In a $\Delta ABC, 2x + 3y + 1 = 0 , x + 2y - 2 = 0 $ are the perpendicular bisectors of its sides $AB$ and $AC$ respectively and if $A = (3,2)$, then the equation of the side $BC$ is
Given that, equation of $A B$ is $3 x-2 y+c=0$ passes through $A(3,2)$ $\Rightarrow C=-5$
$\therefore$ Equation becomes $3 x-2 y-5=0$ Here, $D=(1,-1)$ Since, $D$ is mid-point of $A B$. Let coordinates of $B$ are $(\alpha, \beta)$. $\therefore \frac{3+\alpha}{2}=1$ and $\frac{2+\beta}{2}=-1$ $\rightarrow 3+\alpha=2$ and $2+\beta=-2$ $\Rightarrow \alpha=-1$ and $\beta=-4$ $\therefore B$ is $(-1,-4)$. Similarly, equation of $A C$ is $2 x-y+c=0$ is passes through $(3,2)$ $ \Rightarrow C=-4 \Rightarrow 2 x-y-4=0$ Here, $E$ is a point of intersection of $x+2 y-2=0$ and $2 x-y-4=0$. $\therefore E=(2,0)$ Since, $E$ is mid-point of $A C$. Let coordinates of $C$ are $\left(\alpha_{1}, \beta_{1}\right)$ $\therefore \frac{3+\alpha_{1}}{2}=2$ and $ \frac{2+\beta_{1}}{2}=0 $ $ \Rightarrow \alpha_{1}=4-3 $ and $ \beta_{1}=-2$ $ \Rightarrow \alpha_{1}=1 $ and $ \beta_{1}=-2 $ $\therefore C$ is $(1,-2) $ $\therefore$ Required equation of $B C$ is $(y+2) =\frac{-2+4}{1+1}(x-1) $ $\Rightarrow y+2 =\frac{2}{2}(x-1)$ $\Rightarrow y+2 =(x-1) $ $\Rightarrow x-y -3=0$
A straight line is a figure created when two points A (x1, y1) and B (x2, y2) are connected with a minimum distance between them, and both the ends are extended to immensity (infinity). With variables x and y, the standard form of a linear equation is: ax + by = c, where a, b, and c are constants and x, and y are variables.
Point Slope Form:
The equation of a straight line whose slope is m and passes through a point (x1, y1) is formed or created using the point-slope form. The equation of the point-slope form is:
