Question

The points A (1,5), B (5,5) and C (5, 1) are 3 corners of a square ABCD
What are the coordinates of D, the 4th corner?​

Answer 1

Answer:

(1, 1)

Step-by-step explanation:

Let D be (x, y). As ABCD is a square, diagonals bisect each other(intersect at their mid points). It means:

=> mid point of AC = mid point of BD

Using, mid point formula*:

Mid point of AC = (1+5/2 , 5+1/2) = (3, 3)

Mid point of BD = (5+x/2 , 5+y/2)

As both must be equal:

=> (3, 3) = (5+x/2 , 5+y/2)

=> 3 = (5 + x)/2 and 3 = (5 + y)/2

=> 6 = 5 + x and 6 = 5 + y

=> 1 = x and 1 = y

Coordinates of D are (1, 1).

Mid point formula: *if there are two points (a, b) and (x, y), then their mid point is ( (a+x)/2 , (b+y)/2 ).

Answer 2

Answer:

Given :-

• The points A(1 , 5) , B(5 , 5) and C(5 , 1) are 3 corners of a square ABCD.

To Find :-

• What are the co-ordinates of D i.e, 4th corners.

Formula Used :-

$\clubsuit$ Mid-point Formula :

$\mapsto \sf\boxed{\bold{\pink{\bigg\lgroup \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}\bigg\rgroup}}}$

where,

• x₁ , x₂ = Co-ordinators of the X-axis
• y₁ , y₂ = Co-ordinates of the Y-axis

Solution :-

Let,

$\mapsto$ Co-ordinates of D i.e, 4th corners be (x , y)

$\mapsto$ ABCD is a square, and it is diagonally bisected as each other, then :

$\leadsto \sf\bold{Mid-point\: of\: AC =\: Mid-point\: of\: BD}$

${\small{\bold{\purple{\underline{\bigstar\: In\: case\: of\: mid-point\: of\: AC\: :-}}}}}$

Given points :

$\bullet$ A = (1 , 5)

$\bullet$ C = (5 , 1)

where,

• x₁ = 1
• x₂ = 5
• y₁ = 5
• y₂ = 1

According to the question by using the formula we get,

$\implies \sf Mid-point\: of\: AC =\: \bigg\lgroup \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}\bigg\rgroup$

$\implies \sf Mid-point\: of\: AC =\: \bigg\lgroup \dfrac{1 + 5}{2} , \dfrac{5 + 1}{2}\bigg\rgroup$

$\implies \sf Mid-point\: of\: AC =\: \bigg\lgroup \dfrac{\cancel{6}}{\cancel{2}} , \dfrac{\cancel{6}}{\cancel{2}}\bigg\rgroup$

$\implies \sf Mid-point\: of\: AC =\: \bigg\lgroup \dfrac{3}{1} , \dfrac{3}{1}\bigg\rgroup$

$\implies \sf\bold{\green{Mid-point\: of\: AC =\: (3 , 3)}}$

Again,

${\small{\bold{\purple{\underline{\bigstar\: In\: case\: of\: mid-point\: of\: BD\: :-}}}}}$

Given points :

$\bullet$ B = (5 , 5)

$\bullet$ D = (x , y)

where,

• x₁ = 5
• x₂ = x
• y₁ = 5
• y₂ = y

According to the question by using the formula we get,

$\implies \sf Mid-point\: of\: BD =\: \bigg\lgroup \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}\bigg\rgroup$

$\implies \sf Mid-point\: of\: BD =\: \bigg\lgroup \dfrac{5 + x}{2} , \dfrac{5 + y}{2}\bigg\rgroup$

$\implies \sf\bold{\green{Mid-point\: of\: BD =\: \bigg\lgroup \dfrac{5 + x}{2} , \dfrac{5 + y}{2}\bigg\rgroup}}$

Now,

$\leadsto \sf\bold{\orange{Mid-point\: of\: AC =\: Mid-point\: of\: BD}}$

We have :

$\bullet\: \: \sf Mid-point\: of\: AC =\: (3 , 3)$

$\bullet\: \: \sf Mid-point\: of\: BD =\: \bigg\lgroup \dfrac{5 + x}{2} , \dfrac{5 + y}{2}\bigg\rgroup$

According to the question,

$\longrightarrow \sf (3 , 3) =\: \bigg\lgroup \dfrac{5 + x}{2} , \dfrac{5 + y}{2}\bigg\rgroup$

At first ,

$\longrightarrow \sf 3 =\: \dfrac{5 + x}{2}$

By doing cross multiplication we get,

$\longrightarrow \sf 5 + x =\: 2(3)$

$\longrightarrow \sf 5 + x =\: 2 \times 3$

$\longrightarrow \sf 5 + x =\: 6$

$\longrightarrow \sf x =\: 6 - 5$

$\longrightarrow \sf\bold{\red{x =\: 1}}$

Again,

$\longrightarrow \sf 3 =\: \dfrac{5 + y}{2}$

By doing cross multiplication we get,

$\longrightarrow \sf 5 + y =\: 2(3)$

$\longrightarrow \sf 5 + y =\: 2 \times 3$

$\longrightarrow \sf 5 + y =\: 6$

$\longrightarrow \sf y =\: 6 - 5$

$\longrightarrow \sf\bold{\red{y =\: 1}}$

$\therefore$ The co-ordinates of D i.e, 4th corner is (1 , 1).