How do you find the value of a given the points (4,-1), (a,5) with a distance of #10#?

Distance between two points is given by the formula :

#d = sqrt((x_2-x_1)^2 + (y_2 - y_1)^2)#

#"Given " (x_1, y_1) = (4,-1), (x_2,y_2) = (a,5), d = 10, " To find " a#

#10 = sqrt ((a - 4)^2 + (5+1)^2)#

#(a-4)^2 + 6^2 = 10^2, " squaring both sides"#

#(a - 4)^2 = 10^2 - 6^2 = 64 = 8^2#

#a - 4 = +- 8 , " taking root on both sides"#

#color(brown)(a =) +-8 + 4 = color(brown)(12, -4#

First, let's take a look at the distance formula where #d# is the distance

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

Now you choose which point is point 2 (includes #y_2# and #x_2#) and which point is point 1 (includes #y_1# and #x_1#)

Point 2: #(a,5)->y_2 = 5# and #x_2 =a#
Point 1: #(4,-1)->y_1 = -1# and #x_1 = 4#

Now plug these values into the equation

#d=sqrt((a-4)^2+(5-(-1))^2)#

We also know that the distance #d# is 10 so we can plug that in

#10=sqrt((a-4)^2+(5-(-1))^2)#

Simplify

#10=sqrt((a-4)^2+(5-(-1))^2)#

#10=sqrt((a-4)^2+(5+1)^2)#

#10=sqrt((a-4)^2+(6)^2)#

#10=sqrt((a-4)^2+36)#

Now square both sides to get rid of the radical

#10^2=(sqrt((a-4)^2+36))^2#

#100=(a-4)^2+36#

FOIL the expression #(a-4)^2#

#(a-4)^2#

#(a-4)(a-4)#

#a^2+a(-4)+a(-4)+(-4)(-4)#

#a^2-4a-4a+16#

#a^2-8a+16#

Now plug this expression back in for #(a-4)^2#

#100=(a-4)^2+36#

#100=a^2-8a+16+36#

#100=a^2-8a+52#

Subtract 100 from both sides

#a^2-8a+52-100=100-100#

#a^2-8a-48=0#

Now factor this expression to find the roots/zeros

#a^2-8a-48=0#

#(a-12)(a+4)=0#

So the roots are

#a=12, a=-4#