The line on which the lines ax + by = 1 and b | vedclass

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(D) Given lines are: $ax + by = 1$ ... $(i)$ $bx + ay = 1$ ... (ii) Subtracting (ii) from $(i)$ multiplied by $a$ and $(i)$ from (ii) multiplied by $b

Vedclass · Accepted Answer

$x = y$

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(D) Given lines are: $ax + by = 1$ ... $(i)$ $bx + ay = 1$ ... (ii) Subtracting (ii) from $(i)$ multiplied by $a$ and $(i)$ from (ii) multiplied by $b$ is not necessary,let's solve by elimination: Multiply $(i)$ by $a$ and (ii) by $b$: $a^2x + aby = a$ $b^2x + aby = b$ Subtracting the two equations: $(a^2 - b^2)x = a - b$ $x(a - b)(a + b) = a - b$ Since $a 
eq b$,we get $x = \frac{1}{a + b}$. Similarly,multiplying $(i)$ by $b$ and (ii) by $a$: $abx + b^2y = b$ $abx + a^2y = a$ Subtracting: $(b^2 - a^2)y = b - a$ $y = \frac{b - a}{b^2 - a^2} = \frac{1}{a + b}$. Thus,the point of intersection is $(\frac{1}{a + b}, \frac{1}{a + b})$. Since the $x$-coordinate equals the $y$-coordinate,the intersection point always lies on the line $x = y$.

The line on which the lines $ax + by = 1$ and $bx + ay = 1$ (with $a \neq 0 \neq b$) intersect for any real values of $a$ and $b$ is

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