Question

A line is drawn through the point (-1,-1), parallel to the line 2x-3y+9=0. Where will it cross the x-axis?

Answer 1

Answer:

2x-3y-1=0

I hope I'm correct.

Step-by-step explanation:

$2x-3y+9=0\\Rearrange -in-form-of : y = mx+b\\-3y=0-9-2x\\-3y=-2x-9\\\frac{-3y=-2x-9}{-3} \\y =\frac{2}{3} \\\\For- parallelism- , m1=m2\\One -point- form (-1,-1)\\x1 = -1\\y1 = -1 \\m = \frac{2}{3}\\ y -y1=m(x-x1)\\y -(-1)=\frac{2}{3} (x-(-1))\\y+1=\frac{2}{3}(x+1)\\y+1=\frac{2}{3}x + \frac{2}{3}\\\\y = \frac{2}{3}x + \frac{2}{3}-1\\\\y = \frac{2}{3}x - \frac{1}{3}\\Multiply- through-: by 3 \\3[y = \frac{2}{3}x - \frac{1}{3}]\\\\3y = 2x -1\\Answer = 2x-3y-1=0$

Answer 2

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