Question

A(-1;1) B(5;3) are opposite vertices of a square in xy plane. Find the equation of the other diagonal of the square ​

Answer 1

Answer:  The required equation of the other diagonal is $3x+y=8.$

Step-by-step explanation:  As shown in the attached figure below, A(-1, 1) and B(5, 3) are opposite vertices of the square ACBD. AB and CD are the two diagonals of the square intersecting at the point O.

We are to find the equation of diagonal CD.

We know that the diagonals of a square bisect each other perpendicularly.

So, AB is perpendicular to CD and O is the midpoint of both AB and CD.

The co-ordinates of point O are given by

$\left(\dfrac{-1+5}{2},\dfrac{1+3}{2}\right)=\left(\dfrac{4}{2}+\dfrac{4}{2}\right)=(2,2).$

And, the slope of the diagonal AB is given by

$m=\dfrac{3-1}{5-(-1)}=\dfrac{2}{6}=\dfrac{1}{3}.$

We know that the product of the slopes of two perpendicular lines is -1.

So, the slope of diagonal CD (m') is calculated as follows :

$m'\times m=-1\\\\\\\Rightarrow m'\times\left(\dfrac{1}{3}\right)=-1\\\\\Rightarrow m'=-3.$

Therefore, the equation of diagonal CD with slope m' and passing through the point O(2, 2) is given by

$y-2=m'(x-2)\\\\\Rightarrow y-2=-3(x-2)\\\\\Rightarrow y-2=-3x+6\\\\\Rightarrow 3x+y=8.$

Thus, the required equation of the other diagonal is $3x+y=8.$

Answer 2

Answer:

Equation of Second Diagonal is 3x + y - 8 = 0

Step-by-step explanation:

Given Coordinates of opposite vertices of square, A( -1 , 1 ) and B( 5 , 3 )

We need to find the Equation of other diagonal.

We know that Diagonal of Square are perpendicular bisector.

Means other diagonal will pass through mid point of given diagonal and product of slopes of both diagonal will be -1.

Mid point of given diagonal = $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

                                              = $(\frac{-1+5}{2},\frac{1+3}{2})$

                                              = $(2,2)$

Slope of the given diagonal = $\frac{y_2-y_1}{x_2-x_1}=\frac{1-3}{-1-5}=\frac{-2}{-6}=\frac{1}{3}$

let m be the slope of second diagonal.

We know that Product of slopes of perpendicular line = -1

Product of slopes of diagonal = -1

m × 1/3 = -1

m = -3

Using Slope-Point form,

we get equation of second diagonal.

$y-y_1=m(x-x_1)$

y - 2 = -3 ( x - 2 )

y - 2 = -3x + 6

3x + y - 8 = 0

Therefore, Equation of Second Diagonal is 3x + y - 8 = 0