Question

find the equation of straight line joining (4,-3) and the intersection of the straight lines 2x-y+7=0 and x+y-1=0

Answer 1

The given equations are :

$$\begin{gathered} 2x-y=-7......\left(1\right) \\ x+y=1.......\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 3x=-6 \\ \Rightarrow x=-2 \\ \mathrm{From}\left(2\right),\mathrm{we}\mathrm{get} \\ y=1-x=1+2=3 \\ \mathrm{So},\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\left(1\right)\mathrm{and}\left(2\right)\mathrm{is}\left(-2,3\right). \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{straight}\mathrm{line}\mathrm{joining}\left(x_1,y_1\right)\mathrm{and}\left(x_2,y_2\right)\mathrm{is} \\ \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} \\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{straight}\mathrm{line}\mathrm{joining}\left(4,-3\right)\mathrm{and}\left(-2,3\right)\mathrm{is}, \\ \frac{y-\left(-3\right)}{x-4}=\frac{3-\left(-3\right)}{-2-4} \\ \Rightarrow \frac{y+3}{x-4}=\frac{6}{-6} \\ \Rightarrow \frac{y+3}{x-4}=-1 \\ \Rightarrow y+3=4-x \\ \Rightarrow \boxed{x+y=1} \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{line}. \end{gathered}$$