Question

# Mid-points of the sides AB and AC of a $$\bigtriangleup$$ABC are (3, 5) and (-3, -3) respectively, then the length of the side BC is :

A
10
B
20
C
15
D
30

Solution

## The correct option is B $$20$$$$Given-\\ ABC\quad is\quad a\quad triangle.\\ The\quad mid\quad point\quad of\quad AC\quad is\quad P\quad and\quad that\quad of\quad AB\quad is\quad Q.\\ to\quad find\quad out-\quad \\ the\quad length\quad of\quad PQ=?\\ Solution-\\ We\quad shall\quad use\quad the\quad mid\quad point\quad theorem\quad which\quad says,\\ the\quad line\quad joining\quad the\quad mid\quad points\quad of\quad the\quad two\quad sides\\ of\quad a\quad triangle,\quad is\quad parallel\quad to\quad and\quad half\quad the\quad third\quad side.\\ \therefore \quad 2PQ=BC.\\ Now\quad P({ x }_{ 1 },{ y }_{ 1 })=(3,5)\quad and\quad Q({ x }_{ 2 },{ y }_{ 2 })=(-3,-3).\\ \therefore \quad Using\quad distance\quad formula,\\ PQ=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } \\ =\sqrt { { \left( 3+3 \right) }^{ 2 }+{ \left( 5+3 \right) }^{ 2 } } units=10units.\\ \therefore \quad BC=2PQ=2\times 10units=20units.\\ Ans-\quad Option\quad B.\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \\$$Mathematics

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