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Question
The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is ______.
Options
`130/(17sqrt(29))`
`13/(7sqrt(29))`
`130/7`
None of these
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Solution
The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is `130/(17sqrt(29))`.
Explanation:
Given equations are: 2x – 3y + 5 = 0 .....(i)
3x + 4y = 0 ......(ii)
From equation (ii) we get,
4y = – 3x
⇒ y = `(-3)/4 x` .....(iii)
Putting the value of y in eq. (i) we have
`2x - 3((-3)/4 x) + 5` = 0
⇒ 8x + 9x + 20 = 0
⇒ 17x + 20 = 0
⇒ x = `(-20)/17`
Putting the value of x in equation (iii) we get
y = ` (-3)/4((-20)/17)`
⇒ y = `15/17`
∴ Point of intersection is `(- 20/17, 15/17)`.
Now perpendicular distance from the point `(- 20/17, 15/17)` to the given line 5x – 2y = 0 is
`|(5(- 20/17) - 2(15/17))/sqrt(25 + 4)| = |((-100)17 - 30/17)/sqrt(29)|`
= `130/(17sqrt(29))`
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