Question

If $L_1:x+y-1=0$ and $L_2:2x-y+4=0$ and $S(2,1)$ then find the x-intercept of the line $L=0$.

Answer 1

$L_1=x+4-1=0$ ...(i)

$L_2=2x-y+4=0$ ...(ii)

intersection $S(2,1)$

Slope of $L_1=x+y-1=0$

$y=-x+1$

the $m_1=-1$

Similarly, $m_2=1$

Perpendicular slope of a line $=\frac{-1}{m_1}$

So, slop of a line $DS=1$ and $ES=\frac{-1}{2}$

The equation of $SD$ is

$y-1=1(x-2)$

$y=x-2+1$

$y=x-1$ ...(iii)

The eqn $SE$

$y-1=\frac{-1}{2}(x-2)$

$2y+x-4=0$ ...(iv)

Here $D$ is midpoint of $AB$

Similarly, (i) and (ii) $y=0,x=1$

So, $D$ is $(1,0)$

Similarly, $E$ is midpoint of $AB$

Similarly, (ii) and (iv) $x=\frac{-4}{5}$ and $y=\frac{12}{5}$

So, $E$ is $(\frac{-4}{5},\frac{12}{5})$

adding eqn (i) and (ii)

$x+y-1=0$

$2x-y+4=0$

Similarly eqn will get $x=-1,y=2$

So point $A(-1,2)$

So from triangle the midpoint $D$ and $E$ , now we find $C$

$\frac{x-1}{2}=-\frac{4}{5}\Rightarrow x=-\frac{3}{5}$

$\frac{y-2}{2}=\frac{12}{5}\Rightarrow y=\frac{34}{5}$

So, $C(-\frac{3}{5},\frac{34}{5})$

Similarly, $B(3,-2)$

The intersect of $L$

$L:y+2=\frac{4y}{5}(x-3)/1815$

$y+2=\frac{22}{9}(x-3)\Rightarrow \frac{22x}{9}-\frac{16}{9}-2$

$y=\frac{22x}{9}-\frac{84}{9}$ ...(v)

From finding $x$ intersept

Putting $y=0$ in (v)

$y=\frac{22\left(o\right)}{9}-\frac{84}{9}$

$y=\frac{42}{11}$