0 0 votes $\text{ABCD}$ is a rhombus with the diagonals $\text{AC}$ and $\text{BD}$ intersecting at the origin on the $x-y$ plane. The equation of the straight line $\text{AD}$ is $x + y = 1.$ What is the equation of $\text{BC}?$ $x + y = –1$ $x – y = –1$ $x + y = 1$ None of the above Quantitative Aptitude cat2000 quantitative-aptitude geometry + – go_editor 14.2k points 2.4k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
0 0 votes Opposite sides of rhombus are parallel. So their ratios are equal as... ax+by+c=0 Px+qy+r=0 a/p =b/q then lines are parallel a/p =b/q =c/r then lines are coincident Now AD eqn is x+y=1 So it's parallel line BC eqn would be x+y=k Where k ! =1 otherwise AD and BC would be coincident so Option 2 and 3 not possible. And intersecting point of both diagonals is origin so rhombus equally divided by axes Option 1 can be possible khushtak answered May 13, 2016 khushtak 1.1k points comment Share Follow 0 reply Please log in or register to add a comment.