Let \[A_1x+B_1y+C_1=0\] \[A_2x+B_2y+C_2=0\] be any two straight lines in linear form. Then, the equation of the line through the intersection of the given lines is \[A_1x+B_1y+C_1+\lambda(A_2x+B_2y+C_2)=0\] where $\lambda$ is any constant.
Find the equation of the line joining the point of intersection of the lines $x+3y+2=0$, $2x-y-3=0$ to the point $(3,1)$.
Given lines are, \[x+3y+2=0\] \[2x-y-3=0\] Any line through the intersection of the given lines is, \[x+3y+2+\lambda(2x-y-3)=0\text{ __(1)}\]
Since this line passes through $(3,1)$, so, \[3+3+2+\lambda(6-1-3)=0\] \[8+2\lambda=0\] \[\therefore \lambda=-4\] Putting the value of $\lambda$ in $\text{(1)}$ is, \[x+3y+2-4(2x-y-3)=0\] \[x+3y+2-8x+4y+12=0\] \[-7x+7y+14=0\] \[\therefore x-y-2=0\]
Obtain the equation of the straight line which makes equal intercepts on the axes and passes through the point of intersection of the lines $2x-3y+1=0$ and $x+2y=2$.
Given lines are, \[2x-3y+1=0\] \[x+2y-2=0\] Any line through the intersection of the given lines is, \[2x-3y+1+\lambda(x+2y-2)=0\text{ __(1)}\]
Put $x=0$, \[0-3y+1+\lambda(0+2y-2)=0\] \[-3y+1+2\lambda y-2\lambda=0\] \[(2\lambda-3)y=2\lambda-1\] \[\therefore y=\frac{2\lambda-1}{2\lambda-3}=\text{Y-intercept}\] Put $y=0$, \[2x-0+1+\lambda(x+0-2)=0\] \[2x+1+\lambda x-2\lambda=0\] \[(2+\lambda)x=2\lambda-1\] \[\therefore x=\frac{2\lambda-1}{2+\lambda}=\text{X-intercept}\]
Since the line makes equal intercepts on the axes, \[\text{X-intercept}=\text{Y-intercept}\] \[\frac{2\lambda-1}{2+\lambda}=\frac{2\lambda-1}{2\lambda-3}\] \[2\lambda-3=2+\lambda\] \[\therefore \lambda=5\]
Putting the value of $\lambda$ in $\text{(1)}$, we get, \[2x-3y+1+5(x+2y-2)=0\] \[2x-3y+1+5x+10y-10=0\] \[\therefore 7x+7y-9=0\]
Find the equation of the straight line passing through the intersection of $x+3y+2=0$, $2x-y-3=0$ and $\text{slope}=\frac{1}{2}$.
Given lines are, \[x+3y+2=0\] \[2x-y-3=0\] Any line through the intersection of the given lines is, \[x+3y+2+\lambda(2x-y-3)=0\] \[x+3y+2+2\lambda x-2\lambda y-3\lambda=0\] \[(1+2\lambda)x+(3-\lambda)y+(2-3\lambda)=0\] \[\therefore\text{Slope }(m)=-\frac{1+2\lambda}{3-\lambda}\] \[\frac{1}{2}=\frac{-1-2\lambda}{3-\lambda}\] \[3-\lambda=-2-4\lambda\] \[3\lambda=-5\] \[\therefore\lambda=-\frac{5}{3}\]
Hence, the required line is, \[x+3y+2-\frac{5}{3}(2x-y-3)=0\] \[3x+9y+6-10x+5y+15=0\] \[-7x+14y+21=0\] \[\therefore x-2y-3=0\]
Obtain the equations of two lines passing through the intersection of the lines $4x-3y-1=0$ and $2x-5y+3=0$ and equally inclined to the axes.
Given lines are, \[4x-3y-1=0\] \[2x-5y+3=0\] Any line through the intersection of the given lines is, \[4x-3y-1+\lambda(2x-5y+3)=0\] \[4x-3y-1+2\lambda x-5\lambda y+3\lambda=0\] \[(4+2\lambda)x-(3+5\lambda)y+(3\lambda-1)=0\] \[\therefore\text{Slope }(m)=\frac{3+5\lambda}{4+2\lambda}\]
Since the lines are equally inclined to the axes, $\text{Slope }(m)=\pm 1$. \[\therefore\pm 1=\frac{3+5\lambda}{4+2\lambda}\] Taking $m=1$, \[1=\frac{3+5\lambda}{4+2\lambda}\] \[4+2\lambda=3+5\lambda\] \[\therefore\lambda=\frac{1}{3}\]
Hence, required equation of line is, \[4x-3y-1+\frac{1}{3}(2x-5y+3)=0\] \[12x-9y-3+2x-5y+3=0\] \[14x-14y=0\] \[\therefore x-y=0\]
Taking $m=-1$, \[-1=\frac{3+5\lambda}{4+2\lambda}\] \[-4-2\lambda=3+5\lambda\] \[-7=7\lambda\] \[\therefore\lambda=-1\]
Hence, the required equation of line is, \[4x-3y-1-1(2x-5y+3)=0\] \[4x-3y-1-2x+5y-3=0\] \[2x+2y-4=0\] \[\therefore x+y-2=0\]