Intersection of Lines and their Equations - Questions
Find the point of intersection of the lines: \(3x - 2y = - 7\) \(2x + 3y = 4\)
For the system of equations \(2px - 3y = 3p\) and \(3x + y = 4\) to have no solutions, then \({\rm{ }}p{\rm{ = ?}}\)
\begin{align}
&\begin{aligned}
2px-3y &=3p \\
3y &=2px-3 p \\
y &=\frac{2p}{3}x-p
\end{aligned}\\
&\begin{aligned}
3x+y &=4 \\
y &=-3x+4
\end{aligned}\\
&\text{no solutions if } m_{1}=m_{2},\; c_{1} \neq c_{2}\\ &\frac{2 p}{3}=-3 \rightarrow p=-4.5 \quad, 4 \neq 4.5\\
&(\text{Lines are parallel})
\end{align}
For the system of equations: \(ax - 6y = b\) and \(2x - by = 4b\) to have an infinite number of solutions then the values of a and b are?
\begin{align}
&\begin{aligned}
ax-6y&=b & 2x-by&=4b \\
6y&=ax-b & by&=2x-4b \\
y&=\frac{a}{6} x-\frac{b}{6} & y&=\frac{2}{b}x-4
\end{aligned}\\
&\text{For infinite solutions: }m_{1}=m_{2},\; c_{1}=c_{2}\\
&\begin{aligned}
\therefore \quad \frac{a}{6} &=\frac{2}{b} & -\frac{b}{6}=-4 \rightarrow b&=24 \\
a &=\frac{12}{b} & a&=\frac{12}{24}=\frac{1}{2}
\end{aligned}
\end{align}
Given that \(x + 4y = 12,\,\,\,\,\,2x - 3y = 2,\,\,\,\,3x + ay = 26\) are concurrent then the value of \(a\) is?
\begin{align}
&\begin{aligned}
x+4 y &=12 \cdots (1)\\
2x-3y &=2 \;\;\cdots(2) \\
(1) \times 2 \quad 2 x+8 y &=24 \cdots(3) \\
(2) - (3) \;\;\; -11 y &=-22 \rightarrow y=2 \\
\text {In }(1)\qquad\; x+8 &=12 \rightarrow x=4
\end{aligned}\\
&\therefore \text{Substitute }x=4,\; y=2 \text{ into }\\
&\quad\; \begin{aligned}
3x+ay&=26 \\
12+2a&=26 \rightarrow a=7
\end{aligned}
\end{align}
At time \(t\) a particle \({p_1}\), moving on a straight line has co - ordinates \(x = t - 1,\,y = 3t + 2\). At the same time a second particle \({p_2}\), moving along a straight line has co - ordinates \(x = 2t + 1\,,\,\,y = 1 - t\) the co-ordinates of the point of intersection are?
\begin{align}
&\begin{aligned}
\text { For } P_{1}&: t=x+1 \\
y&=3(x+1)+2 \\
y&=3x+5 \cdots (1)
\end{aligned}\\
&\begin{aligned}
\text { For } P_{2}&: t=1-y \\
x &=2(1-y)+1 \\
x &=2-2y+1 \\
x &=3-2y \cdots (2)
\end{aligned}\\
&\text { sub }(2) \text{ into } (1)\\
&\qquad \begin{aligned}
y &=3(3-2y)+5 \\
y &=9-6y+5\\
7y&=14 \rightarrow y=2 \\
\end{aligned}\\
&\text {In }(2)\quad x=3-4 \rightarrow x=-1
\end{align}