Question

The fourth vertex of the rectangle whose three vertices taken in order are (4, 1) (7, 4) (13, -2) is with answer 10 class
Chapter no 7

Answer 1

We know that a Rectangle has four vertices. According to the question, Let the unknown order be (x, y)

• A = (4,1)
• B = (7,4)
• C = (13,-2)
• D = (x, y)

This Problem can be solved using Midpoint Formula,

$\boxed{Mid\:Point\:=\:({\dfrac{x_1\:+x_2}{2}}),({\dfrac{y_1\:+\:y_2}{2}})}$

Midpoint of AC :

$\implies\sf{Mid\:Point\:=\:({\dfrac{x_1\:+\:x_2}{2}}),({\dfrac{y_1\:+\:y_2}{2}})}$

Points,

• $\sf{x_1\:=\:4,\:y_1\:=\:1,\:x_2\:=\:13\:and\:y_2\:=\:-2}$

$\implies\sf{Mid\:Point\:=\:({\dfrac{4\:+\:13}{2}}),({\dfrac{1\:+\:(-2)}{2}})}$

$\implies\sf{Mid\:Point\:=\:({\dfrac{4\:+\:13}{2}}),({\dfrac{1\:-\:2}{2}})}$

$\implies\sf{Mid\:Point\:=\:({\dfrac{17}{2}}),({\dfrac{-1}{2}})}----\:(1)$

Midpoint of BD,

Points,

• $\sf{x_1\:=\:x,\:y_1\:=\:y,\:x_2\:=\:7\:and\:y_2\:=\:4}$

$\implies\sf{Mid\:Point\:=\:({\dfrac{x_1\:+\:x_2}{2}}),({\dfrac{y_1\:+\:y_2}{2}})}$

$\implies\sf{Mid\:Point\:=\:({\dfrac{x\:+\:7}{2}}),({\dfrac{y\:+\:4}{2}})}$

Equalling it with (1),

$\implies\sf{Mid\:Point\:=\:({\dfrac{x\:+\:7}{2}})\:=\:({\dfrac{17}{2}})}$

$\implies\sf{x\:+\:7\:=\:17}$

$\implies\sf{x\:=\:17\:-\:7}$

$\implies\sf{x\:=\:10}$

Then,

$\implies\sf{Mid\:Point\:=\:({\dfrac{y\:+\:4}{2}}),({\dfrac{-1}{2}})}$

$\implies\sf{y\:+\:4\:=\:-1}$

$\implies\sf{y\:=\:-1\:-\:4}$

$\implies\sf{y\:=\:-5}$

Hence, the fourth vertex of the rectangle = (10, -5).

Note :

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