NCERT Solutions for Class 9th Maths Chapter 2 Exercise Set 2.6 — Visualising linear relationships
Book page 36 Updated on2026-09-08
Q1.
Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’. (i) y = 4x, y = 2x, y = x
Answer
Each has b = 0, so plot the origin and one more point.
Line
Points used
a (slope)
b (y-intercept)
y = 4x
(0, 0), (1, 4)
4
0
y = 2x
(0, 0), (2, 4)
2
0
y = x
(0, 0), (4, 4)
1
0
y = 4x, y = 2x and y = x: same y-intercept 0, increasing slopes, increasing steepness.
Role of a and b here:b = 0 for all three, so all three are pinned to the origin — that is what a zero y-intercept means. The only thing left to vary is a, and it controls the tilt: a run of 1 unit lifts y = 4x by 4, y = 2x by 2 and y = x by 1. Since all slopes are positive, all three represent linear growth, and the three lines fan out from the origin without ever being parallel.
Check it yourself: at x = 3 the heights are 12, 6 and 3. The line with the biggest slope is always the highest one to the right of the origin — and the lowest one to the left.
Q1.
Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’. (ii) y = – 6x, y = – 3x, y = – x
Answer
Line
Points used
a (slope)
b
y = −6x
(0, 0), (1, −6)
−6
0
y = −3x
(0, 0), (1, −3)
−3
0
y = −x
(0, 0), (3, −3)
−1
0
All three fall from left to right through the origin; the more negative the slope, the steeper the fall.
Role of a and b here: again b = 0, so all three pass through the origin. Every slope is negative, so all three are pictures of linear decay: raising x by 1 lowers y by 6, 3 and 1 respectively. Steepness is governed by how far a is from 0, not by its sign — y = −6x is the steepest of the three and is exactly as steep as y = 6x would be, only leaning the other way.
Tip: comparing this set with (i): y = −x is the mirror image of y = x in the x-axis, and likewise for the other pairs.
Q1.
Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’. (iii) y = 5x, y = –5x
Answer
Line
Points used
a
b
y = 5x
(0, 0), (1, 5)
5
0
y = −5x
(0, 0), (1, −5)
−5
0
Equal and opposite slopes: the two lines cross at the origin and make equal angles with the x-axis.
Role of a and b here: the two slopes have the same size, 5, and opposite signs, while b = 0 for both. So the lines are equally steep and cross at their common point, the origin. One shows growth, the other decay. Replacing x by −x in y = 5x gives y = −5x, which says the pair is symmetric about the y-axis; replacing y by −y gives the same pair, so they are symmetric about the x-axis too.
Check it yourself: at x = 2 the lines are at y = 10 and y = −10 — equally far from the x-axis, on opposite sides.
Q1.
Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’. (iv) y = 3x – 1, y = 3x, y = 3x + 1
Answer
Line
Points used
a
b
y = 3x − 1
(0, −1), (1, 2)
3
−1
y = 3x
(0, 0), (1, 3)
3
0
y = 3x + 1
(0, 1), (1, 4)
3
1
Same slope 3, intercepts −1, 0 and 1: three parallel lines, each 1 unit above the last.
Role of a and b here: this is the opposite experiment to (i). Now a is held fixed and b is varied, so the tilt cannot change — the lines are parallel. At any chosen x, the vertical gap between y = 3x + 1 and y = 3x is (3x + 1) − 3x = 1, the same everywhere, so the three lines never converge. Changing b slides the whole line vertically; changing a would rotate it.
Tip: two lines are parallel exactly when their slopes are equal and their intercepts are not. Equal slopes and equal intercepts would make them the same line.
Q1.
Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’. (v) y = –2x – 3, y = –2x, y = 2x + 3
Answer
Line
Points used
a
b
y = −2x − 3
(0, −3), (−2, 1)
−2
−3
y = −2x
(0, 0), (2, −4)
−2
0
y = 2x + 3
(0, 3), (1, 5)
2
3
The first two are parallel (slope −2); the third has slope +2, so it crosses both.
Role of a and b here: the first two share the slope −2 and differ only in b, so they are parallel, with a constant vertical gap of 3. The third line has slope +2, a different number, so it is not parallel to them — it cuts each of them exactly once. Solving 2x + 3 = −2x gives 4x = −3, x = −¾, so it meets y = −2x at (−¾, 1½); solving 2x + 3 = −2x − 3 gives x = −½, the point (−½, 2).
Did you know? Sets (i) to (iv) each vary one thing at a time. Set (v) deliberately mixes them, so that you have to decide parallelism by comparing slopes rather than by the look of the printed equations. Had the third line been y = −2x + 3, all three would have been parallel — compare the shape of the answer in that case.