Question:
The point of intersection of diagonals of a square ABCD is at the origin and one of its vertices is at A(4, 2) . What is the equation of the diagonal BD?
The point of intersection of diagonals of a square ABCD is at the origin and one of its vertices is at A(4, 2) . What is the equation of the diagonal BD?
Options:
- 2x + y = 0
- x + 2y = 0
- 2x – y = 0
- x – 2y = 0
- 2x + y = 0
- x + 2y = 0
- 2x – y = 0
- x – 2y = 0
Correct Answer: a)
Explanation:
‘ The equation of diagonal AC is
Y:
(4,2)
x X
D
0-2
‘ y—2=q5(x—4)
‘ y—2=3(z—4)
⇒2y—4=2-4
‘ x=2yorx-2y=0
‘ As we know that diagonals are perpendicular to each other.
.”. Equation of line perpendicular to AC is 2z + y + ¢ = 0, which passes through (0, 0).
‘ 200) + (0) +c=0=c=0
‘ S2x+y=0
‘ The equation of diagonal AC is
Y:
(4,2)
x X
D
0-2
‘ y—2=q5(x—4)
‘ y—2=3(z—4)
⇒2y—4=2-4
‘ x=2yorx-2y=0
‘ As we know that diagonals are perpendicular to each other.
.”. Equation of line perpendicular to AC is 2z + y + ¢ = 0, which passes through (0, 0).
‘ 200) + (0) +c=0=c=0
‘ S2x+y=0