1), and the angle with the straight line L: 5x + 2y-5 = 0 is 45 degrees
Let the equation be (Y-1) = K (X-2) and the slope of l be (- 5 / 2)
There are ± 1 = ± 45 ° = (K + 5 / 2) / (1-5k / 2)
The solution is k = - 3 / 7 or K = 7 / 3
The equation 3x + 7y-13 = 0
Or 7x-3y-11 = 0
Given that the line L: 5x + 2Y + 3 = 0, the angle from L 'to l of the line passing through point P (2,1) is equal to 45 degrees, the general equation of the line L' is obtained
Let the slope of the straight line L: 5x + 2Y + 3 = 0 be K1 = - 52. Let the slope of the straight line L 'be K. from the meaning of the question, we can get: tan45 ° = K1 & nbsp; - K1 + KK1 = 1, that is (− 52) - k = (1 − 52K). We can get k = 73. Let the straight line L' pass through the point P (2,1). From the point oblique formula, we can get: the equation of the straight line L 'is: 7x-3y-11 = 0
If the line L: 5x + 2Y + 3 = 0, the line L 'passes through the point P (2,1) and the angle with L is equal to 45 degrees, then the general equation of the line L' is to help do the next process 3
tg45°=1
Straight line L: y = - 5 / 2x-3 / 2, slope is - 5 / 2
Straight line passing through point (2,1): point oblique straight line formula Y-1 = K (X-2) slope k = - 5 / 2 ± 1
Two linear equations are obtained by calculating the value of K
There are two lines passing through point (2,1) with an angle of 45 ° to line L
Through the angle between C (2,1) and the straight line 5x + 2Y + 3 = 0 is 45 ° to find the linear equation?
Let the linear equation be y = K (X-2) + 1
The slope of the line 5x + 2Y + 3 = 0 is K1 = - 5 / 2
From the angle formula of two straight lines | (k1-k2) / (1 + k1k2) | = Tan θ
|(k+5/2)/(1-5k/2)|=1
The solution is k = ± 3 / 7
The linear equation is y = ± (3 / 7) (X-2) + 1
(2005, Anhui) let a straight line L cross a point (- 2, 0) and be tangent to a circle x2 + y2 = 1, then the slope of L is ()
A. ±1B. ±12C. ±33D. ±3
∵ the straight line L passes through the point (- 2,0) and is tangent to the circle x2 + y2 = 1. From the circle, the three sides of the triangle with the center of the circle (0,0) and the radius of 1 ∵ are: 2,1,3. The angle between the straight line and the x-axis is 30 ° and the inclination angle of X is 30 ° or 150 ° k = ± 33, so C is selected
The line where the point (1,2) is located is tangent to the circle x2 + y2 = 1 to find the slope k of the line
Let the tangent equation be y = K (x-1) + 2 = kx-k + 2
If the distance from the center of the circle to the straight line is equal to the radius, then | - K + 2 | / root sign (k ^ 2 + 1) = 1
k^2-4k+4=k^2+1
k=3/4
In other words, the tangent equation is y = 3 / 4x-3 / 4 + 2 = 3 / 4x + 5 / 4, and the line x = 1 is also a tangent
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