Question

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

Answer 1

Answer:

Equation of the line in question: $y = x - 21$.

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.)

$\left\{\begin{aligned}&x + 2y =-3\\&3x + 4y=7\end{aligned}\right.$.

Add -3 times the first equation to the second:

$\left\{\begin{aligned}&x + 2y =-3\\& -2y=16\end{aligned}\right.$.

Add the second equation $-2y=16$ to the first:

$\left\{\begin{aligned}&x = 13\\&y=-8\end{aligned}\right.$.

Hence the intersection of the two lines will be $(13, -8)$.

Now, find the slope of that straight line. $y - x = 8$ is equivalent to $y = x +8$. 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: $(13, -8)$,
• Slope: $1$.

Equation of the line:

$(y - (-8)) = (x - 13)$.

Simplify to obtain:

$y = x -21$.