# Any Line Through the Intersection of Two Given Lines

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, $4x-3y-1-1(2x-5y+3)=0$ $4x-3y-1-2x+5y-3=0$ $2x+2y-4=0$ $\therefore x+y-2=0$