Find the value(s) of p in (i) to (iv) and p and q in (v) for the following pair of equations:
(i) \(3x – y – 5 = 0\) and \(6x – 2y – p = 0\),
if the lines represented by these equations are parallel.
(ii) \(–x + py = 1\) and \(px – y = 1\),
if the pair of equations has no solution.
(iii) \(– 3x + 5y = 7\) and \(2px – 3y = 1\),
if the lines represented by these equations are intersecting at a unique point.
(iv) \(2x + 3y – 5 = 0\) and \(px – 6y – 8 = 0\), if the pair of equations has a unique solution.
(v) \(2x + 3y = 7\) and \(2px + py = 28 – qy\), if the pair of equations have infinitely many solutions.
(i) Any \(p\neq 10\).
(ii) \(p = 1\).
(iii) All \(p \ne \dfrac{9}{10}\).
(iv) All \(p \ne -4\).
(v) \(p = 4\) and \(q = 8\).
Slopes are equal when
\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\Rightarrow \dfrac{3}{6} = \dfrac{-1}{-2} = \dfrac{1}{2}.\)
This holds for all \(p\).
To avoid coincidence, keep constants ratio different:
\(\dfrac{c_1}{c_2} = \dfrac{-5}{-p} = \dfrac{5}{p} \neq \dfrac{1}{2}\Rightarrow p \neq 10\).
No solution when
\(\dfrac{-1}{p} = \dfrac{p}{-1} \ne \dfrac{-1}{-1}\).
The first equality gives \(\dfrac{-1}{p} = -p\Rightarrow p^2 = 1\Rightarrow p=\pm 1\).
For \(p=1\): coefficient ratios are \(-1\) and constants ratio is \(1\) ⇒ parallel.
For \(p=-1\): all three ratios are \(1\) ⇒ coincident.
Hence, \(p=1\) gives no solution.
Unique point when slopes are different.
Check equality of ratios:
\(\dfrac{-3}{2p} \ne \dfrac{5}{-3} = -\dfrac{5}{3}.\)
Equality would give
\(\dfrac{-3}{2p} = -\dfrac{5}{3}\Rightarrow 3 = \dfrac{10p}{3}\Rightarrow p = \dfrac{9}{10}.\)
So for \(p \ne \dfrac{9}{10}\) the intersection is unique.
Require \(\dfrac{2}{p} \ne \dfrac{3}{-6} = -\dfrac{1}{2}.\)
Equality occurs when \(\dfrac{2}{p} = -\dfrac{1}{2}\Rightarrow p = -4\).
Hence unique solution for all \(p \ne -4\).
Write second in standard form:
\(2px + (p+q)y - 28 = 0\).
Match ratios with first line \(2x + 3y - 7 = 0\).
\(\dfrac{2}{2p} = \dfrac{3}{p+q} = \dfrac{-7}{-28} = \dfrac{1}{4}.\)
From the first equality: \(\dfrac{1}{p} = \dfrac{1}{4}\Rightarrow p = 4\).
Then \(\dfrac{3}{p+q} = \dfrac{1}{4}\Rightarrow p + q = 12\Rightarrow q = 8\).