(i) Let the x - axis represent the weight and the y-axis represent the length

(ii) Consider the point \(\begin{pmatrix} { x }_{ 1 } & { y }_{ 1 } \\ 2 & 3 \end{pmatrix}\begin{pmatrix} { x }_{ 2 } & { y }_{ 2 } \\ 4 & 4 \end{pmatrix}\)
Equation of the straight line using two point form is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\Rightarrow \quad \frac { y-3 }{ 4-3 } =\frac { x-2 }{ 4-2 } \)
\(\Rightarrow \quad \frac { y-3 }{ 1 } =\frac { x-2 }{ 2 } \)
\(\Rightarrow\) 2y - 6 = x-2
 \(\Rightarrow\) x - 2y + 4 = 0
(iii) To find the actual length of the spring,
Put x = 0
\(\Rightarrow\) 0 - 2y + 4 = 0
\(\Rightarrow\) -2y = -4
\(\Rightarrow\) y = 2 cm
(iv) Put y = 9 in (1) we get
\(\Rightarrow\) x - 18 + 4 = 0
\(\Rightarrow\) x - 14 = 0
\(\Rightarrow\) x = 14 kg
\(\therefore\) 14 Kg must be added
(v) Put x = 6 in (1) we get,
\(\Rightarrow\)  6 - 2y + 4 = 0
\(\Rightarrow\) 10 - 2y = 0
\(\Rightarrow\) 10 = 2y 
\(\Rightarrow\) y = 5
\(\therefore\) Strength of the spring = 5 cm