Please solve this for me.Thanks

Accepted Solution

Answer:10y = 13x + 139Step-by-step explanation:Points (-5, -3) and (-15, -16) are the two points through which Line 1 passes.So, the slope of the line 1 is [tex]\frac{-3-(-16)}{-5-(-15)} =\frac{13}{10}[/tex]Now, the slope of the parallel straight line i.e. line 2 will be same as line 1 i.e. [tex]\frac{13}{10}[/tex]Let us assume that the equation of line 2 is [tex]y= \frac{13}{10} x+c[/tex] {Where c is a constant} ....... (1)Now, line 2 passes through the point (-13, 17). Hence, this point will satisfy equation (1).So, Β [tex]17= \frac{13}{10} (-13)+c[/tex]β‡’ [tex]c=\frac{139}{10}[/tex]Therefore, the equation of line 2 will be [tex]y=\frac{13}{10} x+\frac{139}{10}[/tex]β‡’ 10y = 13x + 139 Β (Answer)