નિશ્ચાયકનો ઉપયોગ કરી mathrm{A}(1, 3) અને math | vedclass

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Question

Let $P(x, y)$ be any point on $AB$. Then, area of triangle $ABP$ is zero (Why?). So

$\frac{1}{2}\left|\begin{array}{lll}
0 & 0 & 1 \
1 &

Vedclass · Accepted Answer

$\mp 2$

Vedclass · Answer

Let $P(x, y)$ be any point on $AB$. Then, area of triangle $ABP$ is zero (Why?). So

$\frac{1}{2}\left|\begin{array}{lll}
0 & 0 & 1 \
1 & 3 & 1 \
x & y & 1
\end{array}ight|=0$

This gives      $\frac{1}{2}(y-3 x)=0 	ext { or } y=3 x$

which is the equation of required line $\mathrm{AB}$.

Also, since the area of the triangle $\mathrm{ABD}$ is $3$ sq. units, we have

$\frac{1}{2}\left|\begin{array}{lll}
1 & 3 & 1 \
0 & 0 & 1 \
k & 0 & 1
\end{array}ight|=\pm 3$

This gives, $\frac{-3 k}{2}=\pm 3,$ i.c., $k=\mp 2$

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