Question

Find the distance between lines 8x−15y+5=0 and 16x−30y−12=0.

Answer 1

Answer:

$\dfrac{11}{17}$

Step-by-step explanation:

Lines $8x-15y+5=0$ and $16x-30y-12=0$ are parallel because

$\dfrac{8}{16}=\dfrac{-15}{-30}\neq \dfrac{5}{-12}$

The distance between two lines $a_1x+b_1y+c_1=0$ and $a_1x+b_1y+c_2=0$ can be calculated using formula

$D=\dfrac{|c_1-c_2|}{\sqrt{a_1^2+b_1^2}}$

Divide the equation of the second line by 2:

$8x-15y-6=0$

Hence, the distance between two lines is

$D=\dfrac{|5-(-6)|}{\sqrt{8^2+(-15)^2}}=\dfrac{|5+6|}{\sqrt{64+225}}=\dfrac{11}{\sqrt{289}}=\dfrac{11}{17}$

Answer 2

Answer:

d = $\frac{11}{17}$.

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: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

We need to find the distance between these two parallel lines.

We know that distance formula, between two parallel line:

d =$\frac{| c_1 - c_2 |}{\sqrt{A^{2} +B^{2} } }$

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 =$\frac{| -5 - (+6) |}{\sqrt{8^{2} +(-15)^{2} } }$

d =$\frac{|-11|}{\sqrt{64 +225} }$

d = $\frac{|-11|}{\sqrt{289} }$

d = $\frac{11}{17}$

Therefore, distance between the these two lines are 11/17.