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Find the distance of the point (4,1) from the line `3x-4y+12=0` |
Answer» Clerly, the required distance is the length of perpendicular from the point P(4,1) on the line 3x-4y+12=0 Let the required distance of d. Then, `d=(|3xx4-4xx1+12|)/(sqrt(3^(2)+(-4)^(2)))=(20)/(5)"units=4units"` |
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