SATHEE: Straight Line Question 127

Straight Line Question 127

Question: The co-ordinates of the foot of perpendicular from the point (2, 3) on the line $ x+y-11=0 $ are [MP PET 1986]

Options:

A) $ (-6,5) $

B) $ (5,6) $

C) $ (-5,6) $

D) $ (6,5) $

Show Answer

Answer:

Correct Answer: B

Solution:

Equation of perpendicular on the line $ x+y-11=0 $ is $ x-y+\lambda =0 $ , but it passes through (2, 3), so $ \lambda =1 $ . Equation of perpendicular is $ x-y+1=0 $ . Now the coordinates of the foot of the perpendicular are the intersection point of the lines, hence point is (5, 6).Aliter : Apply the formula given in the theory part of this book, we get required foot as $ ( \frac{1^{2}\times 2-1\times 1\times 3-1\times (-11)}{1^{2}+1^{2}},\frac{1^{2}\times 3-1\times 1\times 2-1(-11)}{1^{2}+1^{2}} ) $ $ =( \frac{2-3+11}{2},\frac{3-2+11}{2} )=(5,6) $ .

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Straight Line Question 127

Question: The co-ordinates of the foot of perpendicular from the point (2, 3) on the line $ x+y-11=0 $ are [MP PET 1986]

Options:

Answer:

Solution:

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