Question

Write an equation for the line that contains the point (-6, -8) and is parallel to the line 3x – 2y = -12.

Answer 1

Since line is parallel to the line 3x – 2y = -12.
The eq. 3x – 2y = -12 can be written as
2y = 3x + 12
i.e. \[i.e. y = \frac{ 3 }{ 2 } x + 6\]
Now comparing it with y = mx + c we have slope
\[m=\frac{ 3 }{ 2 }\]
Since slopes of parallel lines are same, therefore slope required line will also be
\[m=\frac{ 3 }{ 2 }\]
Now since line passes through the points (-6, -8) i.e. x1 = -6, y1 = -8
Eq of line in slope point form is given as
y-y1=m(x-x1)
i.e. y-(-8) =3/2[x-(-6)]
i.e. y+8 =3/2[x+6]
i.e. 2 y+16 =3x+18
i.e. 3x -2y +18-16=0
i.e. 3x -2y +2=0 and this the required eq of line

Answer 2

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