Step-by-step explanation:
Hey there!
Firstly find slope of the given equation.
Given eqaution is: 3x + 2y = 5.......(i)
Now;
[tex]slope(m1) = \frac{ - coeff. \: of \: x}{coeff. \: of \: y} [/tex]
[tex]or \: m1 = \frac{ - 3}{2} [/tex]
Therefore, slope (m1) = -3/2.
As per the condition of parallel lines,
Slope of the 1st eqaution (m1) = Slope of the 2nd eqaution (m2) = -3/2.
The point is; (-2,-3). From the above solution we know that the slope is (-3/2). So, the eqaution of a line which passes through the point (-2,-3) is;
(y-y1) = m2 (x-x1)
~ Keep all values.
[tex](y + 3) = \frac{ - 3}{2} (x + 2)[/tex]
~ Simplify it.
[tex]2(y + 3) = - 3x - 6[/tex]
[tex]2y + 6 = - 3x - 6[/tex]
[tex]3x + 2y + 12 = 0[/tex]
Therefore, the eqaution of the line which passes through the point (-2,-3) and parallel to 3x + 2y= 5 is 3x + 2y +12 =0.
Hope it helps...
