Distance between lines 3x + 4y = 6 and 6x + 8y = 15 is

Question

Distance between lines $3 x + 4 y = 6$ and $6 x + 8 y = 15$ is

Smart Achievers · Accepted Answer

(4)

$l_{1} : 3 x + 4 y = 6 & l_{2} : 2 (3 x + 4 y) = 15 are parallel$

$∴$ $Distance between l_{1} & l_{2} =$ $| \frac{(c_{1} - c_{2})}{\sqrt{{(3)}^{2} + {(4)}^{2}}} |$ $=$ $| \frac{(\frac{15}{2}) - 6}{\sqrt{25}} |$ $= \frac{\frac{3}{2}}{5} = \frac{3}{10}$

Solution

Q.
Distance between lines $3 x + 4 y = 6$ and $6 x + 8 y = 15$ is

1 $\frac{7}{10}$

2 $\frac{3}{5}$

3 $\frac{1}{5}$

4 $\frac{3}{10}$

Ans.
(4)

$l_{1} : 3 x + 4 y = 6 & l_{2} : 2 (3 x + 4 y) = 15 are parallel$

$∴$ $Distance between l_{1} & l_{2} =$ $| \frac{(c_{1} - c_{2})}{\sqrt{{(3)}^{2} + {(4)}^{2}}} |$ $=$ $| \frac{(\frac{15}{2}) - 6}{\sqrt{25}} |$ $= \frac{\frac{3}{2}}{5} = \frac{3}{10}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Distance between lines 3x+4y=6 and 6x+8y=15 is

1 710

2 35

3 15

4 310

Ans. (4) l1:3x+4y=6 & l2:2(3x+4y)=15 are parallel ∴ Distance between l1 & l2=|(c1-c2)(3)2+(4)2|=|(152)-625|=325=310