NCERT Solutions Class 9 Maths Exercise 2.6 — Introduction to Linear Polynomials

NCERT Class 9 Mathematics — Ganita Manjari Part I, Introduction to Linear Polynomials, Exercise Set 2.6. All five parts solved.

Every equation in this exercise has the form $y = ax + b$, and the exercise exists to make two facts obvious by drawing rather than by proof:

• $a$ is the slope. It says how far $y$ moves for each step of $1$ in $x$. A larger $\lvert a \rvert$ makes a steeper line; a negative $a$ makes the line fall as you read left to right.
• $b$ is the $y$-intercept. It is the value of $y$ when $x = 0$, so the line crosses the $y$-axis at $(0, b)$ — and changing $b$ slides the line up or down without tilting it.

To draw any of these lines you need only two points, and the easiest two are usually $x = 0$ and $x = 1$. A third point is worth plotting as a check, since three points that fail to line up mean an arithmetic slip.

Key insight. $a$ and $b$ do independent jobs. Change $a$ and the line pivots about its $y$-intercept; change $b$ and the line slides without changing direction. That is why lines with the same $a$ and different $b$ can never meet — parallel lines — while lines with the same $b$ and different $a$ always meet, at $(0, b)$.

Question 1

Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘$a$’ and ‘$b$’.

Solution. Each part is a set of lines to be drawn on one pair of axes, so that they can be compared. The tables below give three points for each line.

(i) $y = 4x$,   $y = 2x$,   $y = x$

Here $b = 0$ in all three, so all three lines pass through the origin. Only $a$ differs, and it is $a$ that decides the steepness: $y = 4x$ climbs four units for every one to the right, $y = 2x$ climbs two, $y = x$ climbs one. Each is steeper than the next, and all three rise from left to right because $a > 0$.

(ii) $y = -6x$,   $y = -3x$,   $y = -x$

Again $b = 0$, so all three pass through the origin — but now every $a$ is negative, and all three lines fall as you read left to right. The pattern of steepness is the same as in (i), governed by the size of $a$ regardless of its sign: $y = -6x$ is the steepest.

(iii) $y = 5x$,   $y = -5x$

Two lines of equal steepness through the origin, one rising and one falling. They are mirror images of each other in the $y$-axis — and also in the $x$-axis, since replacing $a$ by $-a$ turns every $y$ into $-y$. They cross at the origin, the only point they share.

(iv) $y = 3x – 1$,   $y = 3x$,   $y = 3x + 1$

Now $a$ is fixed at $3$ and only $b$ changes. Every column of the table differs by exactly $1$ from the row above, so the three lines are the same line slid up and down: they are parallel, and they never meet. They cut the $y$-axis at $(0,-1)$, $(0,0)$ and $(0,1)$ respectively — precisely the values of $b$.

(v) $y = -2x – 3$,   $y = -2x$,   $y = 2x + 3$

Read the third equation carefully — it is $y = 2x + 3$, with $a = +2$. So this set is not the mirror of part (iv). The first two lines have $a = -2$ and are parallel to each other, falling, cutting the $y$-axis at $-3$ and $0$. The third has $a = +2$ and rises, cutting the $y$-axis at $3$.

Because the third line is not parallel to the others, it crosses them. Setting $2x + 3 = -2x$ gives $4x = -3$, so it meets $y = -2x$ at $\left( -\tfrac{3}{4}, \tfrac{3}{2} \right)$; setting $2x + 3 = -2x – 3$ gives $4x = -6$, so it meets $y = -2x – 3$ at $\left( -\tfrac{3}{2}, 0 \right)$.

So this part illustrates both lessons at once: two lines with equal $a$ are parallel, and a line with a different $a$ must cut them.

A note on the pattern. Part (iv) gives a clean triple $3x – 1$, $3x$, $3x + 1$, and by analogy part (v) reads as though it were meant to be $-2x – 3$, $-2x$, $-2x + 3$ — three parallel falling lines cutting the $y$-axis at $-3$, $0$ and $3$. As printed the third equation is $y = 2x + 3$, which is a perfectly good question and is what is answered above. If your copy of the book is read the other way, the only change is that all three lines become parallel and none of them meet.

Common mistakes

• Plotting only two points and not checking. Two points always lie on some line, so a slip in one of them goes unnoticed. A third point that misses the line is the only cheap way to catch an arithmetic error.
• Part (ii), plotting $y = -6x$ as rising. At $x = 1$ the value is $-6$, below the axis. A negative slope means the line goes down to the right; drawing a table with a negative $x$ in it makes the direction unmistakable.
• Part (ii), thinking $-6$ is “less steep” than $-3$ because it is a smaller number. Steepness is the size of $a$ regardless of sign, so $y = -6x$ is the steepest of the three.
• Part (iv), drawing the three lines meeting somewhere off the page. Lines with the same slope never meet, however far they are extended. If your three lines converge, the slopes have been drawn unequal.
• Part (iv), reading $b$ as a horizontal shift. $b$ moves the line vertically. The line $y = 3x + 1$ is $y = 3x$ lifted by one unit, not slid one unit to the right. (The two happen to look similar for a sloping line, which is exactly why the $y$-intercept is the reliable way to place it.)
• Part (v), assuming the third line is $-2x + 3$ because the pattern suggests it. Copy the equation as printed and check the sign of the coefficient of $x$ before plotting; here the printed line has a positive slope and therefore crosses the other two.

Practise next

• The End-of-Chapter Exercises, Question 7 — four lines to draw, with their slopes and $y$-intercepts to be identified, and the question “are any of the lines parallel?” which is part (iv) of this exercise turned into a test.
• Exercise Set 2.5 — the reverse skill. There the numbers $a$ and $b$ had to be calculated from two readings; here they are given and their effect on the picture is what matters.