Question

the no. of lines that can be drawn through the point (5,2) at a distance of 5 units from the point (2,-2) ​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The number of lines that can be drawn through the point (5,2) at a distance of 5 units from the point (2,-2)

CALCULATION

Let the equation of the line passing through the point (5,2) is

$\sf{(y - 2) = m(x - 5)} \: \: where \: m \: is \: slope \: of \: the \: line$

$\implies \: \sf{mx - y - 5m + 2 = 0 \: \: \: \: \: ......(1)}$

Now the Perpendicular Distance from the point (2,-2) to the line given by Equation (1) is

$= \displaystyle \: \sf{ \bigg| \frac{(m \times 2) - ( - 2) - 5m + 2}{ \sqrt{ {m}^{2} + 1} } \bigg| \: }$

$= \displaystyle \: \sf{ \bigg| \frac{2m + 2- 5m + 2}{ \sqrt{ {m}^{2} + 1} } \bigg| \: }$

$= \displaystyle \: \sf{ \bigg| \frac{4 - 3m}{ \sqrt{ {m}^{2} + 1} } \bigg| \: }$

By the given condition

$\displaystyle \: \sf{ \bigg| \frac{4 - 3m}{ \sqrt{ {m}^{2} + 1} } \bigg| = 5\: }$

$\implies\displaystyle \: \sf{ \frac{4 - 3m}{ \sqrt{ {m}^{2} + 1} } = \pm \: 5\: }$

Squaring both sides we get

$\implies\displaystyle \: \sf{ {(4 - 3m)}^{2} = 25( {m}^{2} + 1)}$

$\implies\displaystyle \: \sf{ 9 {m}^{2} - 24m + 16 = 25 {m}^{2} + 25 }$

$\implies\displaystyle \: \sf{ 16 {m}^{2} + 24m + 9 = 0 }$

$\implies\displaystyle \: \sf{ {(4m + 3)}^{2} = 0 }$

$\implies\displaystyle \: \sf{ {(4m + 3)} = 0 }$

$\implies\displaystyle \: \sf{ m = - \frac{3}{4} }$

So the required line is

$\implies\displaystyle \: \sf{ (y - 2) = - \frac{3}{4} (x - 5) }$

$\implies\displaystyle \: \sf{ 4y - 8 = - 3x + 15 }$

$\implies\displaystyle \: \sf{ 3x + 4y = 23 }$

RESULT

Hence ONLY ONE LINE can be drawn through the point (5,2) at a distance of 5 units from the point (2,-2)

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Answer 2

Given : lines that can be drawn through the point (5,2) at a distance of 5 units from the point (2,-2) ​

To find  : number of lines :

Solution:

Given points ( 2 ,-2)  & ( 5 , 2)

Distance between points  

= √((5 - 2)² + (2 - (-2))²)

= √9 + 16

= 5

As distance between points is 5

perpendicular distance is shortest distance

Hence line passing though ( 5 , 2) must be perpendicular to line joining points

( 5 , 2) & ( 2 , - 2)

Hence only one line possible

Slope between ( 5 , 2) & ( 2 , - 2)    =  4/3

Slope of line passing though ( 5 , 2) would be -3/4

y - 2 =  -(3/4)(x - 5)

=> 4y - 8 = -3x + 15

=> 3x + 4y = 23

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