Question

the equation of the line which is parallel to 5x+12y+1=0 and 5x+12y+7=0 and lying midway between them is​

Answer 1

SOLUTION

TO DETERMINE

The equation of the line which is parallel to 5x + 12y + 1 = 0 and 5x + 12y + 7 = 0 and lying midway between them

EVALUATION

Here the given equation of the lines are

5x + 12y + 1 = 0 - - - - - - (1)

5x + 12y + 7 = 0 - - - - - (2)

Since both the constants 1 and 7 are positive

So bother the line in same side of the origin

Now the required line is parallel to the given lines

Let required equation of the line is

$\sf{5x + 12y + k = 0 \: \: \: - - - (3)}$

Now the distance between line 1 and line 3

$\displaystyle \sf{ = \bigg| \frac{k - 1}{ \sqrt{ {5}^{2} + {12}^{2} } } \bigg| }$

$\displaystyle \sf{ = \bigg| \frac{k - 1}{ \sqrt{ 25 + 144} } \bigg| }$

$\displaystyle \sf{ = \bigg| \frac{k - 1}{ \sqrt{ 169 } } \bigg| }$

$\displaystyle \sf{ = \bigg| \frac{k - 1}{13} \bigg| }$

Now the distance between line 2 and line 3

$\displaystyle \sf{ = \bigg| \frac{k - 7}{ \sqrt{ {5}^{2} + {12}^{2} } } \bigg| }$

$\displaystyle \sf{ = \bigg| \frac{k - 7}{ \sqrt{ 25 + 144 } } \bigg| }$

$\displaystyle \sf{ = \bigg| \frac{k - 7}{ \sqrt{169} } \bigg| }$

$\displaystyle \sf{ = \bigg| \frac{k - 7}{ 13 } \bigg| }$

So by the given condition

k - 1 = - ( k - 7 )

$\sf{ \implies \: k - 1 = - k + 7}$

$\sf{ \implies \: 2k = 8 }$

$\sf{ \implies \: k = 4 }$

Graph

In the graph

5x + 12y + 1 = 0 : Green Line

5x + 12y + 7 = 0 : Blue line

5x + 12y + 4 = 0 : Red line

FINAL ANSWER

Hence the required value of k = 4

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Answer 2

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