Question

The mid-points of the sides of a triangle are (2, 1), (1, 0) and (-1, 3). Find the coordinates of the vertices of the triangle.

Answer 1

Answer:

${\pink{\green{\sf{\therefore Co-ordinate \: of \: Vertices \: A = ( - 2,2) }}}}$

${\pink{\green{\sf{\therefore Co-ordinate \: of \: Vertices \: B = ( 0,4) }}}}$

${\pink{\green{\sf{\therefore Co-ordinate \: of \: Vertices \: C = ( 4,-2) }}}}$

Step-by-step explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

• In question the information given that in a triangle ABC coordinate of the mid point of each side is given.

• So,we know mid point formula for finding mid point

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: given \\ \to Co-ordinate \:of \: D = (1,0) \\ \to Co-ordinate \:of \: E = (2,1) \\ \to Co-ordinate \: of \: F = ( - 1,3) \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: to \: find \\ \to Co-ordinate \: of \: A= ( x_{1},y_{1}) \\ \to Co-ordinate \: of \: B= ( x_{2}, y_{2}) \\ \to Co-ordinate \: of \: C= ( x_{3},y_{3})$

• D is the mid-point between side AC

• E is the mid-point between side AB

• F is the mid-point between side BC

• According to given question:

$\to For \: side \: AC\\ \to 1 = \frac{ x_{2} + x_{1}}{2} \\ \to x_{2} + x_{1} = 2 - - - - - (1) \\ \\\to For\:side\:AB\\ \to 2 = \frac{x_{3} + x_{1}}{2} \\ \to x_{3} + x_{1} = 4 - - - - - (2) \\ \\ \to For\:side\:BC\\ \to - 1 = \frac{x_{2} + x_{3}}{2} \\ \to x_{2} + x_{3} = - 2 - - - - - (3) \\ \\ subtracting \: eqn \: (3) \: from \: (1) \\ \to x_{2} + x_{1} - x_{2} - x_{3} = 2 - ( - 2) \\ \to x_{1} - x_{3} = 4 - - - - - (4) \\ \\ adding \: (2) \: and (4) \\ \to x_{3} + x_{1} + x_{1} - x_{3} = 4 + 4 \\ \to 2x_{1} = 8 \\ \to x_{1} = 4 \\ \\ putting \: value \: of \: x_{1} in \: (1) \\ \to x_{2} + x_{1} = 2 \\ \to x_{2} + 4 = 2 \\ \to x_{2} = 2 - 4 \\ \to x_{2} = - 2 \\ \\ putting \: value \: of \: x_{2} \: in \: (3) \\ \to x_{2} + x_{3} = - 2 \\ \to - 2 + x_{3} = - 2 \\ \to x_{3} = - 2 + 2 \\ \to x_{3} = 0$

• Similarly

• Making three more eqn where y1,y2 and y3 is unknown from them we find these three ordinates value.

$\to y_{1} = - 2 \\ \to y_{2} = 2 \\ \to y_{3} = 4 \\ \\ \therefore Co-ordinate \: of \: Vertices \: A = ( - 2,2) \\ \therefore Co-ordinate \: of \: Vertices \: B = ( 0,4) \\ \therefore Co-ordinate \: of \: Vertices \: C = ( 4, - 2)$