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Question Number 38996 by Rio Mike last updated on 01/Jul/18

a l i n e w i t h e q u a t i o n

y = 2 x + 5 . t h e p o i n t (1, 7)

l i e o n t h i s l i n e . f i n d t h e

d i s t a n c e b e t w e e n t h i s l i n e

a n d t h e l i n e j o i n i n g

(2, 3) a n d (6, 7) .

Commented by math khazana by abdo last updated on 02/Jul/18

l e t A (2, 3) a n d B (6, 7)

M (x, y) \in (A B) \Leftrightarrow d e t (A \vec{M}, A \vec{B}) = 0 \Leftrightarrow

| \begin{matrix} x - 2 4 \\ y - 3 4 \end{matrix} | = 0 \Leftrightarrow 4 (x - 2) - 4 (y - 3) = 0 \Rightarrow

x - 2 - y + 3 = 0 \Rightarrow x - y + 1 = 0

d (D, D^{^{'}}) = d (E, D) = \frac{∣ 1 - 7 + 1 ∣}{\sqrt{1^{2} + {(- 1)}^{2}}} = \frac{5}{\sqrt{2}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

t h e {e q}^{n} o f l i n e j o i n i n g (2, 3) a n d (6, 7) i s

y - 3 = (\frac{7 - 3}{6 - 2}) (x - 2)

y - 3 = x - 2

x - y + 1 = 0

d i s t a n c e =∣ \frac{1 - 7 + 1}{\sqrt{1^{2} + {(- 1)}^{2}}} ∣

= \frac{5}{\sqrt{2}}