Question

If the a (2,-4) is equidistant from p(3,8)and q (-10,y),find the value of y.also find the distance pq

Answer 1

Hello,

we have that:
ap=pq
so
ap²=pq²

then:
(2-3)²+(-4-8)²=(2-10)²+(-4-y)²;
(-1)²+(-12)²=(-8)²+16+8y+y²;
1+144=64+16+8y+y²;
-y²-8y+145-80=0;
-y²-8y+65=0;
y²+8y-65=0;

we solve the second degree equation:
y²+8y-65=0
Δ=b²-4ac=(8)²-4(-65)=64+260=324;
y₁,₂=-8±√Δ/2=-8±√324/2=-8±18/2
y₁=-8-18/2=-26/2=-13
y₂-8+18/2=10/2=5

The value of y are -13 or 5

therefore
if y=-13 ,
the point  are :
 p(3,8) and q(-10,-13);
then:
pq=√(3+10)²+(8+13)²=√13²+21²=√169+441=√610= 24.69
or
 if y=5 ,
the point  are :
 p(3,8) and q(-10,5);
then:
pq=√(3+10)²+(8-5)²=√13²+3²=√169+9=√178= 13.34

bye :-)