Given that the straight line L passes through the point (- 2,3) and the distance from the origin to the line L is 2, then the equation of the line L is______ .

When the slope of a straight line does not exist, the equation is x = - 2, which satisfies the condition by testing. When the slope of the straight line exists, let the equation of the line be Y-3 = K (x + 2), that is, kx-y + 2K + 3 = 0. From the meaning of the question, 2 = | 0 − 0 + 2K + 3 | K2 + 1,  k = - 512, so the equation of line L is x = - 2, or 5x + 12y-2

If the straight line L passes through the point P (5,10) and the distance from the origin to it is 5, then the equation of the line L is given One is x = 5 and the other is 3x-4y + 25 = 0 What I want to ask is how does the second answer come from?

(1) When K does not exist, the straight line passing through point P (5,10) is x = 5, which satisfies the meaning of the question; (2) when k exists, the equation of the straight line l can be written by the oblique formula: Y-10 = K (X-5), and then the distance formula from the point to the straight line L is d = | 10-5k | / √ (K ﹤ + 1) = 5, that is, | 2-k | / √ (K | 1) = 1 | 2-k |

Given that the straight line L passes through the point P (5,10) and the distance from the origin to it is 5, then the equation of the line L is______ .

When the slope of the line does not exist, the equation of the line is x = 5, which satisfies the condition. When the slope of the line exists, let the equation of the line be Y-10 = K (X-5), that is, kx-y-5k + 10 = 0. From the condition, − 5K + 10 | 1 + K2 = 5, | k = 34, so the equation of line is 3x-4y + 25 = 0

Find the linear equation passing through the point (5,10) and the distance from the origin is 5

Slope does not exist, x = 5
The satisfied distance is 5
Slope exists
y-10=k(x-5)
kx-y+10-5k=0
So the distance = | 0-0 + 10-5k| / √ (k? 2 + 1) = 5
|k-2|=√(k²+1)
square
k²+4k+4=k²+1
k=-3/4
So X-5 = 0 and 3x + 4y-55 = 0

Given that the straight line L passes through the point P (5,10) and the distance from the origin to it is 5, then the equation of the line L is______ .

The intercept of the line L on the Y axis is 10, and the distance from the origin to the straight line is 8

Suppose that the angle between the line L and the Y axis is p, then the angle between the line and the X axis is q = 90 ° - P
∴tanP=√(10²-8²)/8=6/8=3/4
∴tanQ=4/3
The equation of line L is: y = ± (4 / 3) x = = = > 3Y = ± 4x

If l is the intercept of L to the origin of the line, l is the intercept of the line

The intercept of the line L on the y-axis is 10
The straight line passes through the point (0,10), so let the equation of the straight line l be:
y=kx-10
be
kx-y+10=0
8=|10|/√(1+k^2)
k=±3/4
y=±3/4x-10

Let's find the distance from L. to the origin of the line,

∵ the intercept of the line L on the y-axis is 10,
The linear equation can be set as y = KX + 10, that is, kx-y + 10 = 0
And ∵ the distance from the origin to the line L is 8,
ν 10 / √ (K  2 + 1) = 8, that is, K  2 + 1 = 25 / 16, K  2 = 9 / 16, k = ± 3 / 4,
The linear equation is 3x-4y + 40 = 0, or 3x + 4y-40 = 0