Question

If the point a(2,-4) is equidistant from p(3,8) & q(-10,y) then 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 :-)

Answer 2

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 :-)

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