The equation of the line which passes through the point of intersection of the lines $x + 2 y + 2 = 0$ and $2 x + 5 y + 3 = 0$ and which is at the minimum distance from the point $(- 2, - 3)$ is

2 x + y + 7 = 0

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x - 2 y + 6 = 0

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x + 3 y + 1 = 0

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2 x - y + 3 = 0

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Solution

The correct option is D $2 x + y + 7 = 0$
$x + 2 y + 2 = 0$ ----(1)
$2 x + 5 y + 3 = 0$ ----(2)

From 1 & 2
$y = 1, x = - 4$

So, equation of line which passes through $(- 4, 1)$ and which is also at minimum from $(- 2, - 3)$ i.e. also pass through $(- 2, - 3) .$

So, $y - 1 = (\frac{4}{- 2}) (x + 4)$
$y + 2 x = - 7$
$2 x + y + 7 = 0$

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The equation of the line which passes through the point of intersection of the lines x+2y+2=0 and 2x+5y+3=0 and which is at the minimum distance from the point (−2,−3) is

The correct option is D 2x+y+7=0x+2y+2=0 ----(1)2x+5y+3=0 ----(2)From 1 & 2y=1,x=−4So, equation of line which passes through (−4,1) and which is also at minimum from (−2,−3) i.e. also pass through (−2,−3).So, y−1=(4−2)(x+4)y+2x=−72x+y+7=0

The equation of the line which passes through the point of intersection of the lines $x + 2 y + 2 = 0$ and $2 x + 5 y + 3 = 0$ and which is at the minimum distance from the point $(- 2, - 3)$ is