Question

Prove that the following equation represents a pair of coincident lines

Answer 1

\[\sqrt{(x-2)^{2}+y ^{2}}+\sqrt{(x+2)^{2}+y ^{2}}=4\]

Answer 2

\[\sqrt{(x-2)^2+y^2}=4-\sqrt{(x+2)^2+y^2}\]
square both sides

Answer 3

\[(x-2)^2+y^2=(4-\sqrt{(x+2)^2+y^2})^2\]
\[x^2-4x+y^2=16-8\sqrt{x^2+2x+y^2}+x^2+2x+y^2\]
simplify then square again

Answer 4

@Jonask but how do I actually prove if it is a pair of coincident lines ?

Answer 5

well to lines are coincident if they have same slope and y-intercept
so we shud simplify the equation to that form
y in tercept form

Answer 6

did they give two equations

Answer 7

no only this one

Answer 8

any idea @Jonask ?

Answer 9

\[0=6x+16-8\sqrt{x^2+2x+y^2}\]
\[3x+4=4\sqrt{x^2+2x+y^2}\]\[9x^2+24x+16=16(x^2+2x+y^2) \implies 0=7x^2+8x+16y^2-16\]

Answer 10

but @jonask my answer is \[3x ^{2}+8x+4y ^{2}=0\]
I don't know where I've gone wrong

Answer 11

http://www.transtutors.com/math-homework-help/straight-line/pair-of-straight-lines.aspx

Answer 12

i made mistake its \[3x+8=4\sqrt{x^2+2x+y^2} \implies 9x^2+42x+64=4x^2+8x+4y^2\]
\[5x^2+34x-4y^2+64=0\]
this is in the from\[\huge ax^2+by^2+2hxy+2dx+2fy+c\]

Answer 13

if \[h^2=ab\]
hen lines conside this acually means the angle between thwm is zero

Answer 14

Oh now i get it. Thank you so much :)

Answer 15

but the coefficient of xy is 0 so the value of h is 0, but the value of ab is -20 which means the equation does not represent a pair of straight lines ?

Answer 16

yes but i think we need to verify our equation

Answer 17

yeah maybe. but can you check my steps because i am getting a different equation ?

Answer 18

so \[h^2 \ge ab \] implies the lines are distinct

Answer 19

\[0 \ge -20\]

Answer 20

\[\sqrt{(x-2)^{2}+y ^{2}}=4-\sqrt{(x+2)^{2}+y ^{2}}\]
Squaring on both sides
\[(x-2)^{2}+y^{2}=16+(x+2)^{2}+y^{2}-8\sqrt{(x-2)^{2}+y^{2}}\]
\[x^{2}+4-4x+y^{2}=16+x^{2}+4+4x+y^{2}-8\sqrt{(x-2)^{2}+y^{2}}\]
\[x^{2}-x^{2}+4-4+y^{2}-y^{2}=16+4x+4x-8\sqrt{(x+2)^{2}+y^{2}}\]
\[0=16+8x-8\sqrt{(x+2)^{2}+y^{2}}\]
\[16+8x=8\sqrt{(x+2)^{2}+y^{2}}\]
\[2+x=\sqrt{(x+2)^{2}+y^{2}}\]
Squaring on both sides,
\[(x+2)^{2}=(x+2)^{2}+y^{2}\]
\[y^{2}=0\]