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Perpendicular to y=5x 2 and passing through (10,-5)

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Final answer:

To find a line perpendicular to y=5x 2 and passing through the point (10,-5), we need to determine the slope of the given line and then find the negative reciprocal of that slope. The equation of the line perpendicular to y=5x 2 and passing through (10,-5) is y=-(1/5)x-3.

Step-by-step explanation:

To find a line perpendicular to y=5x 2 and passing through the point (10,-5), we need to determine the slope of the given line and then find the negative reciprocal of that slope. The given line is in the form y=mx+b, where m is the slope. In this case, the slope is 5.

The negative reciprocal of 5 is -1/5. We can use the point-slope form of a linear equation, y-y1=m(x-x1), where (x1,y1) is a point on the line and m is the slope, to write the equation of the line perpendicular to y=5x 2 and passing through (10,-5). Plugging in the values, we get y-(-5)=-(1/5)(x-10).

Simplifying the equation, we have y+5=-(1/5)x+2 or y=-(1/5)x-3. Therefore, the equation of the line perpendicular to y=5x 2 and passing through (10,-5) is y=-(1/5)x-3.

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