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How do I make this equal to zero? y-(-3)=7/2 (x-(-1))
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y+3=7/2(x+1), if you want y = to zero, 0=7/2(x+1)-3
\[0 = \frac{7 \times (x+1)}{2} - (y+3)\]
This was the original question. Find an equation of the line that satisfies the given conditions. Through (−1, −3); perpendicular to the line 2x + 7y + 8 = 0
Subtract 8 and 2x from both sides.. 7y=-2x-8 Divide by 7 y=-2/7x-(8/7) Now you want a line with the slope of 7/2 (perpendicular = opposite reciprocal) that goes through (-1,-3) y=-3+7/2(x+1) Distribute, y=-3+7/2x +7/2 Add 7/2 and -3, -1/2 So your line is y = 7/2x +1/2
YES! Thank you :)
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