Question

Find all x such that the distance between (-6,-6) and (x,-5) is 19.

Answer 1

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{x}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{d}{19}=\sqrt{[x-(-6)]^2+[-5-6]^2}\implies 19=\sqrt{(x+6)^2+(-11)^2} \\\\\\ 19^2=(x+6)^2+(-11)^2\implies 361 = (x^2+12x+36)+121 \\\\\\ 361=x^2+12x+157\implies 0=x^2+12x-204 \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+12}x\stackrel{\stackrel{c}{\downarrow }}{-204} \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}$

$\bf x = \cfrac{-12\pm\sqrt{12^2-4(1)(-204)}}{2(1)}\implies x = \cfrac{-12\pm \sqrt{144+816}}{2} \\\\\\ x = \cfrac{-12\pm \sqrt{960}}{2}\implies x = \cfrac{-12\pm \sqrt{8^2\cdot 15}}{2}\implies x = \cfrac{-12\pm 8\sqrt{15}}{2} \\\\\\ x = -6\pm 4\sqrt{15}\implies x \approx \begin{cases} 9.49\\ -21.49 \end{cases}$

Answer 2

Answer:

-6 + 6√10 and -6 - 6√10

Step-by-step explanation:

By the distance formula,

The distance between the points $(x_1, y_1)$ and $(x_2, y_2)$ is,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So, the distance between (-6,-6) and (x,-5) is,

$d=\sqrt{(x+6)^2+(-5+6)^2}$

$=\sqrt{(x+6)^2+1}$

According to the question,

d = 19 units,

$\implies \sqrt{(x+6)^2+1}=19$

$(x+6)^2+1 = 361$

$(x+6)^2=360$

$x+6=\pm \sqrt{360}$

$x = -6\pm 6\sqrt{10}$

Hence, the possible values of x would be,

-6 + 6√10 and -6 - 6√10