Question

If a triangle $ABC$ remain always similar to a given triangle, and if the point $A$ be fixed and the point $B$ always move along a given straight line, find the locus of the point $C$.

Answer 1

Suppose $\Delta ABC$ is similar to $\Delta DEF$

$\therefore \frac{AB}{DE}=\frac{AC}{DF}$

$\Rightarrow \frac{AB}{AC}=k\dots \left(1\right)$ (where $k$ is a constant)

$A$ is a fixed point and locus of point $B$ is a straight line.

Suppose $A$ is $(0,0)$ and any point on the straight of $B$ be $(h,mh+c)$ (where $c$ and $m$ are constant)

So, distance between point $A$ and point $B$ is

$AB=\sqrt{h^2+(mh+c{)}^2}$

From eqn $\left(1\right)$, we get

$h^2+(mh+c{)}^2=(AC{)}^2{k}^2$

$\Rightarrow AC=\sqrt{\frac{h^2}{k}+\frac{(mh+c{)}^2}{k}}$

$\Rightarrow AC=\sqrt{h_1^2+(m{h}_1+c_1{)}^2}$

Locus of $h_1$ is also a straight line where $(h_1,(m{h}_1+c_1\left)\right)$ is a point present on the locus of $C$

$\therefore$ Locus of point $C$ is straight line.