Question

3) A square is drawn on graph paper. One side of the square is the segment with endpoints (2, 5) and (8, 1). Find the other two vertices of the square. There are two possible solutions. Can you find both?​

Answer 1

Answer:

(6,11) / (12,7)

Step-by-step explanation:

the distance between the two points equals :

$\sqrt{6 ^{2} + (- 4) ^{2} } = 2 \sqrt{13}$

then the length of the square equals 2 root 13 cm

and the the 4 sides of the square are equal so the difference between each one of the given points and each one of the other 2 vertics equals 2 root 13

so put (x,y) the first vertics and the other is (z,n) and find them with the difference between two points

Answer 2

Given:

A square with two endpoints (2,5) and (8,1).

To find:

The other two vertices of the square in two possible solutions.

Solution:

We have two vertices of a square (2,5) and (8,1). On plotting these points on a graph sheet, we get a line AB which is one edge of the square. A square has all sides equal and its sides are perpendicular to each other. So, the two possible solutions of the vertices of the square are ABCD and ABC'D' as shown in the figure.

On determining the slope of line AB,

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$x_{1}=2,y_{1}=5, x_{2}=8,y_{2}=1$

$m=\frac{1-5}{8-2}=\frac{-4}{6}=\frac{-2}{3}$

When two lines are perpendicular, the product of their slopes is -1. Since a square has all sides making 90° with each other, the slope of one side is the negative reciprocal of the other side, i.e., D'AD will have a slope which is the negative reciprocal of the slope of AB. Let the slope of D'AD be $m_{1}$

$mm_{1}=-1$

$\frac{-2}{3}m_{1}=-1$

∴ $m_{1}=\frac{3}{2}=\frac{6}{4}$

From point (2,5) move 4 units right and 6 units up, then 6 gets added to y-coordinate 5 and 4 gets added to x-coordinate 2 to give an ordered pair D(6,11).

From point (8,1) move 4 units right and 6 units up, then 6 gets added to y-coordinate 1 and 4 gets added to x-coordinate 8 to give an ordered pair C(12,7).

So, we obtained the first set of possible coordinates, i.e., C(12,7) and D(6,11).

To obtain the second set of solutions, from point (2,5) move 4 units left and 6 units down, then 6 gets subtracted from y-coordinate 5 and 4 gets subtracted from x-coordinate 2 to give an ordered pair D'(-2,-1).

From point (8,1) move 4 units left and 6 units down, then 6 gets subtracted from y-coordinate 1 and 4 gets subtracted from x-coordinate 8 to give an ordered pair C'(4,-5).

So, we obtained the second set of possible coordinates, i.e., C'(4,-5) and D'(-2,-1).

The figure below shows the two possible solutions of the other two vertices: C and D; C' and D'.

The possible solutions of two vertices of a square whose other vertices are given are: C(12,7) and D(6,11); C'(4,-5) and D'(-2,-1).