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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If $A$ is $(2, 5)$,$B$ is $(4, -11)$ and $C$ lies on $9x + 7y + 4 = 0$,then the locus of the centroid of the $\Delta ABC$ is a straight line parallel to the straight line:

If the equation of the locus of a point equidistant from $(a_1, b_1)$ and $(a_2, b_2)$ is $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,find the value of $c$.

If one vertex of an equilateral triangle of side $a$ lies at the origin and another vertex lies on the line $x - \sqrt{3}y = 0$,then the coordinates of the third vertex are:

Two points $A(-a, 0)$ and $B(a, 0)$ are given. If $C$ is a variable point lying on one side of the line $AB$ such that $\angle CAB - \angle CBA = \alpha$,where $\alpha$ is a positive constant,then the locus of the point $C$ is

Let $A(2,3), B(3,-6), C(5,-7)$ be three points. If $P$ is a point satisfying the condition $PA^2+PB^2=2PC^2$,then a point that lies on the locus of $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module