Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: Class 9

Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: Class 9

Question 1.

Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: Class 9

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.

Solution:

p(x) = ax + b ...(i)

q(x) = cx + d ...(ii)

Substituting (2, 3) in (i), we get

p(2) = a(2) + b = 3

∴ 2a + b = 3

∴ b = 3 - 2a ...(iii)

Substituting (6, 11) in (i), we get

p(6) = a(6) + b = 11

∴6 a + b = 11 —(iv)

Substituting b = 3 - 2a in (iv), we get

6a + (3 - 2a) = 11

∴ 6a + 3 - 2a = 11

∴ 4a = 11 - 3

∴ 4a = 8

∴ a = 2

Substituting a = 2 in (iii), we get

b = 3 - 2(2)

∴ b = 3 - 4

∴ b = -1

∴ p(x) = 2x - 1

Substituting (4, -1) in (ii), we get

q(4) = c(4) + d = -1

∴ 4c + d = -1 ...(v)

Graphs of p(x) and q(x) are parallel

∴ Slope of p(x) = Slope of q(x)

∴ a = c

∴ c = 2 ...[∵ a = 2]

Substituting c = 2 in (v), we get

4(2) + d = -1

∴ 8 + d = -1

∴ d = - 9

∴ q(x) = cx + d = 2x + (-9)

∴ q(x) = 2x - 9

The line p(x) meets the x-axis, when p(x) = 0

∴ 2x - 1 = 0

∴ 2x = 1

∴ x = $\frac{1}{2}$

∴ The line p(x) meets the x-axis at ($\frac{1}{2}$, 0)

The line q(x) meets the x-axis, when q (x) = 0

∴ 2x - 9 = 0

∴ 2x = 9

∴ x = $\frac{9}{2}$

∴ The line q(x) meets the x-axis at ($\frac{9}{2}$, 0)