MATH 132: Introduction to Mathematical Thinking Assignment 4

This document provides detailed solutions to Assignment 4 from MATH 132, Introduction to Mathematical Thinking, at Victoria University of Wellington. The assignment focuses on fundamental concepts in linear equations and graphing. The solutions cover finding the equation of a line given two points, determining equations based on gradient and a point, and converting equations into intercept form. Furthermore, the document includes solutions for sketching linear graphs, identifying x and y-intercepts, and understanding the relationship between the equation and the graphical representation. The solutions also cover tutorial exercises, reinforcing the concepts of finding the equation of a line and sketching graphs based on different given conditions.

Introduction to Mathematical Thinking Solutions
Q1.
a) For the points (2,4) and (5,6) ,
slope= dy
dx = 6−4
5−2 = 2
3
→ 2
3 = y−4
x−2
∴ y= 2
3 x + 8
3
b) For the points (4,3) and (-2,5) ,
slope= dy
dx = 5−3
−2−4 =−1
3
→− 1
3 = y−3
x−4
∴ y=−1
3 x + 13
3
Q2.
a) For the point (2,1) with gradient 3:
slope=3= y −1
x−2
∴ y=3 x−5
¿ the form of x
a + y
b =1, the equation becomes x
(5
3 )+ y
3 =1
, where x−intercept , a= 5
3 ∧ y −intercept , b=3
b) For the point (-2,1) with gradient 1
2 :
slope= 1
2 = y −1
x +2
∴ y= 1
2 x +2
¿ the form of x
a + y
b =1, the equation becomes x
−4 + y
2 =1
, where x−intercept , a=−4∧ y−intercept ,b=2

c) For the points (1,1) with y-intercept (0,3) ,
slope= dy
dx = 3−1
0−1 =−2
→−2= y −3
x−0
∴ y=−2 x+ 3
¿ the form of x
a + y
b =1, the equation becomes x
(3
2 )+ y
−2 =1
, where x−intercept , a= 3
2 ∧ y −intercept , b=−2
d) For the points (-1,3) with y-intercept (2,-1) ,
slope= dy
dx = 3+1
−1−2 =−4
3
→− 4
3 = y−3
x +1
∴ y=−4
3 x +5
3
¿ the form of x
a + y
b =1, the equation becomes x
( 5
4 )+ y
( 5
3 )=1
, where x−intercept , a= 5
4 ∧ y−intercept ,b= 5
3
Q3.
a) ¿ the form of x
a + y
b =1, the e quation is x
−1 + y
2 =1
, where x−intercept=a=−1∧ y−intercept =b=2
(as read¿the graph)
b) For the points (-2,1) with y-intercept (0,4) ,
slope= dy
dx = 4−1
0−−2 = 3
2
→ 3
2 = y−4
x−0
∴ y= 3
2 x +4

¿ the form of x
a + y
b =1, the equation becomes x
(−8
3 ) + y
4 =1
c) For y =2
3 x+2 ,¿ the form of x
a + y
b =1, the equation becomes x
(−3 ) + y
2 =1
d) For 2 x+ 3 y=6 ,
For x=0 , y −intercept=2 while for y=0 , x−intercept =3
∴∈the form of x
a + y
b =1 , the equationbecomes x
3 + y
2 =1
e) For y+1
x−2 = 1
2 ,
For x=0 , y −intercept=−2while for y=0 , x−intercept =4
∴∈the form of x
a + y
b =1 , the equationbecomes x
4 + y
−2 =1
Q4.
a) For x
2 + y
3 =1 ,the sketch of the graph is as shown below :
-6 -4 -2 0 2 4 6
-6
-4
-2
0
2
4
6
8
10
12
𝑥/2+ /3=1𝑦
y
Linear (y)
x−intercept ≅ 2∧ y−intercept ≅ 3
(as read¿the graph)

b) For y =2
3 x+1 , the sketch of the graph is as shownbelow :
-6 -4 -2 0 2 4 6
-3
-2
-1
0
1
2
3
4
5
𝑦=2/3 +1𝑥
y
Linear (y)
x−intercept ≅ −1.5∧ y−intercept ≅ 1
(as read¿the graph)
c) For 2 x+ 3 y=6 , the sketch of the graph is as shownbelow :

