The lines x + 3 y = 7 x+3y=7 x + 3 y = 7 and ( 1 + 64 ) x + 5 y + 5 = 0 (\sqrt{1+\sqrt{64}})x+5y+5=0 ( 1 + 64 ) x + 5 y + 5 = 0 are neither perpendicular nor parallel lines
y = − 1 3 x + 7 3 , s l o p e 1 = − 1 3 y=-\dfrac{1}{3}x+\dfrac{7}{3}, slope_1=-\dfrac{1}{3} y = − 3 1 x + 3 7 , s l o p e 1 = − 3 1
y = − 1 + 64 5 x − 1 , s l o p e 2 = − 9 5 y=-\dfrac{\sqrt{1+\sqrt{64}}}{5}x-1, slope_2=-\dfrac{9}{5} y = − 5 1 + 64 x − 1 , s l o p e 2 = − 5 9
s l o p e 1 ≠ s l o p e 2 slope_1\not=slope_2 s l o p e 1 = s l o p e 2
s l o p e 1 ( s l o p e 2 ) = − 1 3 ( − 9 5 ) = 3 5 ≠ − 1 slope_1(slope_2)=-\dfrac{1}{3}(-\dfrac{9}{5})=\dfrac{3}{5}\not=-1 s l o p e 1 ( s l o p e 2 ) = − 3 1 ( − 5 9 ) = 5 3 = − 1 The lines x + 3 y = 7 x+3y=7 x + 3 y = 7 and ( 1 + 64 ) x + 5 y + 5 = 0 (\sqrt{1+\sqrt{64}})x+5y+5=0 ( 1 + 64 ) x + 5 y + 5 = 0 intersect.
We the shortest distance between the given lines is zero.
When two parallel lines (s l o p e 1 = s l o p e 2 slope_1=slope_2 s l o p e 1 = s l o p e 2 ) are given by
a x + b y + c 1 = 0 ax+by+c_1=0 a x + b y + c 1 = 0
a x + b y + c 2 = 0 ax+by+c_2=0 a x + b y + c 2 = 0 the distance between them can be expressed as
d = ∣ c 2 − c 1 ∣ a 2 + b 2 d=\dfrac{|c_2-c_1|}{\sqrt{a^2+b^2}} d = a 2 + b 2 ∣ c 2 − c 1 ∣
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