Question

Show that the line given by (2+k) x +(3-k) y= 7 + 6k for different value of k,pass through a fixed point. Find the co- ordinates of the fixed point

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Answer 1

Given line,

⇒ (2 + k) x + ( 3 - k) y = 7 + 6k

⇒ 2x + kx + 3y - ky - 7 - 6k = 0

⇒ 1 ( 2x + 3y - 7) + k ( x - y - 6) = 0

This is of the form,

$L_1 + \lambda \: L_2 = 0$

So the line represents two lines which intersect at a point. Therefore, For any value of k, The line passes through a fixed point.

Now, The fixed point is the point of intersection of the lines,

$L_1 : 2x + 3y - 7 = 0 \\ L_2 : x - y - 6 = 0$

Solving the lines,

2x + 3y - 7 = 0

2x - 2y - 12 = 0

Adding them gives,

y - 19 = 0

y = 19

Finding the value of x,

x - 19 - 6 = 0

x - 25 = 0

x = 25.

Therefore, The fixed point is (25, 19)

Answer 2

Hey, please refer to the given attachment!