-6 -4 -2 0 2 4 6
-2
-1
0
1
2
3
4
5
6
2 +3 =6𝑥 𝑦
y
Linear (y)
x−intercept ≅ 3∧ y−intercept ≅ 2
(as read¿the graph)
d) For y+1
x−2 = 3+1
4−2 ,the sketch of the graph is as shown below :
-8 -6 -4 -2 0 2 4 6 8
-20
-15
-10
-5
0
5
10
( +1)/( −2)=(3+1)/(4−2)𝑦 𝑥
y
Linear (y)
x−intercept ≅ 2. 5∧ y−intercept ≅−5
(as read¿the graph)

Solutions to Tutorial Exercises
Ex1.
a) For the points (1,2) and (-3,5) ,
slope= dy
dx = 5−2
−3−1 =−3
4
→− 3
4 = y−2
x−1
∴ y=−3
4 x + 11
4
b) For the points (1,4) and (3,1) ,
slope= dy
dx = 4−1
1−3 =−3
2
→− 3
2 = y−3
x−4
∴ y=−3
2 x +9
Ex2.
a) For the point (2,1) with gradient −3
2 :

slope=−3
2 = y−1
x−2
∴ y=−3
2 x + 4
¿ the form of x
a + y
b =1, the equation becomes x
( 8
3 ) + y
4 =1
, where x−inter cept , a= 8
3 ∧ y−intercept , b=4
b) For the point (2,1) with x-intercept (-1,0) ,
slope= dy
dx =1−0
2+1 = 1
3
→ 1
3 = y−0
x +1
∴ y= 1
3 x + 1
3
¿ the form of x
a + y
b =1, the equation becomes x
−1 + y
( 1
3 )=1
, where x−intercept , a=−1∧ y−intercept , b= 1
3
c) For the points (-1,1) with y-intercept (3,-2) ,
slope= dy
dx = 1+2
−1−3 =−3
4
→− 3
4 = y−1
x +1
∴ y=−3
4 x + 1
4
¿ the form of x
a + y
b =1, the equation becomes x
(1
3 )+ y
( 1
4 )=1
, where x−intercept , a= 1
3 ∧ y −intercept , b= 1
4
Ex3.
a) ¿ the form of x
a + y
b =1, the equation is x
−2 + y
4 =1
, where x−intercept=a=−2∧ y−intercept =b=4

b) For the points (3,1) with y-intercept (6,3) ,
slope= dy
dx = 3−1
6−3 = 2
3
→ 2
3 = y−1
x −3
∴ y= 2
3 x−1
¿ the form of x
a + y
b =1, the equation becomes x
(3
2 )+ y
−1 =1
c) For 6 x + y=2 , For x=0 , y −intercept=2 while for y=0 , x−intercept =1
3
∴∈the form of x
a + y
b =1 , the equationbecomes x
( 1
3 ) + y
2 =1
d) For−5 x+ 2 y =2 ,
For x=0 , y −intercept=1 while for y=0 , x−intercept =−2
5
∴∈the form of x
a + y
b =1 , the equationbecomes x
(−2
5 )+ y
1 =1
e) For y−2
x−6 = 2
3 ,
For x=0 , y −intercept=−2while for y=0 , x−intercept =3
∴∈the form of x
a + y
b =1 , the equationbecomes x
3 + y
−2 =1
Ex4.
a) For x
−3 + y
4 =1 , the sketch of the graph is as shownbelow :

-6 -4 -2 0 2 4 6
-4
-2
0
2
4
6
8
10
12
𝑥/(−3)+ /4=1𝑦
y
Linear (y)
x−intercept ≅ −3∧ y−intercept ≅ 4
(as read¿the graph)
b) For y =−1
4 x+2 , the sketch of the graph is as shownbelow :

-15 -10 -5 0 5 10 15
-1
0
1
2
3
4
5
𝑦=−1/4 +2𝑥
y
Linear (y)
x−intercept ≅ 8∧ y−intercept ≅ 2
(as read¿the graph)
c) For−x +3 y=3 , the sketch of the graph is as shown below :
-6 -4 -2 0 2 4 6
-1
-0.5
0
0.5
1
1.5
2
2.5
3
−𝑥+3 =3𝑦
y
Linear (y)
x−intercept ≅−3∧ y−intercept ≅ 1
(as read¿the graph)
d) For y−2
x+3 = 1−2
−4 +3 ,the sketchof the graphis as shown below :

-8 -6 -4 -2 0 2 4 6 8
-4
-2
0
2
4
6
8
10
12
14
( −2)/( +3)=(1−2)/(−4+3)𝑦 𝑥
y
Linear (y)
x−intercept ≅−5∧ y−intercept ≅ 5
(as read¿the graph)