The distance between the lines 3x + 4y = 9 an | vedclass

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(B) The given lines are $3x + 4y = 9$ and $6x + 8y = 15$. To make the coefficients of $x$ and $y$ the same,multiply the first equation by $2$: $6x + 8

Vedclass · Accepted Answer

$\frac{3}{10}$

Vedclass · Answer

(B) The given lines are $3x + 4y = 9$ and $6x + 8y = 15$. To make the coefficients of $x$ and $y$ the same,multiply the first equation by $2$: $6x + 8y = 18$. Now,the equations are $6x + 8y - 18 = 0$ and $6x + 8y - 15 = 0$. The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Here,$A = 6, B = 8, C_1 = -18, C_2 = -15$. $d = \frac{|-18 - (-15)|}{\sqrt{6^2 + 8^2}} = \frac{|-3|}{\sqrt{36 + 64}} = \frac{3}{\sqrt{100}} = \frac{3}{10} 	ext{ units}$. Hence,option $B$ is correct.

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is equal to units.

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