JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 13)
If the system of linear equations
x – 2y + kz = 1
2x + y + z = 2
3x – y – kz = 3
has a solution (x,y,z), z $$ \ne $$ 0, then (x,y) lies on the straight line whose equation is :
x – 2y + kz = 1
2x + y + z = 2
3x – y – kz = 3
has a solution (x,y,z), z $$ \ne $$ 0, then (x,y) lies on the straight line whose equation is :
4x – 3y – 4 = 0
3x – 4y – 1 = 0
4x – 3y – 1 = 0
3x – 4y – 4 = 0
Explanation
x – 2y + kz = 1 ......(1)
2x + y + z = 2 .........(2)
3x – y – kz = 3 ........(3)
for locus of (x, y) add equation (1) + (3)
4x – 3y = 4
$$ \Rightarrow $$ 4x – 3y - 4 = 0
2x + y + z = 2 .........(2)
3x – y – kz = 3 ........(3)
for locus of (x, y) add equation (1) + (3)
4x – 3y = 4
$$ \Rightarrow $$ 4x – 3y - 4 = 0
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