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What is the equation of the line perpendicular to x+5y=-2 passing through (-3,-4) in standard form?
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slope*(-1/5)=-1 slope=5 y+4=5(x+3) y=5x+11
find the gradient of your line.... then the gradient of the perpendicular is -1/m then use point gradient to find the equation..
that is in slope-intercept how do I put y=5x+11 in standard ?
5x-y=-11
so everything on the left hand side.... y - 5x -11 = 0
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moving the 11 to the left hand side would make it positive right?
i think I will go with 5x-y=-11, thanks.
yup thats it my pleasure XD
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