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l1 : x-3ay = 2 and l2 : 2x +

Berikut ini adalah pertanyaan dari idgsminay14 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

L1 : x-3ay = 2 and l2 : 2x + y = 6 paralle but not overlap, to calculate a

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

a = –1/6
 

Problem

l_1 : x-3ay = 2andl_2 : 2x + y = 6 are parallel but not overlap, to calculate a.

Solution

Parallel lines are those ones sharing same gradient/slope value. So, if l_1\parallel l_2andl_1 doesn’t overlap l_2 or vice versa, l_1has the same gradient value asl_2, but they must have different constant value.

We can solve this problem by converting each of both lines into y=mx+c. But, let's do it the other way, by using the original form of both l_1 and l_2, and comparing them side-by-side.

\begin{aligned}&\begin{array}{rl|l}&x-3ay=2&2x+y=6\\&...\Downarrow(l_1\times2)\\\Rightarrow&\!\!2x-6ay=4&2x+y=6\end{array}\\\end{aligned}

Both constants (on the RHS of both equations) are different, so l_1andl_2 won’t overlap.

Solving for ausing the coefficient ofyonl_2, we get:

\begin{aligned}-6a&=1\quad\therefore a&=\boxed{\:\bf{-\frac{1}{6}}\:}\end{aligned}

\blacksquare

a = –1/6 Problem[tex]l_1 : x-3ay = 2[/tex] and [tex]l_2 : 2x + y = 6[/tex] are parallel but not overlap, to calculate [tex]a[/tex].SolutionParallel lines are those ones sharing same gradient/slope value. So, if [tex]l_1\parallel l_2[/tex] and [tex]l_1[/tex] doesn’t overlap [tex]l_2[/tex] or vice versa, [tex]l_1[/tex] has the same gradient value as [tex]l_2[/tex], but they must have different constant value.We can solve this problem by converting each of both lines into [tex]y=mx+c[/tex]. But, let's do it the other way, by using the original form of both l_1 and l_2, and comparing them side-by-side.[tex]\begin{aligned}&\begin{array}{rl|l}&x-3ay=2&2x+y=6\\&...\Downarrow(l_1\times2)\\\Rightarrow&\!\!2x-6ay=4&2x+y=6\end{array}\\\end{aligned}[/tex]Both constants (on the RHS of both equations) are different, so [tex]l_1[/tex] and [tex]l_2[/tex] won’t overlap.Solving for [tex]a[/tex] using the coefficient of [tex]y[/tex] on [tex]l_2[/tex], we get:[tex]\begin{aligned}-6a&=1\quad\therefore a&=\boxed{\:\bf{-\frac{1}{6}}\:}\end{aligned}[/tex][tex]\blacksquare[/tex]

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Last Update: Thu, 29 Sep 22
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