Find the value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units.

Home » Find the value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units.

Find the value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units.

Solution:

The distance between the two points (x₁,y₁) ( x₂,y₂) :

$d=\surd ({{x}_{2}}-{{x}_{1}}){}^\text{2}+({{y}_{2}}-{{y}_{1}}){}^\text{2}$

The distance between A (–3, –14) and B (a, –5):

$=\surd \left[ \left( a+3 \right){}^\text{2}+\left( -5+14 \right){}^\text{2} \right]\text{ }=9$

Now squaring on both the sides L.H.S and R.H.S.

$\left( a+3 \right){}^\text{2}+81=81$

$\left( a+3 \right){}^\text{2}=0$

$\left( a+3 \right)\left( a+3 \right)=0$

$a+3=0$

$a=-3$

i

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