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Two lines $L_{1}: 2 y-3 \times 6=0$ and $L 2: 3 y+x-20 = 0$ intersect at a point $A$.

(a) Find the coordinates of A.
(b) A third line $L_{3}$ is perpendicular to $L_{2}$ at point $A$. Find the equation of $L_{3}$ in the form $y=m x+c$, Where $m$ and $c$ are constants.
(c) Another line $\mathrm{L}_{4}$ is parallel to $\mathrm{L}_{1}$ and passes through $(-1,3)$. Find the $x$ and $y$ intercepts of $L_{4}$
in Mathematics by Platinum (130,522 points) | 185 views

1 Answer

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(a) Find the coordinates of A.
$$
\begin{array}{l}
2 y-3 x=6 \\
3 y+x=20 \\
\hline 2 y-3 x=6 \\
\dfrac{9 y+3 x=60}{11 y=60} \\
y=6 \\
x=20-18 \\
=2
\end{array}
$$
Coordinates of $\mathrm{A}$ are $(2,6)$

(b) A third line $L_{3}$ is perpendicular to $L_{2}$ at point
A. Find the equation of $L_{3}$ in the form $y=m x+$
c, Where $m$ and $c$ are constants. (3 marks)
(b) $\mathrm{L}_{2}: 3 y=-x+20$
$$
y=-\frac{1}{3} x+20
$$
Gradient of perpendicular $=3$
$$
\begin{array}{l}
\dfrac{y-6}{x-2}=3 \\
y=3 x-6+6 \\
y=3 x
\end{array}
$$

 

\begin{array}{l}
\text { (c) Gradient of } L 4=\text { gradient of } L \text { , }\\
=\dfrac{3}{2}\\
\dfrac{y-3}{x+1}=\frac{3}{2}\\
2 y-6=3 s+3\\
2 y-3 x=9\\
\text { When } x=0 \quad y-4.5\\
\text { When } \mathrm{y}-0 \mathrm{x}=-3
\end{array}
by Platinum (130,522 points)

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