Question

The equations of two lines through \[\left( {0,\;a} \right)\]which are at distance \[a\] from the point \[\left( {2a,\;2a} \right)\]are
A) \[\;y - a = 0\;and\;4x - 3y - 3a = 0\]
B) \[\;y - a = 0\;and\;3x - 4y + 3a = 0\]
C) \[\;y - a = 0\;and\;4x - 3y + 3a = 0\]
D) None of these

Answer 1

Hint: Straight line is a set of infinites points in which all points are linear. In this question first calculate intercept of required line by using general equation of line. Distance of \[\left( {2a,\;2a} \right)\] is given from required straight line therefore use formula of distance of a point from straight line.



Formula Used:General equation of straight line:
\[y = mx + c\]
Where
m is slope of straight line
c is y intercept of straight line



Complete step by step solution:Given: Line passing through\[\left( {0,\;a} \right)\], the distance of point \[\left( {2a,\;2a} \right)\]from line is \[a\]
Equation of required line is given by:
\[y = mx + c\]
This line is passes through point \[\left( {0,\;a} \right)\]
Now put \[\left( {0,\;a} \right)\]in \[y = mx + c\]
We get
\[c = a\]
Now equation becomes
\[y = mx + a\]
We know that equation of a point from straight line is given by
\[d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\]
Where
\[({x_1},{y_1})\]is a point which is at distance d from line \[ax + by + c = 0\]
Now equation of line is \[y = mx + a\]which can be written as \[mx - y + a = 0\]
Distance of point \[\left( {2a,\;2a} \right)\]from line \[mx - y + a = 0\]is given by
\[d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\]
\[d = \left| {\dfrac{{m(2a) - (2a) + a}}{{\sqrt {{m^2} + {{( - 1)}^2}} }}} \right|\]
In question distance is given which is equal to a
\[d = \left| {\dfrac{{m(2a) - (2a) + a}}{{\sqrt {{m^2} + {{( - 1)}^2}} }}} \right| = a\]
\[\left| {\dfrac{{m(2a) - (2a)a}}{{\sqrt {{m^2} + {{( - 1)}^2}} }}} \right| = a\]
\[\left| {\dfrac{{m(2a) - (2a) + a}}{{\sqrt {{m^2} + {{( - 1)}^2}} }}} \right| = a\]
\[\left| {\dfrac{{2am - a}}{{\sqrt {{m^2} + 1} }}} \right| = a\]
\[\left| {\dfrac{{2m - 1}}{{\sqrt {{m^2} + 1} }}} \right| = 1\]
\[\dfrac{{2m - 1}}{{\sqrt {{m^2} + 1} }} = 1\]
\[2m - 1 = \pm \sqrt {{m^2} + 1} \]
On squaring both side we get
\[3{m^2} - 4m = 0\]
\[m(3m - 4)\]
\[m = 0\]and \[3m - 4 = 0\]
\[m = \dfrac{4}{3}\]
Now required lines are
\[y - a = 0\] and \[y = \dfrac{4}{3}x + a\]
\[y - a = 0\]and \[4x - 3y + 3a = 0\]



Option ‘C’ is correct



Note: General equation of straight line used here because one point through which line passing is given.
Whenever distance of a point from a line is given use distance formula used for calculating distance of a point from line.
Here we get two slopes because there are two lines which are passing through \[\left( {0,\;a} \right)\]and which are at distance \[a\] from the point \[\left( {2a,\;2a} \right)\]