Question

If the point $(x,y)$ moves along the straight line $3x+4y+5=0$. Then find the minimum value of $x^2+y^2$.

Answer 1

Given the line is $3x+4y+5=0$ __ (1)

Now

$x^2+y^2$

$=x^2+(\frac{-5-3x}{4}{)}^2$ [using (1)]

$=x^2+\frac{(-5-3x{)}^2}{16}=f\left(x\right)$ [Let]

Now differentiating both sides w.r.t

'x' we get

$f\left(x\right)=2x+\frac{1}{8}\left\{\right(5+3x\left)\right\}.3$

or, $f^{\prime \prime }\left(x\right)=(2+\frac{3}{8});$ __ (2)

Now, for $ma{x}^m$ or minimum value

of f(x) we must have,

$f^{\prime }\left(x\right)=0$

or, $2x+\frac{9x}{8}=-15/8$

or, $\frac{25x}{8}=-15/8$

or, x = -3/5

putting x = -3/5 in (1) we get,

y = -4/5 ;

for this values of x & y we have,

$f^{\prime \prime }\left(x\right)>0$ so, for, These values of

x and y we have minimum

value of f(x).

so minimum value of f(x) is

$3^2/25+4^2/25=\frac{25}{25}=1.$