Question

A straight line passes through the point $\left( {5,0} \right)$ and $\left( {0,3} \right)$ . The length of the perpendicular from the point $\left( {4,4} \right)$ on the line is
$A.{\text{ }}\dfrac{{17}}{{\sqrt {34} }} \\\ B.{\text{ }}\sqrt {\dfrac{{17}}{2}} \\\ C.{\text{ }}\dfrac{{15}}{{\sqrt {34} }} \\\ D.{\text{ }}\dfrac{{17}}{5} \\\$

Answer 1

Hint: In order to solve such type of questions firstly we have to apply the equation of straight line passes through the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\left( {Y - {y_1}} \right) = m\left( {X - {x_1}} \right)$ where, $m$ is the slope and formula is $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ and using the condition for two line which are perpendicular to each other is${m_1} \times {m_2} = - 1$.

Complete step-by-step answer:

We have given that,
The equation of straight line passes through the point $\left( {5,0} \right)$ and $\left( {0,3} \right)$
$\left( {Y - {y_1}} \right) = m\left( {X - {x_1}} \right) - - - - - \left( 1 \right)$
Where, $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ .
Here,${x_1} = 5,{y_1} = 0,{x_2} = 0,{y_2} = 3$
So, the slope of line, ${m_1} = \dfrac{{3 - 0}}{{0 - 5}} = \dfrac{{ - 3}}{5}$
From equation (1), we get
$\left( {Y - 0} \right) = \dfrac{{ - 3}}{5}\left( {X - 5} \right)$
$\Rightarrow$ $5Y = - 3X + 15$
$\Rightarrow$ $3X + 5Y = 15 - - - - \left( 2 \right)$
We know that the condition for two lines which are perpendicular to each other is ${m_1} \times {m_2} = - 1$ .
$\dfrac{{ - 3}}{5} \times {m_2} = - 1$ .
$\Rightarrow$ ${m_2} = \dfrac{5}{3}$
So, the equation of perpendicular line passes through $\left( {4,4} \right)$ is,
$\left( {Y - 4} \right) = \dfrac{5}{3}\left( {X - 4} \right)$
$\Rightarrow$ $3Y - 12 = 5X - 20$
$\Rightarrow$ $5X - 3Y = 8 - - - - - - \left( 3 \right)$
Multiplying equation (3) by $3$ , we get
$9X + 15Y = 45 - - - - \left( 4 \right)$
And equation (4) by $5$ ,we get
$25X - 15Y = 40 - - - - - - \left( 5 \right)$
Adding equation (4) and (5), we get
$9X + 15Y + 25X - 15Y = 45 + 40$
$\Rightarrow$ $34X = 85$
$\Rightarrow$ $X = \dfrac{{85}}{{34}}$
$\Rightarrow$ $X = \dfrac{5}{2}$
And substituting the value of (4), we get
$9\left( {\dfrac{5}{2}} \right) + 15Y = 45$
$\Rightarrow$ $\dfrac{{45}}{2} + 15Y = 45$
$\Rightarrow$ $45 + 30Y = 45 \times 2$
$\Rightarrow$ $30Y = 90 - 45$
$\Rightarrow$ $Y = \dfrac{{45}}{{30}}$
$\Rightarrow$ $Y = \dfrac{3}{2}$
So,$\left( {\dfrac{5}{2},\dfrac{3}{2}} \right)$ is the intersection point of both perpendicular lines.
We know that distance between two points is
$\Rightarrow$ $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
$\Rightarrow$ $d = \sqrt {{{\left( {\dfrac{5}{2} - 4} \right)}^2} + {{\left( {\dfrac{3}{2} - 4} \right)}^2}}$
$\Rightarrow$ $d = \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2} + {{\left( {\dfrac{5}{2}} \right)}^2}}$
$\Rightarrow$ $d = \sqrt {\dfrac{9}{4} + \dfrac{{25}}{4}}$
$\Rightarrow$ $d = \sqrt {\dfrac{{34}}{4}}$
$\Rightarrow$ $d = \dfrac{{17}}{{\sqrt {34} }}$
Answer (d) is correct.

Note: Whenever we face these types of questions the key concept is that after solving the equations we will get the intersection points of the both perpendicular lines and after that we have to find out the distance between two points $d$.