Question

the point on the line 3x-2y=1 which is closest to the origin​

Answer 1

The complete question is:

The point on the line 3x−2y=1 which is closest to the origin is

(A) (3/13,−2/13)

(B) (5/11,2/11)

(C) (3/5,2/5)

(D) none of these

Given:

Equation of the line: 3x - 2y =1

To find:

Point o the line which is closest to the origin.

Solution:

As we know that the if the equation of a line is given s ax + by +c =0,

Then its distance (smallest) to any point is |ax1 + by1 + c| / √(a² + b²)

Since we need to find the distance from the origin:

(x1 , y1) = 0

Distance = |-1| / √(3² + 2²)

= 1/ √13 units.

Now we will check the first option:

The distance of (3/13, -2/13) from origin will be:

d = √(3/13-0)² + (-2/13 - 0)²

d = 1/ √13

Therefore the correct option is 1.

Answer 2

SOLUTION

TO DETERMINE

The point on the line 3x-2y = 1 which is closest to the origin

EVALUATION

Here the equation of the line is

$\sf{3x - 2y = 1} \: \: .......(1)$

Let (h, k) be the required point and S be the distance between the point and origin

Then (h, k) is a point on the line (1)

$\therefore \: \sf{3h - 2k = 1}$

$\displaystyle \implies \sf{k = \frac{3h - 1}{2} } \: \: .....(2)$

Also

$\sf{S = \sqrt{ {(h - 0)}^{2} + {(k - 0)}^{2} } }$

$\implies \sf{S = \sqrt{ {h }^{2} + {k }^{2} } }$

$\displaystyle \implies \sf{S = \sqrt{ {h }^{2} + { \bigg( \frac{3h - 1}{2} \bigg) }^{2} } }$

$\displaystyle \implies \sf{{S}^{2} = {h }^{2} + { \bigg( \frac{3h - 1}{2} \bigg) }^{2} }$

Differentiating both sides with respect h we get

$\displaystyle \sf{ \frac{d}{dh}( {S}^{2} ) = 2h + 2 \times { \bigg( \frac{3h - 1}{2} \bigg) } \times \frac{3}{2} }$

$\implies \displaystyle \sf{ \frac{d}{dh}( {S}^{2} ) = 2h + 3 \times { \bigg( \frac{3h - 1}{2} \bigg) } }$

$\implies \displaystyle \sf{ \frac{d}{dh}( {S}^{2} ) = \frac{13h - 3}{2} }$

Again Differentiating both sides with respect h we get

$\implies \displaystyle \sf{ \frac{ {d}^{2} }{d {h}^{2} }( {S}^{2} ) = \frac{13}{2} }$

For minimum value of S we have

$\displaystyle \sf{ \frac{d}{dh}( {S}^{2} ) = 0 }$

$\implies \displaystyle \sf{\frac{13h - 3}{2} = 0 }$

$\implies \displaystyle \sf{ h = \frac{3}{13} }$

$\displaystyle \sf{For \: \: h = \frac{3}{13} \: \: we \: have }$

$\displaystyle \sf{ \frac{ {d}^{2} }{d {h}^{2} }( {S}^{2} ) = \frac{13}{2} > 0 }$

$\displaystyle \sf{ Therefore \: \: at \: \: h = \frac{3}{13} \: the \: distance \: is \: minimum }$

$\displaystyle \sf{When \: \: h = \frac{3}{13} } \: \: we \: \: have \: \: k = - \frac{2}{13}$

$\displaystyle \sf{Hence \: the \: required \: point \: is \: \: \bigg( \frac{3}{13} \: , \: - \frac{2}{13} \bigg) }$

The distance is

$\displaystyle \sf{S = \sqrt{ { \bigg( \frac{3}{13} \bigg) }^{2} + { \bigg( - \frac{2}{13} \bigg) }^{2} } } \: \: unit$

$\displaystyle \sf{ = \sqrt{ \frac{9 + 4}{169}} } \: \: unit$

$\displaystyle \sf{ = \sqrt{ \frac{13}{169}} } \: \: unit$

$\displaystyle \sf{ = \sqrt{ \frac{1}{13}} } \: \: unit$

FINAL ANSWER

The point on the line 3x-2y = 1 which is closest to the origin is

$\displaystyle \sf{ \bigg( \frac{3}{13} \: , \: - \frac{2}{13} \bigg) }$

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