Question

A line is of length 10 and one end is at the point (2, -3); if the abscissa of the other end be 10, prove that its ordinate must be 3 or -9.

Answer 1

Distance $d$ between two points $P(a,b)$ and $Q(c,d)$ is given by

$d=\sqrt{(a-c{)}^2+(b-d{)}^2}$

Given that :

Length of line i.e $d=10$

Abscissa of second point is $10$

And let us assume the ordinate of second point be $y$

Points $(2,-3)$ & $(10,y)$

Now apply distance formula

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

Squaring both sides

$\Rightarrow 100=(-8{)}^2+(-3-y{)}^2$

$\Rightarrow 100=64+(-3-y{)}^2$

$\Rightarrow 100-64=(-3-y{)}^2$

$\Rightarrow 36=(-3-y{)}^2$

Taking square root both sides

$\Rightarrow \pm 6=(-3-y)$

or $\Rightarrow \pm 6=(3+y)$ (taking - sign common on both sides)

$\Rightarrow \pm 6-3=y$

Taking both cases

$y=+6-3=3$

$y=-6-3=-9$