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| 1. |
For what value of k do the equation 3x+2ky_2=0 and 2x+5y+1 =0 are parallel |
| Answer» 3x + 2ky = 2=> 2ky = -3x + 2=>y = (-3/2k)x + (1/k)°slope(m1)=(-3/2k)also,2x + 5y + 1 = 0=> 5y = -2x - 1=> y = (-2/5)x - (1/5)°slope(m2) = (-2/5)since, lines are paralleltherefore, m1=m2that is, (-3/2k) = (-2/5)this applies: on further solving:{k = 15/4} | |