# How do you find the value of a given the points (-2,-5), (a,7) with a distance of 13?

Mar 30, 2017

a can be either -7 or 3

#### Explanation:

Distance between two points
1. $\left({x}_{1} , {y}_{1}\right)$ &
2. $\left({x}_{2} , {y}_{2}\right)$
is given by $\sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$.

In this case,

distance = 13;
$\left({x}_{1} , {y}_{1}\right) = \left(- 2 , - 5\right)$;
$\left({x}_{2} , {y}_{2}\right) = \left(a , 7\right)$;

$\therefore$ $13 = \sqrt{{\left(- 2 - a\right)}^{2} + {\left(- 5 - 7\right)}^{2}}$

squaring both sides,

$\implies$ ${13}^{2} = \left({a}^{2} + 4 \cdot a + 4\right) + 144$
$\implies$ $169 = {a}^{2} + 4 \cdot a + 148$
$\implies$ ${a}^{2} + 4 \cdot a + 148 = 169$
$\implies$ ${a}^{2} + 4 \cdot a - 21 = 0$ which is a quadratic equation

$\because$ solution of a quadratic equation of the form $p \cdot {x}^{2} + q \cdot x + r = 0$ ;
where $p , q , r$ are constants is given by:-
$x = \frac{- q \pm \sqrt{{q}^{2} - 4 \cdot p \cdot r}}{2 \cdot}$

in our case $x$ becomes $a$ and constants $p , q , r$ become $1 , 4 , - 21$ respectively

$\therefore$ $a = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot \left(- 21\right) \cdot 1}}{2 \cdot 1}$
$\implies$ $a = \frac{- 4 \pm \sqrt{100}}{2}$
$\implies$ $a = \frac{- 4 \pm 10}{2}$
$\implies$ $a = - \frac{14}{2} , \frac{6}{2}$
$\implies$ $a = - 7 \mathmr{and} 3$
$\therefore$ a can be either -7 or 3.