How long is the perpendicular line segment between the parallel lines 4x-3y+12=0 and 
4x-3y-18=0?"

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1 Answer

4x-3y+12=0 konvert tu y=(4/3)x +4

4x-3y=-18 become y=(4/3)x -6

vertikal distans tween 2 lines=10

slope=4/3=1.333

yuze invers tangent, slope=53.130 deg

draw perpend line tu 1 sloped line & yu get rite triangel

distans=diagonal*kosine(53.12 deg)=10*kosine(53.12)=10* 0.60=6
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