Question

किसी वर्ग के दो सम्मुख शीर्ष (-1,2) और (3,2) हैं। वर्ग के अन्य दोनों शीर्ष ज्ञात कीजिए।
NCERT​

Answer 1

Answer:

(-2,1)and (2,3)

Step-by-step explanation:

eanter change the digites

Answer 2

The other two vertices of square are (1,0) and (1,4) or (1,4) and (1,0)

Step-by-step explanation:

Let other two opposite vertices of the square are B (x,y) and  D(x',y').

Let two given vertices of square are A(-1,2) and C (3,2).

We know that all sides of square are equal.

Mid-point formula:

$x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

Using mid-point formula

Mid- point of AC

$x=\frac{-1+3}{2}=1$

$y=\frac{2+2}{2}=2$

Mid-point of AC=(1,2)

Distance formula:$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between A(-1,2) and C (3,2)=$\sqrt{(3+1)^2+(2-2)^2}=4 units$

Distance between A (-1,2) and B (x,y)=AB=$\sqrt{(x+1)^2+(y-2)^2}$

BC=$\sqrt{(3-x)^2+(2-y)^2}$

AB=BC

$AB^2=BC^2$

$(x+1)^2+(y-2)^2=(3-x)^2+(2-y)^2$

$x^2+2x+1+y^2-4y+4=9+x^2-6x+4-4y+y^2$

Using the formula

$(a+b)^2=a^2+b^2+2ab$

$2x-4y+5=-6x-4y+13$

$2x+6x-4y+4y=13-5$

$8x=8$

$x=\frac{8}{8}=1$

$AB^2+BC^2=AC^2$

$(x+1)^2+(y-2)^2+(3-x)^2+(2-y)^2=16$

Substitute the value of x

$(1+1)^2+(y-2)^2+(3-1)^2+(y-2)^2=16$

$4+2(y^2-4y+4)+4=16$

$2y^2-8y+8+8=16$

$2y^2-8y=16-16$

$y^2-4y=0$

$y(y-4)=0$

$y=0 or y-4=0\implies y=4$

The coordinates of B are (1,0) or (1,4)

Using mid-point formula and using B (1,0)

$1=\frac{1+x'}{2},2=\frac{0+y'}{2}$

$1+x'=2,4=y'$

$x'=2-1=1, y'=4$

Using mid-point formula and using B (1,4)

$1=\frac{1+x'}{2},2=\frac{4+y'}{2}$

$1+x'=2,4=4+y'$

$x'=2-1=1, y'=4-4=0$

The other two vertices of square are (1,0) and (1,4) or (1,4) and (1,0)

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