+2448 Write an equation for the line that contains (10,0) and is perpendicular to 5x+2y=1
+6251 \(5x+2y=1\\ y = \dfrac{1-5x}{2} = -\dfrac 5 2 x + \dfrac 1 2\\ \text{the slope }m = -\dfrac 5 2\\ \text{a line perpendicular to this will have slope }-\dfrac 1 m = \dfrac 2 5\\ \text{a line with slope }m=\dfrac 2 5 \text{ and passing through }(10,0) \text{ is given in point slope form}\\ (y-0) = \dfrac 2 5(x-10) \text{ or }\\ y = \dfrac 2 5 x - 4\\ 5y - 2x = -20\)
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