Answer:
[tex]\large\boxed{\sf y = 5x +16}[/tex]
Step-by-step explanation:
Here it is given that a line c has a equation of ,
[tex]\sf\qquad\longrightarrow y = 4x + 9[/tex]
And there is another line d which is parallel to line c and passes through the point (-4,-4) . And we need to find out the equation of the line .
Firstly we know that the slope of two parallel lines is same . So on comparing the given line to the slope intercept form of the line which is y = mx + c , we have ;
[tex]\sf\qquad\longrightarrow m = 4 [/tex]
Therefore the slope of the parallel line will be ,
[tex]\sf\qquad\longrightarrow m_{||}= 4 [/tex]
On using the point slope form of the line , we have ;
[tex]\sf\qquad\longrightarrow y - y_1 = m(x-x_1)\\ [/tex]
Substitute the values ,
[tex]\sf\qquad\longrightarrow y - (-4) = 5\{ x -(-4)\} [/tex]
Simplify ,
[tex]\sf\qquad\longrightarrow y +4 = 5(x +4)[/tex]
Open the brackets ,
[tex]\sf\qquad\longrightarrow y + 4 = 5x + 20 [/tex]
Subtract 4/on both sides ,
[tex]\sf\qquad\longrightarrow y = 5x +20-4 [/tex]
Simplify ,
[tex]\sf\qquad\longrightarrow \pink{ y = 5x + 16} [/tex]
Hence the equation of the line is y = 5x + 16