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Let the sides of a parallelogram be \mathrm{u = p, u = q, v = r } and \mathrm{v = s}  where \mathrm{u = lx + my + n} and \mathrm{v=1'x+m'y+n'}.  Show that the equation of the diagonal through the point of intersection of \mathrm{u = p} and \mathrm{v = r} and \mathrm{u = q} and \mathrm{v = s}, is given by

 

Option: 1

\mathrm{\quad\left|\begin{array}{lll} u & v & 1 \\ p & r & 1 \\ q & s & 1 \end{array}\right|=0 }
 


 


Option: 2

\mathrm{\quad\left|\begin{array}{lll} u & v & 1 \\ p & r & 1 \\ q & s & 1 \end{array}\right|=0 }


Option: 3

\mathrm{\left|\begin{array}{lll} u & v & 1 \\ p & r & 2 \\ q & s & 3 \end{array}\right|=0}


Option: 4

\mathrm{\left|\begin{array}{lll}u & v & 1 \\ p & \gamma & 1 \\ q & s & 1\end{array}\right|=1}


Answers (1)

best_answer

Equation of line through point of intersection D of lines \mathrm{u-p=0 \: and \: v-r=0} is (\mathrm{u}-\mathrm{p})+\lambda(\mathrm{v}-\mathrm{r})=0          ..............(i)

it is also passing through \mathrm{u=q \: and \: \mathrm{v}=\mathrm{s}}

then (i) becomes

\mathrm{ \begin{array}{ll} & (\mathrm{q}-\mathrm{p})+\lambda(\mathrm{s}-\mathrm{r})=0 \\ \end{array} }

\mathrm{ \begin{array}{ll} \therefore \quad & \lambda=\frac{-(\mathrm{q}-\mathrm{p})}{(\mathrm{s}-\mathrm{r})} \end{array} }             .........................(ii)

from (1) and (2) we get

\mathrm{ (u-p)-\frac{(q-p)}{(s-r)}(v-r)=0 }

\mathrm{\Rightarrow u(s-r)-p(s-r)-v(q-p)+r(q-p)=0 }

\mathrm{\Rightarrow u(r-s)-v(p-q)+p s-q r=0 }

\mathrm{\Rightarrow \left|\begin{array}{lll} u & v & 1 \\ p & r & 1 \\ q & s & 1 \end{array}\right|=0}


Hence option 2 is correct.

Posted by

jitender.kumar

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