Find the distance of the line 4x + 7y + 5 = 0 | vedclass

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Question

(A) The given lines are:$2x - y = 0$ ..... $(1)$$4x + 7y + 5 = 0$ ..... $(2)$Point $A$ is $(1, 2)$.Let $B$ be the point of intersection of lines $(1)$

Vedclass · Accepted Answer

$\frac{23 \sqrt{5}}{18}$

Vedclass · Answer

(A) The given lines are:$2x - y = 0$ ..... $(1)$$4x + 7y + 5 = 0$ ..... $(2)$Point $A$ is $(1, 2)$.Let $B$ be the point of intersection of lines $(1)$ and $(2)$.From $(1)$,$y = 2x$. Substituting this into $(2)$:$4x + 7(2x) + 5 = 0$$4x + 14x + 5 = 0$$18x = -5 \implies x = -\frac{5}{18}$Then $y = 2(-\frac{5}{18}) = -\frac{5}{9}$.So,point $B$ is $(-\frac{5}{18}, -\frac{5}{9})$.The distance $AB$ is given by the distance formula:$AB = \sqrt{(1 - (-\frac{5}{18}))^2 + (2 - (-\frac{5}{9}))^2}$$AB = \sqrt{(\frac{18+5}{18})^2 + (\frac{18+5}{9})^2}$$AB = \sqrt{(\frac{23}{18})^2 + (\frac{23}{9})^2}$$AB = \sqrt{(\frac{23}{18})^2 + (\frac{46}{18})^2}$$AB = \frac{23}{18} \sqrt{1^2 + 2^2} = \frac{23 \sqrt{5}}{18}$ units.

Find the distance of the line $4x + 7y + 5 = 0$ from the point $(1, 2)$ along the line $2x - y = 0$.

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