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the point-slope form of the equation of the line that passes through (–5, –1) and (10, –7) is y 7 = (x – 10). what is the standard form of the equation for this line? 2x – 5y = –15 2x – 5y = –17 2x 5y = –15 2x 5y = –17

Respuesta :

texaschic101
(-5,-1)(10,-7)
slope(m) = (-7 -(-1) / (10 - (-5) = (-7 + 1) / (10 + 5) = -6/15 = - 2/5

y - y1 = m(x - x1)
slope(m) = -2/5
(10,-7)...x1 = 10 and y1 = -7
now we sub
y - (-7) = -2/5(x - 10) =
y + 7 = -2/5(x - 10)
y + 7 = -2/5x + 4
y = -2/5x + 4 - 7
y = -2/5x - 3....multiply both sides by 5
5y = -2x - 15
2x + 5y = -15 <== standard form



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