The value of x is 1.
The value of y is 4.
Solution:
Given TQRS is a rhombus.
Property of rhombus:
Diagonals bisect each other.
In diagonal TR
⇒ 3x + 2 = y + 1
⇒ 3x – y = –1 – – – – (1)
In diagonal QS
⇒ x + 3 = y
⇒ x – y = –3 – – – – (2)
Solve (1) and (2) by subtracting
⇒ 3x – y – (x – y) = –1 – (–3)
⇒ 3x – y – x + y = –1 + 3
⇒ 2x = 2
⇒ x = 1
Substitute x = 1 in equation (2), we get
⇒ 1 – y = –3
⇒ –y = –3 – 1
⇒ –y = –4
⇒ y = 4
The value of x is 1.
The value of y is 4.