Find the value of K for which equation x – Ky = 2, 3x + 2y = 5 has unique solution.
Concept:
Consider the following pair of linear equations:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
For a unique solution:
a₁/a₂ ≠ b₁/b₂
For infinite solutions:
a₁/a₂ = b₁/b₂ = c₁/c₂
For no solution:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Calculation:
The given equations are:
x − Ky = 2
3x + 2y = 5
Here,
a₁ = 1, b₁ = −K, a₂ = 3, b₂ = 2
For a unique solution,
1/3 ≠ −K/2
⇒ 2 ≠ −3K
⇒ K ≠ −2/3
∴ The required value of K should not be equal to −2/3.
Important Questions from Co-ordinate Geometry
The graphs of the linear equations 4x - 2y = 10 and 4x + ky = 2 intersect at a point (a, 4). The value of k is equal to:
In which ratio the point (-3, p) divides the line segment joining the points (-5, -4) and (-2, 3)?
The area (in sq. units) of the triangle formed by the graphs of 8x + 3y = 24, 2x + 8 = y and the x-axis is:
In which quadrant both abscissa and ordinate are negative?
Find the slope of the line joining the points (3, -4) and (5, 2).
