a line with equation y = 2x + 5.the point (1,7) lie on this line.find the distance between this line and the line joining (2,3) and (6,7).


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
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

	$t h e {e q}^{n} o f l i n e j o i n i n g (2, 3) a n d (6, 7) i s$ $y - 3 = (\frac{7 - 3}{6 - 2}) (x - 2)$ $y - 3 = x - 2$ $x - y + 1 = 0$ $d i s t a n c e =∣ \frac{1 - 7 + 1}{\sqrt{1^{2} + {(- 1)}^{2}}} ∣$ $= \frac{5}{\sqrt{2}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum