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# Verify the following: (i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle. (ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, –6) are vertices of a right-angled triangle. (iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram. (iv) (5, –1, 1), (7, –4,7), (1, –6,10) and (–1, – 3,4) are the vertices of a rhombus.

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## (i) Let A(0, 7, $-$10) , B(1, 6, $-$6) , C(4, 9, $-$6) be the vertices of $△ABC$.Then, AB = $\sqrt{{\left(1-0\right)}^{2}+{\left(6-7\right)}^{2}+{\left(-6+10\right)}^{2}}$ $=\sqrt{{1}^{2}+{\left(-1\right)}^{2}+{4}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+1+16}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$ BC = $\sqrt{{\left(4-1\right)}^{2}+{\left(9-6\right)}^{2}+{\left(-6+6\right)}^{2}}$ $=\sqrt{{3}^{2}+{3}^{2}+0}\phantom{\rule{0ex}{0ex}}=\sqrt{9+9}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$ CA= $\sqrt{{\left(0-4\right)}^{2}+{\left(7-9\right)}^{2}+{\left(-10+6\right)}^{2}}$ $=\sqrt{16+4+16}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=6$ Clearly, AB = BC Thus, the given points are the vertices of an isosceles triangle. (ii) Let A(0,7,10) , B( $-$1,6,6) and C( $-$4,9,6) be the vertices of $△ABC$. Then , AB = $\sqrt{{\left(-1-0\right)}^{2}+{\left(6-7\right)}^{2}+{\left(6-10\right)}^{2}}$ $=\sqrt{{\left(-1\right)}^{2}+{\left(-1\right)}^{2}+{\left(-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+1+16}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$ BC = $\sqrt{{\left(-4+1\right)}^{2}+{\left(9-6\right)}^{2}+{\left(6-6\right)}^{2}}$ $=\sqrt{{\left(-3\right)}^{2}+{\left(3\right)}^{2}+0}\phantom{\rule{0ex}{0ex}}=\sqrt{9+9}\phantom{\rule{0ex}{0ex}}=\sqrt{18}\phantom{\rule{0ex}{0ex}}=3\sqrt{2}$ AC = $\sqrt{{\left(-4-0\right)}^{2}+{\left(9-7\right)}^{2}+{\left(6-10\right)}^{2}}$ $=\sqrt{{\left(-4\right)}^{2}+{\left(2\right)}^{2}+{\left(-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{16+4+16}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=6$ $A{C}^{2}$$=A{B}^{2}+B{C}^{2}$ Thus, the given points are the vertices of a right-angled triangle. (iii) Let A($-$1, 2, 1) , B(1, $-$2, 5) , C(4, $-$7, 8), D(2, $-$3, 4) be the vertices of quadrilateral $\square ABCD$ $AB=\sqrt{{\left(1+1\right)}^{2}+{\left(-2-2\right)}^{2}+{\left(5-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+16+16}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=6\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(4-1\right)}^{2}+{\left(-7+2\right)}^{2}+{\left(8-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+25+9}\phantom{\rule{0ex}{0ex}}=\sqrt{43}\phantom{\rule{0ex}{0ex}}CD=\sqrt{{\left(2-4\right)}^{2}+{\left(-3+7\right)}^{2}+{\left(4-8\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+16+16}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=6\phantom{\rule{0ex}{0ex}}DA=\sqrt{{\left(-1-2\right)}^{2}+{\left(2+3\right)}^{2}+{\left(1-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+25+9}\phantom{\rule{0ex}{0ex}}=\sqrt{43}\phantom{\rule{0ex}{0ex}}\therefore AB=CD\mathrm{and}BC=DA\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ Since, each pair of opposite sides are equal. Thus, quadrilateral $\square ABCD$ is a parallelogram. (iv) Let A(5, $-$1, 1) , B(7, $-$4, 7) , C(1, $-$6, 10), D($-$1, $-$3, 4) be the vertices of quadrilateral $\square ABCD$ $\mathrm{AB}=\sqrt{{\left(5-7\right)}^{2}+{\left(-1+4\right)}^{2}+{\left(1-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+9+36}\phantom{\rule{0ex}{0ex}}=\sqrt{49}\phantom{\rule{0ex}{0ex}}=7\phantom{\rule{0ex}{0ex}}\mathrm{BC}=\sqrt{{\left(7-1\right)}^{2}+{\left(-4+6\right)}^{2}+{\left(7-10\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{36+4+9}\phantom{\rule{0ex}{0ex}}=\sqrt{49}\phantom{\rule{0ex}{0ex}}=7\phantom{\rule{0ex}{0ex}}\mathrm{CD}=\sqrt{{\left(1+1\right)}^{2}+{\left(-6+3\right)}^{2}+{\left(10-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+9+36}\phantom{\rule{0ex}{0ex}}=\sqrt{49}\phantom{\rule{0ex}{0ex}}=7\phantom{\rule{0ex}{0ex}}\mathrm{DA}=\sqrt{{\left(-1-5\right)}^{2}+{\left(-3+1\right)}^{2}+{\left(4-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{36+4+9}\phantom{\rule{0ex}{0ex}}=\sqrt{49}\phantom{\rule{0ex}{0ex}}=7\phantom{\rule{0ex}{0ex}}\therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ Since, all the sides are equal. Thus, quadrilateral ABCS is a rhombus.  Suggest Corrections  0      Similar questions  Related Videos   Distance Formula
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