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The coordinates of three points are A(1, 1), B(-1, 4)

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

The coordinates of three points are A(1, 1), B(-1, 4) and C(h, k). Given that the gradients of AB, AC and BC are -3a, 3a and a respectively, find the values of h, k and a.​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

h = 5, k = 7, a = ½

 

Explanation/Penjelasan

The coordinates of three points are A(1, 1), B(–1, 4) and C(h, k).

Given that the gradients of AB, AC and BC are –3a, 3a and a respectively, we can get

\begin{aligned}&&m_{\rm AB}&=\frac{y_{\rm B}-y_{\rm A}}{x_{\rm B}-x_{\rm A}}\\\\&\Leftrightarrow&-3a&=\frac{4-1}{-1-1}\\&&&=\frac{3}{-2}\ =\ -3\cdot\frac{1}{2}\\\\&\therefore&a&=\bf\frac{1}{2}\\\\&\therefore&m_{\rm AC}&=3a={\frac{3}{2}}=\frac{y_{\rm C}-y_{\rm A}}{x_{\rm C}-x_{\rm A}}\\&\Leftrightarrow&\frac{3}{2}&=\frac{k-1}{h-1}\\&\Leftrightarrow&\!\!3h-3&=2k-2\quad...(i)\end{aligned}

\begin{aligned}&&m_{\rm BC}&=a={\frac{1}{2}}=\frac{y_{\rm C}-y_{\rm B}}{x_{\rm C}-x_{\rm B}}\\&\Leftrightarrow&\frac{1}{2}&=\frac{k-4}{h-(-1)}\ =\ \frac{k-4}{h+1}\\&\Leftrightarrow&h+1&=2k-8\\&\Leftrightarrow&h&=2k-9\quad...(ii)\end{aligned}

\begin{aligned}&&(ii)\to(i):\\&&\!\!\!\!3(2k-9)-3&=2k-2\\&\Leftrightarrow&6k-27-3&=2k-2\\&\Leftrightarrow&6k-30&=2k-2\\&\Leftrightarrow&6k-2k&=-2+30\\&\Leftrightarrow&4k&=28\\&&\therefore\ k&=\bf7\\\\&&(ii)\implies h&=2(7)-9\\&&&=14-9\\&&\therefore\ h&=\bf5\end{aligned}

 

CONCLUSION/KESIMPULAN

∴  h = 5, k = 7, a = ½

 

h = 5, k = 7, a = ½ Explanation/PenjelasanThe coordinates of three points are A(1, 1), B(–1, 4) and C(h, k).Given that the gradients of AB, AC and BC are –3a, 3a and a respectively, we can get[tex]\begin{aligned}&&m_{\rm AB}&=\frac{y_{\rm B}-y_{\rm A}}{x_{\rm B}-x_{\rm A}}\\\\&\Leftrightarrow&-3a&=\frac{4-1}{-1-1}\\&&&=\frac{3}{-2}\ =\ -3\cdot\frac{1}{2}\\\\&\therefore&a&=\bf\frac{1}{2}\\\\&\therefore&m_{\rm AC}&=3a={\frac{3}{2}}=\frac{y_{\rm C}-y_{\rm A}}{x_{\rm C}-x_{\rm A}}\\&\Leftrightarrow&\frac{3}{2}&=\frac{k-1}{h-1}\\&\Leftrightarrow&\!\!3h-3&=2k-2\quad...(i)\end{aligned}[/tex][tex]\begin{aligned}&&m_{\rm BC}&=a={\frac{1}{2}}=\frac{y_{\rm C}-y_{\rm B}}{x_{\rm C}-x_{\rm B}}\\&\Leftrightarrow&\frac{1}{2}&=\frac{k-4}{h-(-1)}\ =\ \frac{k-4}{h+1}\\&\Leftrightarrow&h+1&=2k-8\\&\Leftrightarrow&h&=2k-9\quad...(ii)\end{aligned}[/tex][tex]\begin{aligned}&&(ii)\to(i):\\&&\!\!\!\!3(2k-9)-3&=2k-2\\&\Leftrightarrow&6k-27-3&=2k-2\\&\Leftrightarrow&6k-30&=2k-2\\&\Leftrightarrow&6k-2k&=-2+30\\&\Leftrightarrow&4k&=28\\&&\therefore\ k&=\bf7\\\\&&(ii)\implies h&=2(7)-9\\&&&=14-9\\&&\therefore\ h&=\bf5\end{aligned}[/tex] CONCLUSION/KESIMPULAN∴  h = 5, k = 7, a = ½  

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Last Update: Thu, 30 Jun 22
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