How do you find the possible values for a if the points (a,6), (-6,2) has a distance of $4 \sqrt{10}$ $4sqrt10$ ?

Question

How do you find the possible values for a if the points (a,6), (-6,2) has a distance of $4 \sqrt{10}$ $4sqrt10$ ?

2 Answers

Impact of this question

4197 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-09T15:25:50

See a solution process below:

Explanation:

The formula for calculating the distance between two points is:

$d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ $d = sqrt((color(red)(x_2) - color(blue)(x_1))^2 + (color(red)(y_2) - color(blue)(y_1))^2)$

Substituting the values for the distance and from the points in the problem gives:

$4 \sqrt{10} = \sqrt{{(- 6 - a)}^{2} + {(2 - 6)}^{2}}$ $4sqrt(10) = sqrt((color(red)(-6) - color(blue)(a))^2 + (color(red)(2) - color(blue)(6))^2)$

We can now solve for $a$ $a$ :

First, square both sides of the equation to eliminate the radicals while keeping the equation balanced:

${(4 \sqrt{10})}^{2} = {(\sqrt{{(- 6 - a)}^{2} + {(2 - 6)}^{2}})}^{2}$ $(4sqrt(10))^2 = (sqrt((color(red)(-6) - color(blue)(a))^2 + (color(red)(2) - color(blue)(6))^2))^2$

$16 \cdot 10 = {(- 6 - a)}^{2} + {(2 - 6)}^{2}$ $16 * 10 = (color(red)(-6) - color(blue)(a))^2 + (color(red)(2) - color(blue)(6))^2$

$160 = {(- 6 - a)}^{2} + {(- 4)}^{2}$ $160 = (color(red)(-6) - color(blue)(a))^2 + (-4)^2$

$160 = {(- 6 - a)}^{2} + 16$ $160 = (color(red)(-6) - color(blue)(a))^2 + 16$

$160 = {(- 6)}^{2} + 6 a + 6 a + a^{2} + 16$ $160 = (-6)^2 + 6a + 6a + a^2 + 16$

$160 = 36 + 12 a + a^{2} + 16$ $160 = 36 + 12a + a^2 + 16$

$160 = a^{2} + 12 a + 52$ $160 = a^2 + 12a + 52$

$160 - 160 = a^{2} + 12 a + 52 - 160$ $160 - color(red)(160) = a^2 + 12a + 52 - color(red)(160)$

$0 = a^{2} + 12 a - 108$ $0 = a^2 + 12a -108$

We can factor this as:

#0 = (a + 18)(a - 6)

Now, equate each term on the right side of the equation to $0$ $0$ and solve for $a$ $a$ :

Solution 1)

$a + 18 = 0$ $a + 18 = 0$

$a + 18 - 18 = 0 - 18$ $a + 18 - color(red)(18) = 0 - color(red)(18)$

$a + 0 = - 18$ $a + 0 = -18$

$a = - 18$ $a = -18$

Solution 3)

$a - 6 = 0$ $a - 6 = 0$

$a - 6 + 6 = 0 + 6$ $a - 6 + color(red)(6) = 0 + color(red)(6)$

$a - 0 = 6$ $a - 0 = 6$

$a = 6$ $a = 6$

The solution is, a could be $- 18$ $-18$ or $6$ $6$

Answer 2 · 2017-07-09T15:41:53

$a = - 18$ $a=-18$
or
$a = 6$ $a=6$

Explanation:

Distance between two points, $(x_{1}, y_{1})$ $(x_1,y_1)$ and $(x_{2}, y_{2})$ $(x_2,y_2)$

= $\sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}}$ $sqrt((x_1-x_2)^2+(y_1-y_2)^2$

In this case
$(a, 6)$ $(a,6)$ and $(- 6, 2)$ $(-6,2)$

$4 \sqrt{10} = \sqrt{{[a - (- 6)]}^{2} + {(6 - 2)}^{2}}$ $4sqrt(10)=sqrt([a-(-6)]^2+(6-2)^2)$
${(4 \sqrt{10})}^{2} = {\sqrt{{(a + 6)}^{2} + {(4)}^{2}}}^{2}$ $(4sqrt(10))^2={sqrt((a+6)^2+(4)^2)}^2$
$16 \cdot 10 = {(a + 6)}^{2} + 16$ $16*10=(a+6)^2+16$
$160 = a^{2} + 6 x + 6 x + 36 + 16$ $160=a^2+6x+6x+36+16$
$a^{2} + 12 x - 108 = 0$ $a^2+12x-108=0$

Carry out factorisation
$a^{2} + 12 x - 108 = 0$ $a^2+12x-108=0$
$(a + 18) (a - 6) = 0$ $(a+18)(a-6)=0$

Divide RHS $(0)$ $(0)$ with $(a + 18)$ $(a+18)$ and $(a - 6)$ $(a-6)$
You will have
$(a + 18) = 0$ $(a+18)=0$ and $(a - 6) = 0$ $(a-6)=0$ which give you
$a = - 18$ $a=-18$ or $a = 6$ $a=6$ respectively

Science

Math

Humanities

... and beyond

How do you find the possible values for a if the points (a,6), (-6,2) has a distance of $4 \sqrt{10}$ $4sqrt10$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the possible values for a if the points (a,6), (-6,2) has a distance of 4√104sqrt10?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you find the possible values for a if the points (a,6), (-6,2) has a distance of $4 \sqrt{10}$ $4sqrt10$ ?