Question

if the distance between the points (3 2) and(-1 x) is 5 then x is equal to

Answer 1

$\sf \pink{Given}\begin{cases}&\sf{Coordinates\:of\:point\:P=\bf{(3,2).}} \\ \\ &\sf{Coordinates\:of\:point\:Q=\bf{(-1,x).}} \\ \\ &\sf{Distance\:between\:point\:P\:and\:Q=\bf{5\:units.}}\end{cases}$

To FinD:-

The value of x.

SolutioN:-

We know that distance formula is,

$\normalsize{\purple{\underline{\boxed{\sf{Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}}}}$

$\sf {Here}\begin{cases}&\sf{x_2=\bf{-1}} \\ &\sf{x_1=\bf{3}} \\ &\sf{y_2=\bf{x}} \\ &\sf{y_1=\bf{2}} \\ &\sf{Distance=\bf{5}}\end{cases}$

• Putting the values,

$\\ :\normalsize\implies{\sf{5=\sqrt{((-1)-3)^2+(x-2)^2}}}$

$\\ :\normalsize\implies{\sf{5=\sqrt{(-1-3)^2+(x-2)^2}}}$

$\\ :\normalsize\implies{\sf{5=\sqrt{(-4)^2+(x^2-2.x.2+2^2)}(Identity=a^2-2ab+b^2)}}$

$\\ :\normalsize\implies{\sf{5=\sqrt{16+x^2-4x+4}}}$

$\\ :\normalsize\implies{\sf{5=\sqrt{20+x^2-4x}}}$

• Squaring both the sides,

$\\ :\normalsize\implies{\sf{5^2=20+x^2-4x}}$

$\\ :\normalsize\implies{\sf{25=20+x^2-4x}}$

$\\ :\normalsize\implies{\sf{25-20=x^2-4x}}$

$\\ :\normalsize\implies{\sf{5=x^2-4x}}$

$\\ :\normalsize\implies{\sf{0=x^2-4x-5}}$

$\\ :\normalsize\implies{\sf{x^2-4x-5=0}}$

• Splitting the middle term,

$\\ :\normalsize\implies{\sf{x^2+x-5x-5=0}}$

$\\ :\normalsize\implies{\sf{x(x+1)-5(x+1)=0}}$

$\\ :\normalsize\implies{\sf{(x-5)(x+1)=0}}$

$\\ :\normalsize\implies{\sf{x-5=0}}$

$\normalsize\therefore\boxed{\sf{\mathfrak{\purple{x=5}}}}$

$\\ :\normalsize\implies{\sf{x+1=0}}$

$\normalsize\therefore\boxed{\sf{\mathfrak{\purple{x=-1}}}}$

Value of x is (5) or (-1) .