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 = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ \stackrel{d}{19}=√([x-(-6)]^2+[-5-6]^2)\implies 19=√((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√(12^2-4(1)(-204))}{2(1)}\implies x = \cfrac{-12\pm √(144+816)}{2} \\\\\\ x = \cfrac{-12\pm √(960)}{2}\implies x = \cfrac{-12\pm √(8^2\cdot 15)}{2}\implies x = \cfrac{-12\pm 8√(15)}{2} \\\\\\ x = -6\pm 4√(15)\implies x \approx \begin{cases} 9.49\\ -21.49 \end{cases}`

Answer 2

-6 + 6√10 and -6 - 6√10

Explanation:

By the distance formula,

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

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

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

`d=√((x+6)^2+(-5+6)^2)`

`=√((x+6)^2+1)`

According to the question,

d = 19 units,

`\implies √((x+6)^2+1)=19`

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

`(x+6)^2=360`

`x+6=\pm √(360)`

`x = -6\pm 6√(10)`

Hence, the possible values of x would be,

-6 + 6√10 and -6 - 6√10