
The triangle formed by the points $\left( {0,7,10} \right)$ $\left( { - 1,6,6} \right)$ and $\left( { - 4,9,6} \right)$ is
A. Equilateral
B. Isosceles
C. Right - angled
D. Right-angled isosceles
Answer
162.9k+ views
Hint:
In this question we need to recognize the shape of a triangle with provided coordinates. To find distance between any two coordinates we are going to use this formula for distance if coordinates are given. Distance=$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $. Then, we compare the characteristics of the polygon with the information we have obtained and given.
Formula Used:
Distance between two points$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Complete step-by-step solution:
Given: The points of triangle are given as $\left( {0,7,10} \right)$ $\left( { - 1,6,6} \right)$ and $\left( { - 4,9,6} \right)$
Let these three points be denoted by A, B, C, respectively
Length of AB
$ AB = \sqrt {{{\left( {0 + 1} \right)}^2} + {{\left( {7 - 6} \right)}^2} + {{\left( {10 - 6} \right)}^2}}\\
= \sqrt {{{\left( 1 \right)}^2} + {{\left( { 1} \right)}^2} + {{\left( 4 \right)}^2}} \\
= \sqrt {1 + 1 + 16} = \sqrt {18}\\
AB = 3\sqrt 2 …(1)$
Length of BC
$BC = \sqrt {{{\left( {-1+4} \right)}^2} + {{\left( {6 - 9} \right)}^2} + {{\left( { 6-6} \right)}^2}}\\
= \sqrt {{{\left( 3 \right)}^2} + {{\left( -3 \right)}^2} + {{\left( 0 \right)}^2}} \\
= \sqrt {9 + 9} = \sqrt {18} \\
BC = 3\sqrt 2…(2)$
Length of AC
$AC=\sqrt {{{\left({0 + 4} \right)}^2} + {{\left( {7 - 9} \right)}^2} + {{\left( { 10-6} \right)}^2}}\\
= \sqrt {{{\left( {4} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( {4} \right)}^2}} \\
= \sqrt {16 + 4 + 16} = \sqrt {36} \\
AC = 6…(3)$
From the above equation (1), (2) and (3) we can conclude that
$AB = BC \ne AC$
As two sides are equal we can say that it is isosceles triangle. Now to find weather it’s a right-angled or not we are going to use Pythagoras theorem.
$A{C^2} = A{B^2} + B{C^2}$
${\left( {3\sqrt 2 } \right)^2} + {\left( {3\sqrt 2 } \right)^2} = {6^2}$
$\therefore 36 = 36$
As the Pythagoras theorem works we can say that it is a right- angled triangle.
Hence the correct option is D (Right-angled isosceles).
So, option D is correct.
Note:The length of the line segment is the distance that separates two points, so using distance formula we are calculating the length of the sides of triangle. For such question we must know the properties of triangle, Pythagoras Theorem.
In this question we need to recognize the shape of a triangle with provided coordinates. To find distance between any two coordinates we are going to use this formula for distance if coordinates are given. Distance=$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $. Then, we compare the characteristics of the polygon with the information we have obtained and given.
Formula Used:
Distance between two points$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Complete step-by-step solution:
Given: The points of triangle are given as $\left( {0,7,10} \right)$ $\left( { - 1,6,6} \right)$ and $\left( { - 4,9,6} \right)$
Let these three points be denoted by A, B, C, respectively
Length of AB
$ AB = \sqrt {{{\left( {0 + 1} \right)}^2} + {{\left( {7 - 6} \right)}^2} + {{\left( {10 - 6} \right)}^2}}\\
= \sqrt {{{\left( 1 \right)}^2} + {{\left( { 1} \right)}^2} + {{\left( 4 \right)}^2}} \\
= \sqrt {1 + 1 + 16} = \sqrt {18}\\
AB = 3\sqrt 2 …(1)$
Length of BC
$BC = \sqrt {{{\left( {-1+4} \right)}^2} + {{\left( {6 - 9} \right)}^2} + {{\left( { 6-6} \right)}^2}}\\
= \sqrt {{{\left( 3 \right)}^2} + {{\left( -3 \right)}^2} + {{\left( 0 \right)}^2}} \\
= \sqrt {9 + 9} = \sqrt {18} \\
BC = 3\sqrt 2…(2)$
Length of AC
$AC=\sqrt {{{\left({0 + 4} \right)}^2} + {{\left( {7 - 9} \right)}^2} + {{\left( { 10-6} \right)}^2}}\\
= \sqrt {{{\left( {4} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( {4} \right)}^2}} \\
= \sqrt {16 + 4 + 16} = \sqrt {36} \\
AC = 6…(3)$
From the above equation (1), (2) and (3) we can conclude that
$AB = BC \ne AC$
As two sides are equal we can say that it is isosceles triangle. Now to find weather it’s a right-angled or not we are going to use Pythagoras theorem.
$A{C^2} = A{B^2} + B{C^2}$
${\left( {3\sqrt 2 } \right)^2} + {\left( {3\sqrt 2 } \right)^2} = {6^2}$
$\therefore 36 = 36$
As the Pythagoras theorem works we can say that it is a right- angled triangle.
Hence the correct option is D (Right-angled isosceles).
So, option D is correct.
Note:The length of the line segment is the distance that separates two points, so using distance formula we are calculating the length of the sides of triangle. For such question we must know the properties of triangle, Pythagoras Theorem.
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