Question

The equation of straight line which is equidistant from the points $A(2,–2)$, $B(6,1)$ and $C(–3,4)$ can be

• A. $2x+6y-5=0$
• B. $12x+10y-43=0$
• C. $6x-8y-11=0$
• D. $6x-8y+11=0$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $2x+6y-5=0$
B $12x+10y-43=0$
D $6x-8y+11=0$

Let $m_1,m_2$ and $m_3$ be slopes of $BC,CA$ and $AB$ respectively.
Then, $m_1=\frac{4-1}{-3-6}=\frac{-1}{3}$
$m_2=\frac{-2-4}{2+3}=\frac{-6}{5}$
and $m_3=\frac{1+2}{6-2}=\frac{3}{4}$
Let $AD$ be the altitude from $A$ to $BC$.
Then equation of $AD$ is
$(y+2)=3(x-2)\Rightarrow 3x-y-8=0$
which intersects $BC$ i.e., $x+3y-9=0$ at $D(\frac{33}{10},\frac{19}{10})$

Now, mid point of $AD$ is $D^{\prime }(\frac{53}{20},\frac{-1}{20})$
Equation of straight line parallel to $BC$ and passing through $D^{\prime }$ is $2(x+3y)-5=0$

Similarly, the equation of a straight line parallel to $AC$ and $AB$ and passing through mid-point of $B$ and $C$ respectively, will be $12x+10y-43=0$ and $6x-8y+11=0$

Alternate :
Find perpendicular distance of points $A,B$ and $C$ from the given lines in options. If distance is same, then that option is correct.