A rectangle ABCD has its side AB parallel to line y = x and vertices A, B and D lie on y = 1, x = 2 and x = - 2, respectively. Locus of vertex ‘C’ is

Question

A rectangle ABCD has its side AB parallel to line y = x and vertices A, B and D lie on y = 1, x = 2 and x = - 2, respectively. Locus of vertex ‘C’ is (A) x=5 (B) x - y = 5 (C) y=5 (D) x + y = 5

Tardigrade Admin · Accepted Answer

Let the equation of side B C be y=x+a. ⇒ A=(1-a, 1), B=(2,2+a) Equation of side A D is y-1=-[(x-(1-a)] ⇒ D ≡(-2,4-a) Let C ≡(h, k) ⇒ h+1-a=2-2 ⇒ h=a-1 and k+1=2+a+4-a ⇒ k=5 Thus, the locus of C is y=5

Q. A rectangle $A BC D$ has its side $A B$ parallel to line $y = x$ and vertices $A$ , $B$ and $D$ lie on $y = 1$ , $x = 2$ and $x = - 2$ , respectively. Locus of vertex ‘ $C$ ’ is

$x = 5$

$x - y = 5$

$y = 5$

$x + y = 5$

Q. A rectangle ABCD has its side AB parallel to line y=x and vertices A, B and D lie on y=1, x=2 and x=−2, respectively. Locus of vertex ‘C’ is

x=5

x−y=5

y=5

x+y=5

Let the equation of side BC be y=x+a. ⇒A=(1−a,1),B=(2,2+a) Equation of side AD is y−1=−[(x−(1−a)] ⇒D≡(−2,4−a) Let C≡(h,k)⇒h+1−a=2−2 ⇒h=a−1 and k+1=2+a+4−a⇒k=5 Thus, the locus of C is y=5

Q. A rectangle $A BC D$ has its side $A B$ parallel to line $y = x$ and vertices $A$ , $B$ and $D$ lie on $y = 1$ , $x = 2$ and $x = - 2$ , respectively. Locus of vertex ‘ $C$ ’ is

$x = 5$

$x - y = 5$

$y = 5$

$x + y = 5$