13. Let and be two linear polynomials such that: (i) The graph of passes through the points and . (ii) The graph of passes through the point . (iii) The graph of is parallel to the graph of . Find the polynomials and . Also, find the coordinates of the point where these lines meet the -axis.

Understand the Question A linear polynomial represents a straight line with slope . If the graph passes through given points, substituting their coordinates allows us to solve for unknown coefficients. Parallel lines have identical slopes (). The coordinates where a line meets the -axis correspond to points where (the zeros of the polynomials). Step 1 · Find the Polynomial Let the equation of the line representing be: Since it passes through and : Subtracting equation from equation : Substituting into equation : Therefore, the polynomial is: Step 2 · Find the Polynomial Let the equation of the line representing be: Since the graph of is parallel to the graph of , their slopes are equal: So, . Since the graph of passes through : Therefore, the polynomial is: Step 3 · Find the Coordinates Where the Lines Meet the -axis A line intersects the -axis when . For : Thus, meets the -axis at . For : Thus, meets the -axis at . Answer > The polynomials are and . The lines meet the -axis at and respectively. Common Mistakes Parallel Lines Condition: Forgetting that parallel linear polynomials share the exact same coefficient of (slope). Intercept Confusion: Setting (which gives the -intercept) instead of setting (to find the point where the line meets the -axis). Coordinate Notation: Writing only the value of instead of writing the full point coordinates .

Introduction to Linear Polynomials | EOT

Question 13

*13. 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.

Solution

Understand the Question

A linear polynomial $y = mx + c$ represents a straight line with slope $m$ .
If the graph passes through given points, substituting their coordinates allows us to solve for unknown coefficients.
Parallel lines have identical slopes ( $c = a$ ).
The coordinates where a line meets the $x$ -axis correspond to points where $y = 0$ (the zeros of the polynomials).

Step 1 · Find the Polynomial $p(x)$

Let the equation of the line representing $p(x)$ be: $y = ax + b$ Since it passes through $(2, 3)$ and $(6, 11)$ : $3 = 2a + b \quad \dots (1)$ $11 = 6a + b \quad \dots (2)$

Subtracting equation $(1)$ from equation $(2)$ :

\begin{aligned} (6a + b) - (2a + b) &= 11 - 3 \\ 4a &= 8 \\ a &= 2 \end{aligned}

Substituting $a = 2$ into equation $(1)$ :

\begin{aligned} 2(2) + b &= 3 \\ 4 + b &= 3 \\ b &= -1 \end{aligned}

Therefore, the polynomial is: $p(x) = 2x - 1$

Step 2 · Find the Polynomial $q(x)$

Let the equation of the line representing $q(x)$ be: $y = cx + d$

Since the graph of $q(x)$ is parallel to the graph of $p(x)$ , their slopes are equal: $c = a = 2$

So, $q(x) = 2x + d$ .

Since the graph of $q(x)$ passes through $(4, -1)$ :

\begin{aligned} -1 &= 2(4) + d \\ -1 &= 8 + d \\ d &= -9 \end{aligned}

Therefore, the polynomial is: $q(x) = 2x - 9$

Step 3 · Find the Coordinates Where the Lines Meet the $x$ -axis

A line intersects the $x$ -axis when $y = 0$ .

For $p(x)$ :

\begin{aligned} 2x - 1 &= 0 \\ 2x &= 1 \\ x &= \dfrac{1}{2} \end{aligned}

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

For $q(x)$ :

\begin{aligned} 2x - 9 &= 0 \\ 2x &= 9 \\ x &= \dfrac{9}{2} \end{aligned}

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

Answer

The polynomials are $p(x) = 2x - 1$ and $q(x) = 2x - 9$ .

The lines meet the $x$ -axis at $\left(\dfrac{1}{2}, 0\right)$ and $\left(\dfrac{9}{2}, 0\right)$ respectively.

Common Mistakes

Parallel Lines Condition: Forgetting that parallel linear polynomials share the exact same coefficient of $x$ (slope).
Intercept Confusion: Setting $x = 0$ (which gives the $y$ -intercept) instead of setting $y = 0$ (to find the point where the line meets the $x$ -axis).
Coordinate Notation: Writing only the value of $x$ instead of writing the full point coordinates $(x, 0)$ .

Question 13

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