Question

19. If A(3, 6) and C(-1, 2) are two vertices of a rhombus ABCD, then find the equation
of straight line that lies along the diagonal BD.

Answer 1

HELLO DEAR,



WE KNOW THAT:-

→In any rhombus two diagonals bisect each other
→ they are perpendicular to each other.



Midpoint of AC = Midpoint of BD

= (x₁ + x₂)/2 , (y₁ + y₂)/2

= (3 - 1)/2 , (6 + 2)/2

= (2/2),(8/2)

= (1 , 4)



$\large{\mathbf{NOW}}$


Slope of AC = (y₂ - y₁)/(x₂ - x₁)

= (2-6)/(-1-3)

= -4/-4

= 1



Slope of BD = -1/1

= -1



Equation for BD


(y - y₁)/(x - x₁) = m


(y - y₁) = m(x - x₁)

(y - 4) = -1 (x - 1)

y - 4 = - x + 1

x + y - 4 - 1 = 0

x + y - 5 = 0

x + y = 5



I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Solution:-

given by:- A(3, 6) and C(-1, 2) are two vertices of a rhombus ABCD, then find the equation
of straight line that lies along the diagonal BD.

we have:-

In any rhombus two diagonals bisect each otherand they are perpendicular to each other.

》Midpoint of AC = Midpoint of BD

》midpoint = (x₁ + x₂)/2 , (y₁ + y₂)/2
 
》       = (3 - 1)/2 , (6 + 2)/2

》            = (2/2),(8/2)

》            = (1 , 4)

》Slope of AC = (y₂ - y₁)/(x₂ - x₁)

》                = (2-6)/(-1-3)

》                = -4/(-4)

》                = 1

》Slope of BD = -1/1

》                = -1

Equation of BD 

》(y - y₁) = m(x - x₁)

》(y - 4) = -1 (x - 1)

》y - 4 = - x + 1

》x + y - 4 - 1 = 0

》x + y - 5 = 0


the equation of straight line (x+y-5)

☆i hope its help☆