find the equation of straight line which passes through the point of intersection of straight lines.x+2y+3=0 and 3x+4y=7 and parallel to the straight line y-x=8

Accepted Solution

Answer:Equation of the line in question: [tex]y = x - 21[/tex].Step-by-step explanation:Start by finding the intersection of the two straight lines. The equation for both lines shall hold at their intersection. (Using the idea of the Gaussian Elimination.)[tex]\left\{\begin{aligned}&x + 2y =-3\\&3x + 4y=7\end{aligned}\right.[/tex].Add -3 times the first equation to the second:[tex]\left\{\begin{aligned}&x + 2y =-3\\& -2y=16\end{aligned}\right.[/tex].Add the second equation [tex]-2y=16[/tex] to the first:[tex]\left\{\begin{aligned}&x = 13\\&y=-8\end{aligned}\right.[/tex].Hence the intersection of the two lines will be [tex](13, -8)[/tex].Now, find the slope of that straight line. [tex]y - x = 8[/tex] is equivalent to [tex]y = x +8[/tex]. The slope of that line is equal to 1. So will be the slope of the line in question. Apply the point-slope form of a line on a Cartesian plane:Point: [tex](13, -8)[/tex],Slope: [tex]1[/tex].Equation of the line:[tex](y - (-8)) = (x - 13)[/tex].Simplify to obtain:[tex]y = x -21[/tex].