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Question: Find the foot of perpendicular from the point \(\left( { - 3, - 4} \right)\) on the line \(4(x + 2) ...

Find the foot of perpendicular from the point (−3,−4)\left( { - 3, - 4} \right) on the line 4(x+2)=3(y−4)4(x + 2) = 3(y - 4).
A) (−315,−85)\left( { - \dfrac{{31}}{5}, - \dfrac{8}{5}} \right)
B) (315,85)\left( {\dfrac{{31}}{5},\dfrac{8}{5}} \right)
C) (−275,−135)\left( { - \dfrac{{27}}{5}, - \dfrac{{13}}{5}} \right)
D) (275,85)\left( {\dfrac{{27}}{5},\dfrac{8}{5}} \right)

Explanation

Solution

Hint: For any two lines being perpendicular to each other with slope m1m_1 and m2m_2 respectively, then the product of their slope is equal to -1. (m1m2=−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)4(x + 2) = 3(y - 4)
4x+8=3y−124x + 8 = 3y - 12
4x−3y=−204x - 3y = - 20

Let us consider the foot of perpendicular from the given point M(−3,−4)\left( { - 3, - 4} \right) on the line is N(a,b)\left( {a,b} \right).
Slope of given line is 43\dfrac{4}{3}
∴\therefore Slope of line MN is = −34 - \dfrac{3}{4} (m1m2=−1m_1m_2 = - 1 for perpendicular lines)---(1)
Now, calculating slope using point form method,
Slope of line MN = y2−y1x2−x1\dfrac{{y_2 - y_1}}{{x_2 - x_1}}
Slope of line MN = b+4a+3\dfrac{{b + 4}}{{a + 3}}---(2)
Now using equation (1) and (2)
b+4a+3=−34\dfrac{{b + 4}}{{a + 3}} = - \dfrac{3}{4}
4b+16=−3a−94b + 16 = - 3a - 9
3a+4b=−253a + 4b = - 25---(3)
Also, the point N(a,b)\left( {a,b} \right) lies on the given line 4x−3y=−204x - 3y = - 20
∴\therefore 4a−3b=−204a - 3b = - 20---(4)
Using (3) and (4) we obtain,
a=−315,b=−85a = - \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.