Question

if two opposite vertices of a square are 5,4 and 1, -6 find the coordinates of its remaining two vertices​

Answer 1

Answer:

sorry

Step-by-step explanation:

but I don't know this answer

Answer 2

Given Question :-

• If the two opposite vertices of a square are (5,4) and (1, -6), find the coordinates of its remaining two vertices.

$\large\underline{\bold{Solution-}}$

Given :-

• Two opposite vertices of a square are (5,4) and (1, -6).

To find :-

• The coordinates of its remaining two vertices.

Formula Used :-

Distance Formula :-

Let us assume a line segment joining the points A and B, then distance between A and B is given by

$\rm\:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$\tt \: where \: coordinates \: are \: A(x_1,y_1) \: and \: B(x_2,y_2).$

Calculation :-

• Let ABCD be a square and let A(5, 4) and C(1, - 6) be the given angular points.

• Let the coordinates of vertex B be (x, y).

Since,

• ABCD is a square.

• ⇛ AB = BC

On squaring both sides, we get

• ⇛AB² = BC²

So, by using distance formula, we get

$\rm {(x - 5)}^{2} + {(y - 4)}^{2} = {(x - 1)}^{2} + {(y + 6)}^{2}$

$\rm \: {(x - 5)}^{2} - {(x - 1)}^{2} = {(y + 6)}^{2} - {(y - 4)}^{2}$

$\rm \: (x - 5 + x - 1)(x - 5 - x + 1) = (y + 6 + y - 4)(y + 6 - y + 4)$

$\rm \: (2x - 6)( - 4) = (2y + 2)(10)$

$\rm \: - 8x + 24 = 20y + 20$

$\rm \: 8x + 20y = 4$

$\rm \: 2x + 5y = 1$

$\therefore \: x \: = \: \dfrac{1 - 5y}{2} - - - (1)$

Again,

• In right triangle ABC,

Using Pythagoras Theorem, we have

• AB² + BC² = AC²

Therefore,

By using Distance Formula, we get

$\rm {(x - 5)}^{2} + {(y - 4)}^{2}+ {(x - 1)}^{2} + {(y + 6)}^{2} = {(5 - 1)}^{2} + {(4 + 6)}^{2}$

$\rm \: {2x}^{2} + {2y}^{2} - 12x + 4y + 78 = 116$

$\rm \: {x}^{2} + {y}^{2} - 6x + 2y - 19 = 0$

• On substituting the value of x evaluated in equation (1), we get

$\rm \: \bigg(\dfrac{ 1- 5y}{2} \bigg) ^{2} + {y}^{2} - 6\bigg(\dfrac{1 - 5y}{2} \bigg) + 2y - 19 = 0$

$\rm \: \dfrac{ {25y}^{2} + 1 - 10y }{4} + {y}^{2} + 15y - 3 + 2y - 19 = 0$

$\rm \: \dfrac{ {25y}^{2} + 1 - 10y }{4} + {y}^{2} + 17y - 22 = 0$

$\rm \: \dfrac{ {25y}^{2} + 1 - 10y + {4y^{2} + 68y - 88} }{4} = 0$

$\rm \: {29y}^{2} + 58y - 87 = 0$

$\rm \: 29( {y}^{2} + 2y - 3) = 0$

$\rm \: {y}^{2} + 3y - y - 3 = 0$

$\rm \: y(y + 3) - 1(y + 3) = 0$

$\rm \: (y + 3)(y - 1) = 0$

$\bf :\implies\:y \: = \: - \: 3 \: \: \: or \: \: \: \: y \: = \: 1$

So,

On substituting the values of y in equation (1), we get

$\begin{gathered}\boxed{\begin{array}{c|c} \bf y & \bf x = \dfrac{1 - 5y}{2} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 3 & \sf 8 \\ \\ \sf 1 & \sf - 2 \end{array}} \\ \end{gathered}$

So,

• The opposite remaining vertices of the square be

$\begin{gathered}\begin{gathered}\bf \:Opposite \: vertices \: are \: \begin{cases} &\sf{(8, \: - \: 3)} \\ &\sf{( - \: 2, \: 1)} \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

1. Section Formula :-

Let us assume a line segment joining the points A and B and let C (x, y) be any point which divides the line segment joining the points A and B internally in the ratio m : n, then coordinates of C is given by

${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$

2. Midpoint Formula :-

Let us assume a line segment joining the points A and B and let C (x, y) be Midpoint of line segment joining the points A and B, then coordinates of midpoint C is given by

${\underline{\boxed{ \rm(x, \: y) = \rm{ \bigg(\quad \dfrac{x_1 + x_2}{2} \; ,\; \dfrac{y_1 + y_2}{2} \quad \bigg)}}}}$