Question

the distance of the point (2,3) from the line 8x+15y+k = 0 is 5 units,then the positive value of k is


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Answer 1

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Answer 2

The positive value of k is 14

Given :

The distance of the point (2,3) from the line 8x + 15y + k = 0 is 5 unit

To find :

The positive value of k

Solution :

Step 1 of 2 :

Form the equation

The distance of the point (2,3) from the line 8x + 15y + k = 0 is

$\displaystyle \sf{ = \bigg| \: \frac{(8 \times 2) + (15 \times 3) + k}{ \sqrt{ {8}^{2} + {15}^{2} } } \: \bigg| }$

By the given condition

$\displaystyle \sf{ \bigg| \: \frac{(8 \times 2) + (15 \times 3) + k}{ \sqrt{ {8}^{2} + {15}^{2} } } \: \bigg| = 5}$

Step 2 of 2 :

Find positive value of k

$\displaystyle \sf{ \bigg| \: \frac{(8 \times 2) + (15 \times 3) + k}{ \sqrt{ {8}^{2} + {15}^{2} } } \: \bigg| = 5}$

$\displaystyle \sf{ \implies \bigg| \: \frac{16 + 45 + k}{ \sqrt{ 64 + 225} } \: \bigg| = 5}$

$\displaystyle \sf{ \implies \bigg| \: \frac{61 + k}{ \sqrt{ 289} } \: \bigg| = 5}$

$\displaystyle \sf{ \implies \bigg| \: \frac{61 + k}{ 17 } \: \bigg| = 5}$

$\displaystyle \sf{ \implies | \: 61 + k| = 85}$

$\displaystyle \sf{ \implies 61 + k= \pm \: 85}$

61 + k = 85 gives k = 14

61 + k = - 85 gives k = - 146

Hence positive value of k is 14