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

1 Answer
Mar 30, 2017

Answer:

a can be either -7 or 3

Explanation:

Distance formula:-

Distance between two points
1. #(x_1, y_1)# &
2. #(x_2, y_2)#
is given by #sqrt((x_1-x_2)^2 + (y_1-y_2)^2)#.

In this case,

distance = 13;
#(x_1, y_1) = (-2, -5)#;
#(x_2, y_2) = (a, 7)#;

#therefore# #13 = sqrt((-2-a)^2 + (-5-7)^2) #

squaring both sides,

#implies# #13^2 = (a^2 + 4*a + 4) + 144#
#implies# #169 = a^2 + 4*a + 148#
#implies# #a^2 + 4*a + 148 = 169#
#implies# #a^2 + 4*a -21 = 0# which is a quadratic equation

#because# solution of a quadratic equation of the form #p*x^2 + q*x + r =0# ;
where #p, q, r# are constants is given by:-
#x = (-q +- sqrt(q^2 - 4*p*r))/(2*)#

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

#therefore# # a = (-4 +- sqrt(4^2 - 4*(-21)*1))/(2*1) #
#implies# # a = (-4 +- sqrt100)/2 #
#implies# #a=(-4 +- 10)/2#
#implies# # a = -14/2 , 6/2#
#implies# # a = -7 or 3#
#therefore# a can be either -7 or 3.