Question

Find the equation of the straight line which passes through the midpoint of the line segment joining $ \left( {4,2} \right) $ and $ \left( {3,1} \right) $ whose angle of inclination is $ {30^ \circ } $ .

Answer 1

Hint: In order to find the equation of the line passing through the midpoint of another line, initiate with calculating the midpoint of the line using Section Formula, the move to calculate the slope using the angle of inclination, then from the one-point slope form that is $ y - {y_1} = m\left( {x - {x_1}} \right) $ , substitute all the obtained values, solve and get the equation.
Formula used:
One-point slope form: $ y - {y_1} = m\left( {x - {x_1}} \right) $
Section Formula: $ \left( {{x_3},{y_3}} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right) $

Complete step by step solution:
We are given two points joining to form a straight line, which is $ \left( {4,2} \right) $ and $ \left( {3,1} \right) $ .
Comparing the point $ \left( {4,2} \right) $ by $ \left( {{x_1},{y_1}} \right) $ , we get:
 $
  {x_1} = 4 \\
  {y_1} = 2 \\
  $
Similarly, Comparing the point $ \left( {3,1} \right) $ by $ \left( {{x_2},{y_2}} \right) $ , we get:
 $
  {x_2} = 3 \\
  {y_2} = 1 \;
  $
Since, another line is passing through the mid-point of this line made by joining the points, we need to calculate the midpoint first.
Supposing the midpoint to be $ \left( {{x_3},{y_3}} \right) $ .
From midpoint formula we know that:
 $ \left( {{x_3},{y_3}} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right) $
Substituting the values of $ \left( {{x_1},{y_1}} \right) $ and $ \left( {{x_1},{y_1}} \right) $ , in the above equation, we get:
 $
  \left( {{x_3},{y_3}} \right) = \left( {\dfrac{{4 + 3}}{2},\dfrac{{2 + 1}}{2}} \right) \\
   \Rightarrow \left( {{x_3},{y_3}} \right) = \left( {\dfrac{7}{2},\dfrac{3}{2}} \right) \\
  $
Therefore, the midpoint of $ \left( {4,2} \right) $ and $ \left( {3,1} \right) $ is $ \left( {\dfrac{7}{2},\dfrac{3}{2}} \right) $ .
Since, we know that slope of a line is nothing but the tangent function of the angle of inclination, that is numerically written as:
 $ m = \tan \left( {{{30}^ \circ }} \right) $
From trigonometric formulas, we know that $ \tan \left( {{{30}^ \circ }} \right) = \dfrac{1}{{\sqrt 3 }} $ .Substituting this in the upper value, we get:
 $
  m = \tan \left( {{{30}^ \circ }} \right) = \dfrac{1}{{\sqrt 3 }} \\
   \Rightarrow m = \dfrac{1}{{\sqrt 3 }} \;
  $
Now, for the line that is crossing from the midpoint, we have the following details:
 $ slope\left( m \right) = \dfrac{1}{{\sqrt 3 }} $
One point on the line: $ \left( {{x_3},{y_3}} \right) = \left( {\dfrac{7}{2},\dfrac{3}{2}} \right) $
Since, there is one point and slope given, so from the one-point slope formula we know that the equation of line is:
 $ \left( {y - {y_3}} \right) = m\left( {x - {x_3}} \right) $
Substituting the values of $ \left( {{x_3},{y_3}} \right) $ and $ m $ , on the above equation, we get:
 $ \left( {y - \dfrac{3}{2}} \right) = \dfrac{1}{{\sqrt 3 }}\left( {x - \dfrac{7}{2}} \right) $
Multiplying both the sides by $ \sqrt 3 $ , we get:
 $
  \sqrt 3 \left( {y - \dfrac{3}{2}} \right) = \dfrac{1}{{\sqrt 3 }} \times \sqrt 3 \left( {x - \dfrac{7}{2}} \right) \\
   \Rightarrow \sqrt 3 \left( {y - \dfrac{3}{2}} \right) = \left( {x - \dfrac{7}{2}} \right) \;
  $
Multiplying and dividing $ x,y $ inside the brackets by $ 2 $ , in order to have a common denominator:
 $ \sqrt 3 \left( {\dfrac{2}{2}y - \dfrac{3}{2}} \right) = \left( {\dfrac{2}{2}x - \dfrac{7}{2}} \right) $
Solving the parenthesis, we get:
 $ \Rightarrow \sqrt 3 \left( {\dfrac{{2y - 3}}{2}} \right) = \left( {\dfrac{{2x - 7}}{2}} \right) $
Multiplying both the sides by $ 2 $ :
 $
   \Rightarrow \sqrt 3 \left( {\dfrac{{2y - 3}}{2}} \right) \times 2 = \left( {\dfrac{{2x - 7}}{2}} \right) \times 2 \\
   \Rightarrow \sqrt 3 \left( {2y - 3} \right) = \left( {2x - 7} \right) \;
  $
Opening the brackets, we get:
 $ \Rightarrow 2\sqrt 3 y - 3\sqrt 3 = 2x - 7 $
Adding both the sides by $ 3\sqrt 3 $ and subtracting by $ 2\sqrt 3 y $ , we get:
 $ \Rightarrow 2\sqrt 3 y - 3\sqrt 3 - 2\sqrt 3 y + 3\sqrt 3 = 2x - 7 - 2\sqrt 3 y + 3\sqrt 3 $
On solving it can be written as:
 $
   \Rightarrow 0 = 2x - 7 - 2\sqrt 3 y + 3\sqrt 3 \\
   \Rightarrow 2x - 2\sqrt 3 y + 3\sqrt 3 - 7 = 0 \;
  $
Which is the required equation.
Therefore, the equation of the straight line which passes through the midpoint of the line segment joining $ \left( {4,2} \right) $ and $ \left( {3,1} \right) $ whose angle of inclination is $ {30^ \circ } $ is: $ 2x - 2\sqrt 3 y + 3\sqrt 3 - 7 = 0 $ .
So, the correct answer is “ $ 2x - 2\sqrt 3 y + 3\sqrt 3 - 7 = 0 $ ”.

Note: If instead of the angle of inclination, we have two points for the new line, then we would have used the two-point slope form in which we would have calculated the slope using the formula \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\] and then substitute it in the formula $ y - {y_1} = m\left( {x - {x_1}} \right) $ by taking any one point.