Find the equation of the line which passes through the point of inters

1.	Find the equation of the line which passes through the point of intersection of lines x + y – 3 = 0, 2x – y + 1 = 0 and which is parallel to X-axis.
Answer» Let u ≡ x + y – 3 = 0 and v ≡ 2x – y + 1 = 0 Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0. (x + y – 3) + k(2x – y + 1) = 0 …..(i) x + y – 3 + 2kx – ky + k = 0 x + 2kx + y – ky – 3 + k = 0 (1 + 2k)x + (1 – k)y – 3 + k = 0 But, this line is parallel to X-axis Its slope = 0 ⇒ \(\frac {-(1+2k)}{1-k} =0\) ⇒ 1 + 2k = 0 ⇒ k = -1/2 Substituting the value of k in (i), we get (x + y – 3) + -1/2 (2x – y + 1) = 0 ⇒ 2(x + y – 3) – (2x – y + 1 ) = 0 ⇒ 2x + 2y – 6 – 2x + y – 1 = 0 ⇒ 3y – 7 = 0, which is the equation of the required line.

Find the equation of the line which passes through the point of intersection of lines x + y – 3 = 0, 2x – y + 1 = 0 and which is parallel to X-axis.

Answer»

Let u ≡ x + y – 3 = 0 and v ≡ 2x – y + 1 = 0

Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.

(x + y – 3) + k(2x – y + 1) = 0 …..(i)

x + y – 3 + 2kx – ky + k = 0

x + 2kx + y – ky – 3 + k = 0

(1 + 2k)x + (1 – k)y – 3 + k = 0

But, this line is parallel to X-axis Its slope = 0

⇒ \(\frac {-(1+2k)}{1-k} =0\)

⇒ 1 + 2k = 0

⇒ k = -1/2

Substituting the value of k in (i), we get

(x + y – 3) + -1/2 (2x – y + 1) = 0

⇒ 2(x + y – 3) – (2x – y + 1 ) = 0

⇒ 2x + 2y – 6 – 2x + y – 1 = 0

⇒ 3y – 7 = 0, which is the equation of the required line.