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This is one of our many quizzes on Line segments. Click on this to access more such competitive quizzes on Geometry.

Line segments Quiz 2

mathematics

 Find the midpoint of the given line segment: $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6cd74277-c10b-4e2a-b956-c371147a1f548591185963701537963.png"/>

 Apply the formula for the midpoint of a line segment $\left( {x,y} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$. 

  Midpoint of line segment, $ \\ \left( { - 6,3} \right){\text{ to }}\left( {4, - 1} \right)   \\  \left( {x,y} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)   \\   \Rightarrow \left( {\dfrac{{ - 6 + 4}}{2},\dfrac{{3 - 1}}{2}} \right)   \\   \Rightarrow \left( {\dfrac{{ - 2}}{2},\dfrac{2}{2}} \right)   \\   \Rightarrow \left( { - 1,1} \right)   \\  $

 Find the centre of a circle of diameter with endpoints $\left( { - 2, - 7} \right)$ and $\left( {7, - 2} \right)$.

 Apply the formula for the midpoint of a line segment $\left( {x,y} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$.

  Midpoints for the diameter $ \\ \left( {x,y} \right) = \left( {\dfrac{{7 - 2}}{2},\dfrac{{ - 2 - 7}}{2}} \right)   \\   \Rightarrow \left( {\dfrac{5}{2}, - \dfrac{9}{2}} \right)   \\  $ So the centre of the circle is at $\left( {\dfrac{5}{2}, - \dfrac{9}{2}} \right).$

 Find the length of the given line segment $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4bb7decb-da2d-4ea1-acdd-6a7a162b14715108146929364827695.png"/>

 Apply the formula for the length of a line segment: $\ell  = \left( {x,y} \right) = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} - {{\left( {{y_2} - {y_1}} \right)}^2}} $. 

Length of the line segment is given by: $ \\   \ell  = \left( {x,y} \right) = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}    \\   \Rightarrow \ell  = \sqrt {{{\left( {2 + 6} \right)}^2} + {{\left( { - 6 - 4} \right)}^2}}    \\   \Rightarrow \ell  = \sqrt {{{\left( 8 \right)}^2} + {{\left( { - 10} \right)}^2}}    \\   \Rightarrow \ell  = \sqrt {64 + 100}    \\   \Rightarrow \ell  = \sqrt {164}    \\   \Rightarrow \ell  = 2\sqrt {41}    \\  $

 Find the point where diagonals of a square intersect with each other, we have given endpoints of one diagonal i.e. $\left( {4,2} \right)$ and $\left( { - 6, - 4} \right)$.

 Apply the formula for to find the midpoint of a line segment  $\left( {x,y} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$ because diagonals of square bisect each other.

 Diagonals intersect each other at midpoints so we have to find the midpoint of the diagonals i.e. $ \\   \left( {x,y} \right) = \left( {\dfrac{{4 - 6}}{2},\dfrac{{2 - 4}}{2}} \right)   \\   \Rightarrow \left( {x,y} \right) = \left( {\dfrac{{ - 2}}{2},\dfrac{{ - 2}}{2}} \right)   \\   \Rightarrow \left( {x,y} \right) = \left( { - 1, - 1} \right)   \\  $ So, the diagonals intersect each other at $\left( { - 1, - 1} \right).$

 Find the length of the line with endpoints $\left( {2, - 6} \right)$ and $\left( {4, - 12} \right)$. 

 Apply the formula to find  the length of a segment of a line: $\ell  = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} - {{\left( {{y_2} - {y_1}} \right)}^2}} $

 Length of the line is given by $ \\   \ell  = \left( {x,y} \right) = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} - {{\left( {{y_2} - {y_1}} \right)}^2}}    \\   \Rightarrow \ell  = \sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( { - 12 + 6} \right)}^2}}    \\   \Rightarrow \ell  = \sqrt {{{\left( 2 \right)}^2} + {{\left( 6 \right)}^2}}    \\   \Rightarrow \ell  = \sqrt {4 + 36}    \\   \Rightarrow \ell  = \sqrt {40}    \\   \Rightarrow \ell  = 2\sqrt {10}    \\  $