
Putting x = 4 in eq. (i)
y = -2/(4 - 3) = -2
Then, point = (4, – 2)
Again putting, x = 2 in eq. (i)
y = -2/(2 - 3) = 2
Then, point = (2, 2)
Now, tangent at point (4, – 2),
y – (- 2) = 2 (x – 4)
⇒ y + 2 = 2x – 8
⇒ 2x – y – 10 = 0
Again, tangent at point (2, 2),
y – 2 = 2 (x – 2)
⇒ y – 2 = 2x – 4
⇒ 2x – y – 2 = 0
Hence, required equations are 2x – y – 10 = 0 and 2x – y – 2 = 0.