CBSE Class 11-science Answered
Find the distance between the lines 4x-3y+5=0 and 8x-6y+7=0..
Asked by Suryavanshikatoch | 11 Feb, 2019, 19:37: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/7bbf396f34838d22e1a97c1f9f5ae0bb5c61a31f696371.01397456feb1102.png)
lines 4x-3y+5=0 and 8x-6y+7 =0 are parallel lines with slope 4/3 as shown in figure.
Point of intersection of both lines with y-axis is obtained by substituting x=0 in the equations.
Points of intersection A (0, 5/3) and B(0,7/6) are shown in figure.
Perpendiculr line to both the lines is in the form 3x+4y = k . This perpendicular line passes through A(0, 5/3)
by substituting values for x and y in the equation we get k = 20/3
point of intersection is obtained by solving the equations 8x-6y+7=0 and 3x+4y = (20/3)
we get point of intersection as C ( 0.24, 1.5)
distance between the lines = ![begin mathsize 12px style square root of open parentheses 0.24 close parentheses squared plus open parentheses 0.16 close parentheses squared end root space almost equal to space 0.29 end style](https://images.topperlearning.com/topper/tinymce/cache/6825654f1d33260cdaa6e5c447f82b30.png)
![begin mathsize 12px style square root of open parentheses 0.24 close parentheses squared plus open parentheses 0.16 close parentheses squared end root space almost equal to space 0.29 end style](https://images.topperlearning.com/topper/tinymce/cache/6825654f1d33260cdaa6e5c447f82b30.png)
Answered by Thiyagarajan K | 11 Feb, 2019, 22:24: PM
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