Search any question & find its solution
Question:
Answered & Verified by Expert
The points $(1,3,4),(-1,6,10),(-7,4,7)$ and $(-5,1,1)$ are the vertices of a
Options:
Solution:
1750 Upvotes
Verified Answer
The correct answer is:
rhombus
Let $(1,3,4),(-1,6,10),(-7,4,7)$ and $(-5,1,1)$ be the
coordinates of points $A, B, C$ and $D$ respectively.
$\begin{aligned} A B &=\sqrt{(-1-1)^{2}+(6-3)^{2}+(10-4)^{2}} \\ &=\sqrt{4+9+36}=7 \\ B C &=\sqrt{(-7+1)^{2}+(4-6)^{2}+(7-10)^{2}} \\ &=\sqrt{36+4+9}=7 \\ C D &=\sqrt{(-5+7)^{2}+(1-4)^{2}+(1-7)^{2}} \\ &=\sqrt{4+9+36}=7 \\ D A &=\sqrt{(1+5)^{2}+(3-1)^{2}+(4-1)^{2}} \\ &=\sqrt{36+4+9}=7 \\ A C &=\sqrt{(-7-1)^{2}+(4-3)^{2}+(7-4)^{2}} \\ &=\sqrt{64+1+9}=\sqrt{74} \end{aligned}$
and $B D=\sqrt{(-5+1)^{2}+(1-6)^{2}+(1-10)^{2}}$
$=\sqrt{16+25+81}=\sqrt{122}$
$\therefore A B=B C=C D=D A$
But $B D \neq A C$
\therefore Points $A$, $B, C$ and $D$ are the vertices of a rhombus
coordinates of points $A, B, C$ and $D$ respectively.
$\begin{aligned} A B &=\sqrt{(-1-1)^{2}+(6-3)^{2}+(10-4)^{2}} \\ &=\sqrt{4+9+36}=7 \\ B C &=\sqrt{(-7+1)^{2}+(4-6)^{2}+(7-10)^{2}} \\ &=\sqrt{36+4+9}=7 \\ C D &=\sqrt{(-5+7)^{2}+(1-4)^{2}+(1-7)^{2}} \\ &=\sqrt{4+9+36}=7 \\ D A &=\sqrt{(1+5)^{2}+(3-1)^{2}+(4-1)^{2}} \\ &=\sqrt{36+4+9}=7 \\ A C &=\sqrt{(-7-1)^{2}+(4-3)^{2}+(7-4)^{2}} \\ &=\sqrt{64+1+9}=\sqrt{74} \end{aligned}$
and $B D=\sqrt{(-5+1)^{2}+(1-6)^{2}+(1-10)^{2}}$
$=\sqrt{16+25+81}=\sqrt{122}$
$\therefore A B=B C=C D=D A$
But $B D \neq A C$
\therefore Points $A$, $B, C$ and $D$ are the vertices of a rhombus
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.