Question

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 vertices of a rhombus.

Answer 1

Here, A(0, 7, -10), B(1, 6, -6), C(4, 9, -6)

AB = $\sqrt{(1-0{)}^2+(6-7{)}^2+(-6+10{)}^2}$

=$\sqrt{1+1+16}$

=$3\sqrt{2}$ units

BC = $\sqrt{(4-1{)}^2+(9-6{)}^2+(-6+6{)}^2}$

=$\sqrt{9+9}$

=$3\sqrt{2}$ units

CA = $\sqrt{(0-4{)}^2+(7-9{)}^2+(-10+6{)}^2}$

=$\sqrt{16+4+16}$

=6 units

Since, AB = BC

So, $△ABC$ is an isosceles $△$

(ii) Here, A(0, 7, 10), B(-1, 6, 6), C(-4, 9, 6)

AB = $\sqrt{(-1-0{)}^2+(6-7{)}^2+(6-10{)}^2}$

=$\sqrt{1+1+16}$

=$3\sqrt{2}$ units

BC=$\sqrt{(-4+1{)}^2+(9-6{)}^2+(6-6{)}^2}$

=$\sqrt{9+9+0}$

$3\sqrt{2}$ units

AC=$\sqrt{(-4-0{)}^2+(9-7{)}^2+(6-10{)}^2}$

=$\sqrt{16+4+16}$

=$\sqrt{36}$ units

Since, $(AB{)}^2+(BC{)}^2=(AC{)}^2$

So, $△ABC$ is a right triangle.

(iii) Here, A(-1, 2, 1), B(1, -2, 5), C(4, -7, 8), D(2, -3, 4)

AB=$\sqrt{(1+1{)}^2+(-2-2{)}^2+(5-1{)}^2}$

=$\sqrt{4+16+16}$

=6 units

BC = $\sqrt{(4-1{)}^2+(-7+2{)}^2+(8-5{)}^2}$

=$\sqrt{9+25+9}$

=$\sqrt{43}$ units

CD = $\sqrt{(2-4{)}^2+(-3+7{)}^2+(4-8{)}^2}$

=$\sqrt{4+16+16}$

=6 units

DA = $\sqrt{(1+2{)}^2+(2+3{)}^2+(1-4{)}^2}$

=$\sqrt{9+25+9}$

=$\sqrt{43}$ units

Since, AB = CD and BC = DaA

So, $△ABC$ is a parallelogram

(iv) Here, A(5, -1, 1), B(7, -4, 7), C(1, -6, 10), D(-1, -3, 4)

AB $=\sqrt{(7-5{)}^2+(-4+1{)}^2+(7-1{)}^2}$

=$\sqrt{4+9+36}$ = $\sqrt{49}$ = 7 units

BC =$\sqrt{(1-7{)}^2+(-6+4{)}^2+(10-7{)}^2}$

=$\sqrt{36+4+9}$

=$\sqrt{49}=7$ units

CD = $\sqrt{(-1-1{)}^2+(-3+6{)}^2+(4-10{)}^2}$

= $\sqrt{4+9+36}$=$\sqrt{49}$ = 7 units

DA = $\sqrt{(-1-5{)}^2+(-3+1{)}^2+(4-1{)}^2}$

$\sqrt{36+4+9}=\sqrt{49}$=7 units

$\therefore$ AB = BC = CD = DA

Since all the sides are equal.

Thus quadrilateral ABCD is a rhombus.