Find the equations of two straight lines which are parallel to x+7y+2=0 and at 2 $\sqrt{2}$ distance away from it.[[1]].

Question

Find the equations of two straight lines which are parallel to x+7y+2=0 and at 22 distance away from it.[[1]].

Accepted Answer

Let us consider the equation of the given straight line, x+7y+2=0
Now, we find the equations of the lines parallel to the line x+7y+2=0.
Let us consider the formula, the equations of the lines parallel to the line ax+by+c=0 are of the form ax+by+k=0 where k is some constant.
By using the above formula, the equations of the lines parallel to the line x+7y+2=0 are of the form x+7y+k=0.
Now, we find the distance between the parallel lines x+7y+2=0 and x+7y+k=0.
Let us consider the formula, the distance between two parallel lines ax+by+c1=0 and
ax+by+c2=0 is given by

\frac{|c 1 - c 2|}{a^{2} + b^{2}}

.
By using the above formula, the distance between parallel lines x+7y+2=0 and x+7y+k=0 is given by

= \frac{|2 - k|}{\sqrt{1^{2} + 7^{2}}}

= \frac{|2 - k|}{\sqrt{1 + 49}}

= \frac{|2 - k|}{\sqrt{50}}

= \frac{|2 - k|}{\sqrt{25 \times 2}}

= \frac{|2 - k|}{5 \sqrt{2}}

But, we were given that the distance between the parallel lines is

\sqrt{2}

.
So, we get,

\frac{|2 - k|}{5 \sqrt{2}} = \sqrt{2}

\Rightarrow | 2 - k | = 5 \sqrt{2} \times \sqrt{2}

⇒|2−k|=10
⇒2−k=±10
First, let us consider, 2−k=10, we get,
2−k=10
⇒k=2−10
⇒k=−8
Then we consider, 2−k=−10, we get,
2−k=−10
⇒k=2+10
⇒k=12
So, we got,
k=−8 and k=12
Therefore, the equations of the two straight lines which are parallel to x+7y+2=0 and at

\sqrt{2}

distance away from it are x+7y−8=0 and x+7y+12=0.

Find the equations of two straight lines which are parallel to x+7y+2=0 and at 2 $\sqrt{2}$ distance away from it.[[1]].

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