Question

A straight line L is drawn perpendicular to the line \[5x - y{\text{ = }}1\]. The area of the triangle formed by the line L and coordinates axes is 5, then the distance of line L from the line \[x + 5y = 0\] is:
A) \[\dfrac{7}{{\sqrt 5 }}\]
B) \[\dfrac{5}{{\sqrt {13} }}\]
C) \[\dfrac{7}{{\sqrt {13} }}\]
D) \[\dfrac{5}{{\sqrt 7 }}\]

Answer 1

Hint: Here first we will find the slope of the perpendicular line L using the relation between the slopes of two perpendicular lines and then we will find the value of c using the area of triangle and then we will find its equation using the slope form i.e.
\[y = mx + c\]
Then finally we will use the formula of distance between parallel lines to get the distance between the lines L and \[x + 5y = 0\].

Complete step-by-step answer:

It is given that:
The area of triangle formed by the line L and coordinates axes is 5
Therefore the area of triangle OAB is 5
Now since we know that:
The area of triangle is given by:-
\[area = \dfrac{1}{2} \times base \times height\]
Hence area of triangle ABC is given by:-
\[area = \dfrac{1}{2} \times 5c \times c\] where base=5c, height=c
Putting the value of area we get:-
\[5 = \dfrac{5}{2}{c^2}\]
Solving for the value of c we get:-
\[ \Rightarrow {c^2} = 2\]
Taking square root of both sides we get:-
\[  \sqrt {{c^2}} = \sqrt 2 \]
\[   \Rightarrow c = \pm \sqrt 2 ....................(1) \]
Let us now find the slope of the given line \[5x - y{\text{ = }}1\]
Simplifying it we get:-
\[y = 5x - 1\]
Comparing it with general equation of line i.e.\[y = mx + c\]
We get:-
\[m = 5\]
Let the slope of line L be m
Now it is given that line L is perpendicular to the \[5x - y{\text{ = }}1\]
Therefore we know that the product of slopes of perpendicular lines is -1.
Hence, we get:-
\[m \times 5 = - 1\]
Evaluating for m we get:-
\[m = \dfrac{{ - 1}}{5}\]………………………..(2)
Now we will find the equation of line L using the slope intercept form i.e.
\[y = mx + c\]
Putting values from equation 1 and 2 we get:-
\[y = \left( { - \dfrac{1}{5}} \right)x \pm \sqrt 2 \]
Simplifying it further we get:-

$  y = \dfrac{{ - x \pm 5\sqrt 2 }}{5} $

$  \Rightarrow 5y = - x \pm 5\sqrt 2 $

Hence we get the equation of L as:
\[5y + x \mp 5\sqrt 2 = 0\]
Now we need to find the distance between the lines \[5y + x \mp 5\sqrt 2 = 0\] and \[x + 5y = 0\]
Now since these lines are parallel.
The distance between the parallel lines \[ax + by + {c_1} = 0\& ax + by + {c_2} = 0\]is given by:-
\[d = \left| {\dfrac{{{c_1} - {c_2}}}{{\sqrt {{a^2} + {b^2}} }}} \right|\]
Hence applying this formula for the above lines we get:-
\[d = \left| {\dfrac{{ \mp 5\sqrt 2 - 0}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 5 \right)}^2}} }}} \right|\]
Solving it further we get:-

\[  d = \left| {\dfrac{{ \mp 5\sqrt 2 }}{{\sqrt {1 + 25} }}} \right| \]

\[   \Rightarrow d = \left| {\dfrac{{ \mp 5\sqrt 2 }}{{\sqrt {26} }}} \right| \]

Cancelling the term we get:-
\[d = \left| {\dfrac{{ \mp 5}}{{\sqrt {13} }}} \right|\]
Evaluating it further we get:-
\[d = \dfrac{5}{{\sqrt {13} }}\] units
Hence the distance is \[\dfrac{5}{{\sqrt {13} }}\] units.

Therefore, option (B) is the correct option.

Note: Students should take a note that in parallel lines \[ax + by + {c_1} = 0\] and \[ax + by + {c_2} = 0\], the coefficients of x and y are same.

Also, students should use the slope intercept form to find the equation of line to make the solution easier.