Find coordinates of the foot of perpendicular and equation of perpendicular drawn from the point $(2, 3)$ to the line $y = 3 x - 4$ .

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Solution

$T h e g i v e n e q u a t i o n o f t h e l i n e i s y = 3 x - 4 ⟶ (1) S l o p e o f t h e g i v e n l i n e i s m_{1} = 3 s u p p o s e t h e e q u a t i o n o f t h e l i n e p e r p e n d i c u l a r t o t h e g i v e n l i n e i s y = m_{2} x + c ⟶ (2) S i n c e b o t h t h e l i n e s a r e p e r p e n d i c u l a r t o e a c h o t h e r s o w e h a v e; m_{1} m_{2} = - 1 \Rightarrow 3 m_{2} = - 1 \Rightarrow m_{2} = - \frac{1}{3} F r o m (2) w e h a v e, y = - \frac{1}{3} x + c S i n c e t h e l i n e p a s s e s t h r o u g h t h e p o i n t (2, 3), w e h a v e 3 = - \frac{2}{3} + c \Rightarrow c = \frac{11}{3} ∴ e q u a t i o n o f t h e p e r p e n d i c u l a r t o t h e g i v e n l i n e i s y = - \frac{1}{3} x + \frac{11}{3} \Rightarrow 3 y = - x + 11 ⟶ (3) S o l v i n g e q u a t i o n (1) a n d (3) w e h a v e x = \frac{23}{11}, y = \frac{25}{11} H e n c e, t h e f o o t a n d e q u a t i o n o f t h e p e r p e n d i c u l a r i s (\frac{23}{11}, \frac{25}{11}) a n d 3 y = - x + 11 r e s p e c t i v e l y .$

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\vec{r} = (2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + \hat{j} + 3 \hat{k}) .

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