# coordinate geometry

the line (2-k)x+(1+2k)y-(4+3k)=0 passes through a fixed point P for any value of k. Find the coordinates of P.

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- ☂雨後晴空☀Lv 71 decade agoFavorite Answer
(2-k)x+(1+2k)y-(4+3k)=0

2x + y - 4 - kx + 2ky - 3k = 0

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

Solve :

L1 : 2x + y - 4 = 0....(1)

L2 : x - 2y + 3 = 0....(2)

(1) - (2)*2 :

5y - 10 = 0

y = 2 and hence

x = 1

The coordinates of P is (1 , 2)

Method 2 :

Since the line passes through a fixed point P for any value of k.

When k = 2 , the line becomes (2-2)x + (1+2*2)y - (4+3*2) = 0 ,

5y - 10 = 0 , y = 2 , y-coordinate of P must be 2 ,

when k = -1/2 , the line becomes (2 + 1/2)x + (1 - 2*1/2)y - (4 - 3*1/2) = 0 ,

(5/2)x - 5/2 = 0

x = 1 , the x - coordinate of P must be 1 ,

So the coordinates of P is (1 , 2)

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