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Distance of a point (2,5) from the line 2x-y-4=0 measured parallel to the line 3x-4y + 8 = 0 is : |
Answer» (y-5)=3/4(x-2) 4y-20=3x-6 4y-3x-14=0-(1) Given 2x-y-4=0-(2) From equation 1 and 2 4(2x-4)-3-14=0 8x-3x-30=0 x=6 y=8 A(6,8) AB=`sqrt((6-2)^2+(8-5)^2)` `=sqrt(4^2+3^2)` =5 Units. |
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