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Mathematics

Find the equation of the line which is perpendicular to the line xayb=1\dfrac{x}{a} - \dfrac{y}{b} = 1 at the point where the given line meets y-axis.

Straight Line Eq

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Answer

Let A be the point where the line xayb=1\dfrac{x}{a} - \dfrac{y}{b} = 1 meets y-axis.

So, x-co-ordinate of point A will be zero.

Substituting x = 0 in equation we get,

0ayb=1yb=1y=b\Rightarrow \dfrac{0}{a} - \dfrac{y}{b} = 1 \\[1em] \Rightarrow -\dfrac{y}{b} = 1 \\[1em] \Rightarrow y = -b

A = (0, -b).

The given line equation is,

xayb=1yb=xa1y=bxab\Rightarrow \dfrac{x}{a} - \dfrac{y}{b} = 1 \\[1em] \Rightarrow \dfrac{y}{b} = \dfrac{x}{a} - 1 \\[1em] \Rightarrow y= \dfrac{bx}{a} - b

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

Slope (m) = ba\dfrac{b}{a}

Let slope of perpendicular line be m1.

As product of slope of perpendicular lines is -1,

∴ m × m1 = -1

ba\dfrac{b}{a} × m1 = -1

⇒ m1 = ab-\dfrac{a}{b}

Equation of line through A (0, -b) and slope = ab-\dfrac{a}{b} is :

⇒ y - y1 = m(x - x1)

⇒ y - (-b) = ab-\dfrac{a}{b}(x - 0)

⇒ b(y + b) = -ax

⇒ by + b2 = -ax

⇒ ax + by + b2 = 0

Hence, equation of required line is ax + by + b2 = 0.

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