the problem is: Perpendicuar to y= -5/7x passing through (6/5,0) I know that to be perp it has to be opp/co so itll be 7/5x but i dont know what to do next at all. The answer says: y= 7/5x - 6/5 I have no idea how to get there
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Originally Posted by gurrry the problem is: Perpendicuar to y= -5/7x passing through (6/5,0) I know that to be perp it has to be opp/co so itll be 7/5x but i dont know what to do next at all. The answer says: y= 7/5x - 6/5 I have no idea how to get there The normal will be of the form y = mx + c, and you have already found m = 7/5 So y = (7/5)x + c You also know that (x, y) = (6/5, 0) is a point on the line, so you can use this in the equation of your line 0 = (7/5) x (6/5) + c Solve for c.
Originally Posted by gurrry the problem is: Perpendicuar to y= -5/7x passing through (6/5,0) I know that to be perp it has to be opp/co so itll be 7/5x but i dont know what to do next at all. The answer says: y= 7/5x - 6/5 You have a point and a slope. Use point-slope form.
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