Answer:
[tex]\dfrac{11}{17}[/tex]
Step-by-step explanation:
Lines [tex]8x-15y+5=0[/tex] and [tex]16x-30y-12=0[/tex] are parallel because
[tex]\dfrac{8}{16}=\dfrac{-15}{-30}\neq \dfrac{5}{-12}[/tex]
The distance between two lines [tex]a_1x+b_1y+c_1=0[/tex] and [tex]a_1x+b_1y+c_2=0[/tex] can be calculated using formula
[tex]D=\dfrac{|c_1-c_2|}{\sqrt{a_1^2+b_1^2}}[/tex]
Divide the equation of the second line by 2:
[tex]8x-15y-6=0[/tex]
Hence, the distance between two lines is
[tex]D=\dfrac{|5-(-6)|}{\sqrt{8^2+(-15)^2}}=\dfrac{|5+6|}{\sqrt{64+225}}=\dfrac{11}{\sqrt{289}}=\dfrac{11}{17}[/tex]
Answer:
d = [tex]\frac{11}{17}[/tex].
Step-by-step explanation:
We, have given two line lines 8x-15y+5=0 and 16x−30y−12=0. These two lines are parallel because it follows parallel condition: [tex]\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}[/tex].
We need to find the distance between these two parallel lines.
We know that distance formula, between two parallel line:
d =[tex]\frac{| c_1 - c_2 |}{\sqrt{A^{2} +B^{2} } }[/tex]
We have these two equation 8x−15y+5=0 and 16x−30y−12=0 but its coffiecients are not same, then we will first same coffiecients:
8x−15y+5=0, we can written as 8x-15y = -5
16x−30y−12=0, we can written as 16x-30y = 12, common 2 from these equation: 8x-15y = 12/2 = 6. Now, both lines have same cofficients.
Applying distance formula,
d =[tex]\frac{| -5 - (+6) |}{\sqrt{8^{2} +(-15)^{2} } }[/tex]
d =[tex]\frac{|-11|}{\sqrt{64 +225} }[/tex]
d = [tex]\frac{|-11|}{\sqrt{289} }[/tex]
d = [tex]\frac{11}{17}[/tex]
Therefore, distance between the these two lines are 11/17.
