The equation is given as,
![\begin{gathered} y\text{ = -}(5)/(2)x\text{ - 4} \\ \\ \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/c18bhndp3tebbzo1r6mvxut6svjhho8lny.png)
Converting the given equation to standard form,
![\begin{gathered} y\text{ = }(-5x)/(2)\text{ - }(8)/(2) \\ y\text{ = }(-5x-8)/(2) \\ 2y\text{ = -5x - 8} \\ 5x\text{ + 2y + 8 = 0} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/66lbzvfa951md40p0npz5pfzlw0wluk29v.png)
The slope of the given line is calculated as,
![Slope\text{ = }(-5)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/himddzcefhhg9gp8pj9gsuy44l8t5pmyha.png)
As the required line is parallel to the given line. Therefore slope of the required line is equal to the given line which is -5/2.
The required line passes through the point (5, -4).
The equation of a required line is calculated using the slope point formula.
![(y-y_1)\text{ = m}*\text{\lparen x-x}_1)](https://img.qammunity.org/2023/formulas/mathematics/college/74t5wa29iy1ag8wul9vvr2vky5dthpqrbc.png)
Where,
![\begin{gathered} m\text{ = }(-5)/(2) \\ (x_1,y_1)\text{ = \lparen 5, -4 \rparen} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/dwh6oa5iyjvz5dffbfk2t9ufjny4090z5b.png)
Required equation is calculated as,
![\begin{gathered} (y-(-4))\text{ = }(-5)/(2)\text{ \lparen x- 5\rparen} \\ 2*(y+4)\text{ = -5}*\text{\lparen x - 5\rparen} \\ 2y\text{ + 8 = -5x + 25} \\ 5x\text{ + 2y + 8 - 25 = 0} \\ 5x\text{ + 2y -17 = 0} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6ogma3jvbq1omv9qelfo7gus5yt51ctbgr.png)
Thus the equation of the required line is,
![5x\text{ + 2y - 17 = 0}](https://img.qammunity.org/2023/formulas/mathematics/college/tw3hbd5c3ksxewrimmftmb1s31bcpqakke.png)