Point P has coordinates (k,-3). Point Q has coordinates (2,-3).

Berikut ini adalah pertanyaan dari sksooewj pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

Point P has coordinates (k,-3).Point Q has coordinates (2,-3). The length of PQ is 6.5 units and k < 0 Find the value of k.

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

The value of k is –4.5.

Explanation

Distance of Two Points

It is given that:

Point P has coordinates (k, –3).
Point Q has coordinates (2, –3).
The length of PQ is 6.5 units.
k < 0

Question: Find the value of k.

SIMPLER METHOD

Looking at both ordinates, we can get that point P and point Q are collinear, on the line of y = –3. It means the distance between point P and point Q is determined by the difference of their abscissas. Considering that k < 0, we get:
$\begin{aligned}d_{PQ}&=x_Q-x_P\\\Rightarrow 6.5&=2-k\\k&=2-6.5\\\therefore\ k&=\bf{-}4.5\end{aligned}$

DISTANCE FORMULA METHOD

To find the value of k, we can also use the distance of two points formula derived from the Pythagorean theorem.

$\begin{aligned}&d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\end{aligned}$

Rewriting that formula based on known facts we have, we get:

$\begin{aligned}d_{PQ}&=\sqrt{\left(x_Q-x_P\right)^2+\left(y_Q-y_P\right)^2}\\\Rightarrow\;{d_{PQ}}^2&=\left(x_Q-x_P\right)^2+\left(y_Q-y_P\right)^2\\\Rightarrow {(6.5)}^2&={(2-k)}^2+\left(-3-(-3)\right)^2\\&={(2-k)}^2+0^2\\\Rightarrow\quad\;6.5&={}\pm\sqrt{(2-k)^2}\\&={}\pm(2-k)\\&=\begin{cases}2-k\\k-2\end{cases}\\\Rightarrow\qquad k&=\begin{cases}2-6.5\\6.5+2\end{cases}\\\Rightarrow\qquad k&=\begin{cases}\bf{-}4.5\\\bf8.5\end{cases}\end{aligned}$