Question

If $z_1=3+4i$ and $z_2=-3-4i$ then the equation $|z-z_1|+|z-z_2|=10$ represents

• A. pair of straight lines
• B. circle
• C. ellipse
• D. line segment joining $z_1$ and $z_2$

Answer 1

The correct option is A pair of straight lines

Let z = x + iy.

$z_1=3+4i$

$z_2=-3-4i$

$|x+iy-(3+4i)-(3+4i\left)\right|+|x+iy-(-3-4i\left)\right|=10$

$\sqrt{(x-3{)}^2+(y-4{)}^2}+\sqrt{(x+3{)}^2+(y+4{)}^2}=10$

$\sqrt{(x-3{)}^2+(y-4{)}^2}=10-\sqrt{(x+3{)}^2+(y+4{)}^2}$

squaring both sides

$(x-3{)}^3+(y-4{)}^2=100-20\sqrt{(x+3{)}^2+(y+4{)}^2}+(x+3{)}^2+(y+4{)}^2$

$x^2-6x+9+y^2-8y+19=100-20\sqrt{(x+3{)}^2+(y+4{)}^2}+x^2+6x+9+y^2+8y+16$

$20\sqrt{(x+3{)}^2+(y+4{)}^2}=100+12x+16y$

$5\sqrt{(x+3{)}^2+(y+4{)}^2}=25+3x+4y$

squaring again

$25\left[\right(x+3{)}^2+(y+4{)}^2]={25}^2+q{x}^2+16y^2+2(75x+12xy+100y)$

$(a+b+c{)}^2=a^2+b^2+c^2+2(abc+bc+ac)$

$25[x^2+6x+9+y^2+8y+16]=625+9x^2+16y^2+152x24xy+200y$

$\therefore 16x^2+9y^2+150x+200y+625=625+150+200y+24xy$

$\therefore 16x^2+(-24xy)+9y^2=0$

$\therefore 16x^2-24xy+9y^2=0$

Represents pair of straight lines.