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(A) The given line is $2x + 3y = 6$,which has a slope $m_1 = -2/3$. Since the pair of lines passing through the origin forms an isosceles right-angled

Vedclass · Accepted Answer

$x - 5y = 0, 5x + y = 0$

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(A) The given line is $2x + 3y = 6$,which has a slope $m_1 = -2/3$. Since the pair of lines passing through the origin forms an isosceles right-angled triangle with this line,the lines must make an angle of $45^{\circ}$ with the given line. Let the slope of the required lines be $m$. The angle between the lines is given by $	an 	heta = |(m - m_1) / (1 + m \cdot m_1)|$. Setting $	heta = 45^{\circ}$ and $m_1 = -2/3$: $1 = |(m - (-2/3)) / (1 + m(-2/3))| = |(3m + 2) / (3 - 2m)|$. This gives two cases: Case $1$: $(3m + 2) / (3 - 2m) = 1$ $\Rightarrow 3m + 2 = 3 - 2m$ $\Rightarrow 5m = 1$ $\Rightarrow m = 1/5$. The equation of the line is $y = (1/5)x \Rightarrow x - 5y = 0$. Case $2$: $(3m + 2) / (3 - 2m) = -1$ $\Rightarrow 3m + 2 = -3 + 2m$ $\Rightarrow m = -5$. The equation of the line is $y = -5x \Rightarrow 5x + y = 0$. Thus,the lines are $x - 5y = 0$ and $5x + y = 0$.

If a pair of lines drawn through the origin forms an isosceles right-angled triangle with the line $2x + 3y = 6$,then those lines are

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