Given A(1, 1) and AB is any line through it c | vedclass

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

Question

(A) Let the slope of line $AB$ be $m$. The equation of line $AB$ passing through $A(1, 1)$ is $y - 1 = m(x - 1)$.Since $AB$ cuts the $x-$ axis at $B$,

Vedclass · Accepted Answer

$x + y = 1$

Vedclass · Answer

(A) Let the slope of line $AB$ be $m$. The equation of line $AB$ passing through $A(1, 1)$ is $y - 1 = m(x - 1)$.Since $AB$ cuts the $x-$ axis at $B$,set $y = 0$: $-1 = m(x - 1) \implies x - 1 = -\frac{1}{m} \implies x = 1 - \frac{1}{m}$. So,$B = (1 - \frac{1}{m}, 0)$.Since $AC \perp AB$,the slope of $AC$ is $-\frac{1}{m}$. The equation of line $AC$ is $y - 1 = -\frac{1}{m}(x - 1)$.Since $AC$ meets the $y-$ axis at $C$,set $x = 0$: $y - 1 = -\frac{1}{m}(-1) = \frac{1}{m} \implies y = 1 + \frac{1}{m}$. So,$C = (0, 1 + \frac{1}{m})$.Let $P(h, k)$ be the midpoint of $BC$. Then $h = \frac{(1 - \frac{1}{m}) + 0}{2} = \frac{1}{2} - \frac{1}{2m}$ and $k = \frac{0 + (1 + \frac{1}{m})}{2} = \frac{1}{2} + \frac{1}{2m}$.Adding the two equations: $h + k = (\frac{1}{2} - \frac{1}{2m}) + (\frac{1}{2} + \frac{1}{2m}) = 1$.Replacing $(h, k)$ with $(x, y)$,the locus is $x + y = 1$.

Given $A(1, 1)$ and $AB$ is any line through it cutting the $x-$ axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y-$ axis in $C$,then the equation of the locus of the midpoint $P$ of $BC$ is

Similar Questions

If the sum of distances from a point $P$ to two mutually perpendicular straight lines is $1$ unit,then the locus of $P$ is

Given points $A(6,0)$,$B(0,4)$,and $O$ as the origin,find the locus of a point $P(x, y)$ such that the area of $\triangle POB$ is $2$ times the area of $\triangle POA$.

If the equation of a curve remains unchanged by replacing $x$ with $y$ and $y$ with $x$,then the curve is

$A$ point starts moving from $(1, 2)$ and its projections on $x$ and $y$-axes are moving with velocities of $3 \ m/s$ and $2 \ m/s$ respectively. Its locus is

Consider the triangles with vertices $A(2,1)$,$B(0,0)$,and $C(t,4)$,where $t \in [0,4]$. If the maximum and the minimum perimeters of such triangles are obtained at $t=\alpha$ and $t=\beta$ respectively,then $6\alpha + 21\beta$ is equal to $.........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module