Question

The plane which bisects the line joining the points $(4,-2,3)$and $(2,4,-1)$ at right angles also passes through the point

• A. $(0,-1,1)$
• B. $(4,0,1)$
• C. $(4,0,-1)$
• D. $(0,1,-1)$

Answer 1

The correct option is C

$(4,0,-1)$

Explanation for the correct option:

Find the equation of the plane

As the plane bisects the line joining the points $A=(4,-2,3)$and $B=(2,4,-1)$.

Let the point $P=(\alpha ,\beta ,\gamma )$lies on a given plane.

So,

$$\begin{gathered} PA=PB \\ \Rightarrow {\left(PA\right)}^2={\left(PB\right)}^2 \\ \Rightarrow {(\alpha -4)}^2+{(\beta +2)}^2+{(\gamma -3)}^2={(\alpha -2)}^2+{(\beta -4)}^2+{(\gamma +1)}^2 \\ \Rightarrow {\alpha }^2+16-8\alpha +{\beta }^2+4+4\beta +{\gamma }^2+9-6\gamma ={\alpha }^2+4-4\alpha +{\beta }^2+16-8\beta +{\gamma }^2+1+2\gamma \\ \Rightarrow -4\alpha +12\beta -8\gamma =-8 \end{gathered}$$

$\Rightarrow 2x-6y+4z=4.....\left(\mathrm{i}\right)$ [ Replace $(\alpha ,\beta ,\gamma )$ with $(x,y,z)$ ]

Option (C): By substituting $(x,y,z)=(4,0,-1)$ in the L.H.S. of the equation (i) we get,

$$\begin{gathered} L.H.S.=2\left(4\right)-6\left(0\right)+4(-1) \\ =8-0-4 \\ =4 \\ =R.H.S \end{gathered}$$

Hence, option (C) is correct.

Explanation for the incorrect options:

Option (A): By substituting $(x,y,z)=(0,-1,1)$ in the L.H.S. of the equation (i) we get,

$$\begin{gathered} L.H.S.=2\left(0\right)-6(-1)+4\left(1\right) \\ =0+6+4 \\ =10 \\ \ne R.H.S \end{gathered}$$

Hence, option (A) is incorrect.

Option (B): By substituting $(x,y,z)=(4,0,1)$ in the L.H.S. of the equation (i) we get,

$$\begin{gathered} L.H.S.=2\left(4\right)-6\left(0\right)+4\left(1\right) \\ =8-0+4 \\ =12 \\ \ne R.H.S \end{gathered}$$

Hence, option (B) is also incorrect.

Option (D): By substituting $(x,y,z)=(0,1,-1)$ in the L.H.S. of the equation (i) we get,

$$\begin{gathered} L.H.S.=2\left(0\right)-6(-1)+4\left(1\right) \\ =0+6+4 \\ =10 \\ \ne R.H.S \end{gathered}$$

Hence, option (D) is also incorrect.

Therefore, option (C) is the correct answer.