Question

Find the foot of perpendicular from the point $\left( { - 3, - 4} \right)$ on the line $4(x + 2) = 3(y - 4)$.
A) $\left( { - \dfrac{{31}}{5}, - \dfrac{8}{5}} \right)$
B) $\left( {\dfrac{{31}}{5},\dfrac{8}{5}} \right)$
C) $\left( { - \dfrac{{27}}{5}, - \dfrac{{13}}{5}} \right)$
D) \[\left( {\dfrac{{27}}{5},\dfrac{8}{5}} \right)\]

Answer 1

Hint: For any two lines being perpendicular to each other with slope $m_1$ and $m_2$ respectively, then the product of their slope is equal to -1. \[\left( {m_1m_2 = - 1} \right)\]

Complete step-by-step answer:
According to question,
The equation of the given line is $4(x + 2) = 3(y - 4)$
$4x + 8 = 3y - 12$
$4x - 3y = - 20$

Let us consider the foot of perpendicular from the given point M$\left( { - 3, - 4} \right)$ on the line is N$\left( {a,b} \right)$.
Slope of given line is $\dfrac{4}{3}$
$\therefore $ Slope of line MN is = $ - \dfrac{3}{4}$ (\[m_1m_2 = - 1\] for perpendicular lines)---(1)
Now, calculating slope using point form method,
Slope of line MN = $\dfrac{{y_2 - y_1}}{{x_2 - x_1}}$
Slope of line MN = $\dfrac{{b + 4}}{{a + 3}}$---(2)
Now using equation (1) and (2)
$\dfrac{{b + 4}}{{a + 3}} = - \dfrac{3}{4}$
$4b + 16 = - 3a - 9$
$3a + 4b = - 25$---(3)
Also, the point N$\left( {a,b} \right)$ lies on the given line $4x - 3y = - 20$
$\therefore $$4a - 3b = - 20$---(4)
Using (3) and (4) we obtain,
$a = - \dfrac{{31}}{5}, b = - \dfrac{8}{5}$

Note: Whenever such questions appear, then try to draw the diagram and write the slope of line using points in the question and equate it to the slope of line calculated using the details in the question. For perpendicular lines, use the product of slopes of lines and for parallel, equate it to the slope directly.