Question

Find the coordinates of the foot of perpendicular from the point \[\left( { - 1,3} \right)\] to the line \[3x - 4y - 16 = 0\]

Answer 1

Hint: First, convert this equation to slope-intercept form to find the slope of the line. We are given that the two lines are perpendicular to each other and we know that the product of the slope of two perpendicular lines is equal to $ - 1 $ , this way we can find the slope of the perpendicular and find the coordinates of the foot of the perpendicular.

Complete step-by-step answer:
Let the coordinates of the foot of the perpendicular from the point \[\left( { - 1,3} \right)\] to the line \[3x - 4y - 16 = 0\]be $ (a,b) $
Equation of the line is \[3x - 4y - 16 = 0\]
The equation can be rewritten as –
 $
\Rightarrow 4y = 3x - 16 \\
\Rightarrow y = \dfrac{3}{4}x - 4 \;
  $
Comparing it with $ y = mx + c $ , we get the slope of the line, $ {m_1} = \dfrac{3}{4} $
As the line from the point \[\left( { - 1,3} \right)\] is perpendicular to the line\[3x - 4y - 16 = 0\], the angle between these two lines is $ 90^\circ $
Let the slope of the perpendicular be $ {m_2} $ . So,
 $
\Rightarrow {m_1}{m_2} = - 1 \\
\Rightarrow {m_2} = - 1 \times \dfrac{4}{3} = \dfrac{{ - 4}}{3} \;
  $
Also, the slope of the line having coordinates $ ({x_1},{y_1}) $ and $ ({x_2},{y_2}) $ is given by the formula $ m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} $
So the slope of the perpendicular is, $ {m_2} = \dfrac{{b - 3}}{{a - ( - 1)}} = \dfrac{{b - 3}}{{a + 1}} $
We have,
 $
\Rightarrow \dfrac{{ - 4}}{3} = \dfrac{{b - 3}}{{a + 1}} \\
\Rightarrow - 4a - 4 = 3b - 9 \\
\Rightarrow 4a + 3b = 5...(1) \;
  $
As the point $ (a,b) $ lies on the line $ 3x - 4y - 16 = 0 $
So it will satisfy the equation of this line,
 $
\Rightarrow 3a - 4b - 16 = 0 \\
\Rightarrow 3a - 4b = 16...(2) \;
  $
Now, we have two unknown quantities and two equations to find them, using the elimination method –
 $
  4 \times (2) - 3 \times (1) \\
  12a - 16b\,\, = 64 \\
  \underline { - 12a - 9b = - 15} \\
  \underline {\,\,\,\,\,\,\,\,\, - 25b = 49\,\,\,\,} \\
   \Rightarrow b = \dfrac{{ - 49}}{{25}} \;
  $
Put this value of b in (1)
 $
\Rightarrow 4a + 3b = 5 \\
\Rightarrow 4a = 5 - 3(\dfrac{{ - 49}}{{25}}) \\
\Rightarrow 4a = \dfrac{{125 + 147}}{{25}} = \dfrac{{272}}{{25}} \\
 \Rightarrow a = \dfrac{{68}}{{25}} \;
  $
Thus, the coordinates of the foot of the perpendicular are $ (\dfrac{{68}}{{25}},\dfrac{{ - 49}}{{25}}) $
So, the correct answer is “$ (\dfrac{{68}}{{25}},\dfrac{{ - 49}}{{25}}) $ ”.

Note: The slope of a line tells the steepness of a line, it can be defined as a change in y per unit change in x. The slope of a line can be found out by converting it into slope-intercept form, in this form of equations the point at which the line cuts the y-axis is the intercept of the line.