1.

Line L is perpendicular to the line 3y+2x = 5 and passes through (-11,7), find the equation of L.

Answer» Let slope of line L be m.

Since line L is perpendicular to 3y+2x=5,therefore,

mx(-2/3)= -1

m=3/2

The equation of line L passing through (-11,7) will be

y-7=3/2[x-(-11)]          

2y-3x=47

Slope of line 3y + 2x = 5 is m1\(\frac{-2}3\) 

∴ Slope of line perpendicular to it \(m = \frac{-1}{m_1} = \frac32\)

It passes through (-11, 7)

∴ \(y - 7 = \frac32 (x - (-11))\)

⇒ \(y - 7 = \frac32 (x +11)\)

⇒ \(2y - 14 = 3x + 33\)

⇒ \(3x - 2y + 47 = 0\)

∴ Equation of line L is 3x - 2y + 47 = 0.



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