(數學)直線方程問題一問!!!

Line L1: y=√3 x

Line L2: 2√3 x-2y+5=0
2y = 2√3 x+5
y = √3 x + 5/2

Draw a line L3 perpendicular to line L2 passing through the point of origin (0, 0)
Line L3 has a slope -1/√3
slope of L1 乘 slope of L3 = -1
The linear equation L3 is y = (-1/√3) x

Consider lines L1 and line L3, find the point of intersection
√3 x + 5/2 = -1/√3x
√3 x + (1/√3)x = -5/2
[√3 + (1/√3)]x = -5/2
x = (-5√3)/8
y = 5/8
The intersection point is ((-5√3)/8, 5/8)
The point of origin is (0, 0)

The distance between these 2 points = square root of [ (x2 – x1)^2 + (y2 – y1)^2]
[((-5√3)/8 – 0)^2 + (5/8 – 0)^2]^0.5
(75/64 + 25/64)^0.5
(100/64)^0.5
10/8 = 1.25

L1與L2之間的距離 = 1.25


Method 2: Using formula
Length of a perpendicular from a point to a line
The line is: lx +my +n = 0
The point is (x1, y1)

lx1 +my1+ n
distance = -----------------------
(l^2 + m^2)^0.5

l = 2√3, m = -2, n = 5, x1 =0, y1 = 0
the distance =( lx1 +my1+ n)/(l^2 + m^2)^0.5
the distance + 5/((2√3)^2 + (-2)^2)^0.5
the distance + 5/(12 + 4)^0.5
the distance + 5/(16)^0.5
the distance = 5/4 = 1.25