Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) The graph of p(x) passes through the points (2, 3) and (6, 11).
(ii) The graph of q(x) passes through the point (4, –1).
(iii) The graph of q(x) is parallel to the graph of p(x).

Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.

Given:

p(x) = ax + b and q(x) = cx + d

Using condition (i): p(x) passes through (2, 3) and (6, 11)

When x = 2, p(x) = 3:

3 = 2a + b …(I)

When x = 6, p(x) = 11:

11 = 6a + b …(II)

Subtracting (I) from (II):

(6a + b) - (2a + b) = 11 - 3

⇒ 4a = 8

⇒ a = 2

Substituting a = 2 in (I):

3 = 2(2) + b

⇒ 3 = 4 + b

⇒ b = -1

∴ p(x) = 2x - 1.

Using condition (iii): q(x) is parallel to p(x)

Parallel lines have the same slope. So, c = a = 2.

Using condition (ii): q(x) passes through (4, -1)

When x = 4, q(x) = -1:

-1 = c(4) + d

⇒ -1 = 2(4) + d

⇒ -1 = 8 + d

⇒ d = -9

∴ q(x) = 2x - 9.

Finding the coordinates of the points where the lines meet the x-axis:

For p(x) = 2x - 1, when y = 0:

2x - 1 = 0

⇒ 2x = 1

⇒ x = $\dfrac{1}{2}$

So, p(x) meets the x-axis at $\left(\dfrac{1}{2}, 0\right)$.

For q(x) = 2x - 9, when y = 0:

2x - 9 = 0

⇒ 2x = 9

⇒ x = $\dfrac{9}{2}$

So, q(x) meets the x-axis at $\left(\dfrac{9}{2}, 0\right)$.

Hence, p(x) = 2x - 1 and q(x) = 2x - 9. The graph of p(x) meets the x-axis at $\left(\dfrac{1}{2}, 0\right)$ and the graph of q(x) meets the x-axis at $\left(\dfrac{9}{2}, 0\right)$.