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Mathematics

Find the slope of the line which is perpendicular to xy2+3=0x - \dfrac{y}{2} + 3 = 0.

Straight Line Eq

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Answer

Given,

xy2+3=0y2=x+3y=2x+6.\Rightarrow x - \dfrac{y}{2} + 3 = 0 \\[1em] \Rightarrow \dfrac{y}{2} = x + 3 \\[1em] \Rightarrow y = 2x + 6.

Comparing above equation with y = mx + c we get,

Slope = 2.

Let slope of perpendicular line be m.

Since, product of slopes of perpendicular lines = -1.

∴ m × 2 = -1

⇒ m = 12-\dfrac{1}{2}.

Hence, slope of the line perpendicular to the line xy2+3=0 is 12.x - \dfrac{y}{2} + 3 = 0 \text{ is } -\dfrac{1}{2}.

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