Question

Find the distance of the line $$4x + 7y + 5 = 0$$ from the point $$\left( {1,2} \right)$$ along the line $$2x - y = 0$$.
A. $$\dfrac{{23}}{7}\sqrt 5 $$ sq units
B. $$\dfrac{{23}}{{18}}\sqrt 5 $$ sq units
C. $$\dfrac{{23}}{8}\sqrt 5 $$ sq units
D. None of these

Answer 1

Hint: Here in this question, we have to find the distance between the given point from the line $$4x + 7y + 5 = 0$$ along the line $$2x - y = 0$$. For this first we need to find the point of intersection of two given lines by elimination method. Then take the coordinates of two points you want to find the distance between. Call one point. Point 1 $$\left( {{x_1},{y_1}} \right)$$ and make the other Point 2 $$\left( {{x_2},{y_2}} \right)$$. Know the distance formula $$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$ . This formula finds the length of a line that stretches between two points: Point 1 and Point 2.

Complete step by step answer:
The distance between two points is the length of the interval joining the two points. If the two points lie on the same horizontal or same vertical line. In general the distance can be found by subtracting the coordinates that are not the same.
The distance between two points of the $$xy$$ -plane can be found using the distance formula. An ordered pair $$\left( {x,{\text{ }}y} \right)$$ represents co-ordinate of the point, where x-coordinate (or abscissa) is the distance of the point from the centre and y-coordinate (or ordinate) is the distance of the point from the centre.
Formula to find Distance Between Two Points in 2d plane. Consider two points, point 1 $$\left( {{x_1},{y_1}} \right)$$ and point 2 $$\left( {{x_2},{y_2}} \right)$$ on the given coordinate axis.
The distance between these points is given as: $$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$
Now consider the given equations of lines $$4x + 7y + 5 = 0$$ and $$2x - y = 0$$ and we have to calculate distance along the line so the point $$P = \left( {1,2} \right)$$ lies on $$2x - y = 0$$.
$$4x + 7y = - 5$$ -----(1)
$$2x - y = 0$$ -----(2)
Now we have to solve these two equations to find the two line intersecting points.
Multiply equation (2) with 2, then we get
$$4x + 7y = - 5$$
$$4x - 2y = 0$$
Since the coordinates of $$x$$ are the same then no need to change the sign here and we simplify to know the unknown value $$y$$.
$$\underline {\begin{matrix}
   + 4x + 7y = - 5 \\
  \mathop + \limits_{(-)} 4x\mathop - \limits_{(+)} 2y = 0
 \end{matrix}} $$
Now we cancel the $$x$$ term so we have
$$\eqalign{
  & \underline {\begin{matrix}
   + 4x + 7y = - 5 \\
  \mathop + \limits_{( - )} 4x\mathop - \limits_{( + )} 2y = 0
 \end{matrix}} \\
  & 9y = - 5} $$
Divide 9 on both sides, then
$$ \Rightarrow \,\,\,y = - \dfrac{5}{9}$$
We have found the value of $$y$$ now we have to find the value of $$x$$ . So we will substitute the value $$y$$ to any one of the equations (1) or (2) . we will substitute the value of $$y$$ to equation (1).
Therefore, we have $$4x + 7y = - 5$$
 $$ \Rightarrow 4x + 7\left( { - \dfrac{5}{9}} \right) = - 5$$
$$ \Rightarrow 4x - \dfrac{{35}}{9} = - 5$$
Take 9 as LCM in LHS
$$ \Rightarrow \dfrac{{36x - 35}}{9} = - 5$$
Multiply both side by 9
$$ \Rightarrow 36x - 35 = - 45$$
add both side by 35
$$ \Rightarrow 36x = - 45 + 35$$
$$ \Rightarrow 36x = - 10$$
Divide both side by 36
$$ \Rightarrow \,\,x = - \dfrac{{10}}{{36}}$$
Divide both numerator and denominator by 2, then we have
$$ \Rightarrow \,\,x = - \dfrac{5}{{18}}$$
Hence, we get the coordinates $$Q = \left( { - \dfrac{5}{{18}}, - \dfrac{5}{9}} \right)$$
The distance between the points PQ is
$$ \Rightarrow \,\,d = \sqrt {{{\left( {1 - \left( { - \dfrac{5}{{18}}} \right)} \right)}^2} + {{\left( {2 - \left( { - \dfrac{5}{9}} \right)} \right)}^2}} $$
$$ \Rightarrow \,\,d = \sqrt {{{\left( {1 + \dfrac{5}{{18}}} \right)}^2} + {{\left( {2 + \dfrac{5}{9}} \right)}^2}} $$
$$ \Rightarrow \,\,d = \sqrt {{{\left( {\dfrac{{18 + 5}}{{18}}} \right)}^2} + {{\left( {\dfrac{{18 + 5}}{9}} \right)}^2}} $$
$$ \Rightarrow \,\,d = \sqrt {{{\left( {\dfrac{{23}}{9}} \right)}^2}\left( {{{\left( {\dfrac{1}{2}} \right)}^2} + 1} \right)} $$
$$ \Rightarrow \,\,d = \dfrac{{23}}{9}\sqrt {\dfrac{1}{4} + 1} $$
$$ \Rightarrow \,\,d = \dfrac{{23}}{9}\sqrt {\dfrac{{1 + 4}}{4}} $$
$$ \Rightarrow \,\,d = \dfrac{{23}}{9}\sqrt {\dfrac{5}{4}} $$
$$ \Rightarrow \,\,d = \dfrac{{23}}{9}\left( {\dfrac{{\sqrt 5 }}{2}} \right)$$
$$\therefore \,\,d = \dfrac{{23}}{{18}}\sqrt 5 \,\,sq\,units$$
Hence, the required distance is $$\dfrac{{23}}{{18}}\sqrt 5 \,\,sq\,units$$.

So, the correct answer is “Option B”.

Note:
The distance is a length between the two points. In geometry we have a formula to determine the distance between the points. While determining the distance between the points we consider the both values of x and the value of y. Where x and y are the coordinates.