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Standard XII

Mathematics

Equation of a Line Passing through 2 Points

If ax12+by12+...

Question

If $a x_{1}^{2} + b y_{1}^{2} + c z_{1}^{2} = a x_{2}^{2} + b y_{2}^{2} + c z_{2}^{2} = a x_{3}^{2} + b y_{3}^{2} + c z_{3}^{2} = d$ and
$a x_{2} x_{3} + b y_{2} y_{3} + c z_{2} z_{3} = a x_{3} x_{1} + b y_{3} y_{1} + c z_{3} z_{1} = a x_{1} x_{2} + b y_{1} y_{2} + c z_{1} z_{2} = f,$ where $a, b, c, d, f > 0$ and $d > 2 f,$ then the value of $∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & z_{1} x_{2} & y_{2} & z_{2} x_{3} & y_{3} & z_{3} \end{matrix} ∣ ∣ ∣ ∣$ is:

(d - f) {\frac{(d + 2 f)}{a b c}}^{\frac{1}{2}}

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| f - d | {\frac{(d - 2 f)}{a b c}}^{\frac{1}{2}}

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(d - f) {\frac{(d - 2 f)}{a b c}}^{\frac{1}{2}}

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(d - f) {\frac{(d + f)}{a b c}}^{\frac{1}{2}}

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Solution

The correct option is A $(d - f) {\frac{(d + 2 f)}{a b c}}^{\frac{1}{2}}$
Let $D = ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & z_{1} x_{2} & y_{2} & z_{2} x_{3} & y_{3} & z_{3} \end{matrix} ∣ ∣ ∣ ∣$
$∴ D^{2} = D \times D$
$\Rightarrow D^{2} = ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & z_{1} x_{2} & y_{2} & z_{2} x_{3} & y_{3} & z_{3} \end{matrix} ∣ ∣ ∣ ∣ \times ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & z_{1} x_{2} & y_{2} & z_{2} x_{3} & y_{3} & z_{3} \end{matrix} ∣ ∣ ∣ ∣$
$= \frac{1}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} a x_{1} & b y_{1} & c z_{1} a x_{2} & b y_{2} & c z_{2} a x_{3} & b y_{3} & c z_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & z_{1} x_{2} & y_{2} & z_{2} x_{3} & y_{3} & z_{3} \end{matrix} ∣ ∣ ∣ ∣$
$= \frac{1}{a b c} ∣ ∣ ∣ ∣ ∣ \begin{matrix} a x_{1}^{2} + b y_{1}^{2} + c z_{1}^{2} & a x_{1} x_{2} + b y_{1} y_{2} + c z_{1} z_{2} & a x_{3} x_{1} + b y_{3} y_{1} + c z_{3} z_{1} a x_{1} x_{2} + b y_{1} y_{2} + c z_{1} z_{2} & a x_{2}^{2} + b y_{2}^{2} + c z_{2}^{2} & a x_{2} x_{3} + b y_{2} y_{3} + c z_{2} z_{3} a x_{3} x_{1} + b y_{3} y_{1} + c z_{3} z_{1} & a x_{2} x_{3} + b y_{2} y_{3} + c z_{2} z_{3} & a x_{3}^{2} + b y_{3}^{2} + c z_{3}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= \frac{1}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} d & f & f f & d & f f & f & d \end{matrix} ∣ ∣ ∣ ∣$
$C_{1} \to C_{1} + C_{2} + C_{3}, C_{2} \to C_{2} - C_{3}$
$= \frac{1}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} d + 2 f & 0 & f d + 2 f & d - f & f d + 2 f & f - d & d \end{matrix} ∣ ∣ ∣ ∣$
$= \frac{(d + 2 f) (d - f)}{a b c} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & f 1 & 1 & f 1 & - 1 & d \end{matrix} ∣ ∣ ∣ ∣$
On expanding and solving the determinant, we have:

$D^{2} = \frac{(d + 2 f) (d - f)^{2}}{a b c}$
$\Rightarrow D = (d - f) {\frac{(d + 2 f)}{a b c}}^{\frac{1}{2}}$

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Similar questions

Q. If

a x_{1}^{2} + b y_{1}^{2} + c z_{1}^{2} = a x_{2}^{2} + b y_{2}^{2} + c z_{2}^{2} = a x_{3}^{2} + b y_{3}^{2} + c z^{3} = d, a x_{2} x_{3} + b y_{2} y_{3} + c z_{2} z_{3} = a x_{3} x_{1} + b y_{3} y_{1} + c z_{3} z_{1} = a x_{1} x_{2} + b y_{1} y_{2} + c z_{1} z_{2} = f

, then

∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & z_{1} x_{2} & y_{2} & z_{2} x_{3} & y_{3} & z_{3} \end{matrix} ∣ ∣ ∣ ∣ = ?

Q. If

a x_{1}^{2} + b y_{1}^{2} + c z_{1}^{2} = a x_{2}^{2} + b y_{2}^{2} + c z_{2}^{2} = a x_{3}^{2} + b y_{3}^{2} + c z_{3}^{2} = d

and

a x_{2} x_{3} + b y_{2} y_{3} + c z_{2} z_{3} = a x_{3} x_{1} + b y_{3} y_{1} + c z_{3} z_{1} = a x_{1} x_{2} + b y_{1} y_{2} + c z_{1} z_{2} = f

, then prove that

∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & z_{1} x_{2} & y_{2} & z_{2} x_{3} & y_{3} & z_{3} \end{matrix} ∣ ∣ ∣ ∣ = (d - f) {[\frac{d + 2 f}{a b c}]}^{1 / 2} (a, b, c \neq 0)

Show that the centroid of the triangle with vertices A(x1, y1, z1), B(x2, y2, z2) and C(x3, y3, z3) is
((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3).

Show that the coordinates of the centroid of the triangle with vertices $A (x_{1}, y_{1}, z_{1})$ , $B (x_{2}, y_{2}, z_{2})$ and $C (x_{3}, y_{3}, z_{3})$ is $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3})$ .

The coordinates of the mid-point of the line segment joining two points P(X₁, y₁, z₁) and Q(x₂, y₂, z₂) are:

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Equation of a Line Passing through 2 Points

Standard XII Mathematics

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If ax21+by21+cz21=ax22+by22+cz22=ax23+by23+cz23=d and ax2x3+by2y3+cz2z3=ax3x1+by3y1+cz3z1=ax1x2+by1y2+cz1z2=f, where a,b,c,d,f>0 and d>2f, then the value of ∣∣ ∣∣x1y1z1x2y2z2x3y3z3∣∣ ∣∣ is: