If the line segment joining the points (1,0) | vedclass

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Question

(A) Let $P = (h, k)$,$A = (1, 0)$,and $B = (0, 1)$.The slope of $AP$ is $m_1 = \frac{k-0}{h-1} = \frac{k}{h-1}$.The slope of $BP$ is $m_2 = \frac{k-1}

Vedclass · Accepted Answer

$\left(x^2+y^2-1ight)\left(x^2+y^2-2x-2y+1ight)=0, x 
eq 0,1$

Vedclass · Answer

(A) Let $P = (h, k)$,$A = (1, 0)$,and $B = (0, 1)$.The slope of $AP$ is $m_1 = \frac{k-0}{h-1} = \frac{k}{h-1}$.The slope of $BP$ is $m_2 = \frac{k-1}{h-0} = \frac{k-1}{h}$.The angle $	heta$ between $AP$ and $BP$ is $45^{\circ}$.Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$,we have:$	an 45^{\circ} = \left| \frac{\frac{k}{h-1} - \frac{k-1}{h}}{1 + \left(\frac{k}{h-1}ight)\left(\frac{k-1}{h}ight)} ight|$$1 = \left| \frac{kh - (k-1)(h-1)}{h(h-1) + k(k-1)} ight|$$1 = \left| \frac{kh - (kh - k - h + 1)}{h^2 - h + k^2 - k} ight|$$1 = \left| \frac{h + k - 1}{h^2 + k^2 - h - k} ight|$This gives two cases: $h^2 + k^2 - h - k = h + k - 1$ or $h^2 + k^2 - h - k = -(h + k - 1)$.Case $1$: $h^2 + k^2 - 2h - 2k + 1 = 0$.Case $2$: $h^2 + k^2 - 1 = 0$.Combining these,the locus is $(x^2 + y^2 - 1)(x^2 + y^2 - 2x - 2y + 1) = 0$.

If the line segment joining the points $(1,0)$ and $(0,1)$ subtends an angle of $45^{\circ}$ at a variable point $P$,then the equation of the locus of $P$ is

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