Question

Find the value of k for which the line (k-3)x-(4-k^2)y+k^2-7k+6=0 passes through the origin

Answer 1

Answer:

The given equation of line is:

(k-3)x - (4 - k^2)y + k^2 - 7k + 6 = 0

It is given that,

The given line passes through the origin.

ie, the line passes through the point (0,0).

Since,The line passes through the (0,0) thus , the coordinates of the point (0,0)

will Satisfy the equation of the given line.

now, let's put the values abscissa (x-coordinate) and ordinate (y-coordinate) of the point (0,0) in the given equation of line.

ie, put x=0 and y=0 in the given equation of line.

thus, we have,

=> (k-3)•0 - (4 - k^2)•0 + k^2 -7k + 6 = 0

=> k^2 - 7k + 6 = 0

=> k^2 - k - 6k + 6 = 0

=> k(k-1) - 6(k-1) = 0

=> (k-1)(k-6) = 0

=> k = 1 or 6

Thus , the required values of k are :

1 and 6.

Moreover, we will get two different equations corresponding to two different values of k.

Case:(1)

when k =1,

then the equation will reduce to;

=> (k-3)x - (4 - k^2)y + k^2 - 7k + 6 = 0

=> (1-3)x - (4 - 1^2)y + 1^2 - 7•1 + 6 = 0

=> -2x - 3x + 1 - 7 + 6 = 0

=> -2x - 3y = 0

=> 2x + 3y = 0

Case:(2)

when k = 6,

=> (k-3)x - (4 - k^2)y + k^2 - 7k + 6=0

=> (6-3)x - (4 - 6^2)y + 6^2 - 7•6 + 6 = 0

=> 3x - (4 - 36)y + 36 - 42 + 6 = 0

=> 3x + 32y = 0

Answer 2

Answer:

k=1 or 6

Step-by-step explanation:

Given equation;

(k-3)x-(4-k²)y+k²-7k+6=0

we know, that if a line passes through the origin, it will satisfy the condition x=0 and y=0 i.e., (0,0)

so, putting x=0 and y=0 in the given equation,

we get,

(k − 3)x − (4 − k²)y + k² − 7k + 6 = 0

(k − 3)0 − (4 − k²)0 + k² − 7k + 6 = 0

k² − 7k + 6 = 0

k² − 6k − k + 6 = 0

k(k − 6) − 1(k − 6) = 0

k(k − 6) − 1(k − 6) = 0

(k − 1)(k − 6) = 0

So, k = 1 or k = 6