NDA Math Coordinate System and Straight Line

Concept: The equation of a line passing through the point $(x_{1}, y_{1})$ and having the slope ' $m$ ' is given as: $y-y_{1}=m \cdot (x-x_{1})$ Calculations: Given: The point of intersection of diagonals of a square $\text{ABCD}$ is at the originand one of its vertices is at $A(4,2)$. So, the diagonal AC passes through the origin As we know that, the slope of the line joining the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is: $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ The slope of line $\text{A C}$ is givenby $\frac{2-0}{4-0}=\frac{1}{2}$ In square $\text{A B C D}$, the diagonals $\text{A C}$ and $\text{B D}$ are perpendicular to each other. $\Rightarrow$ Slope of $\text{AC} \times$ slope of $\text{BD}=-1$ So, the slope of $\text{B D}$ is $-2$. As we know that, the equation of a line passing through thepoint $(x_{1}, y_{1})$ and having the slope ' $m$ ' is given as: $y - y_{1} = m \cdot (x - x_{1})$ The equation of BD whose slope is - 2 and passes through origin is given by: $y-0=(-2) \cdot (x-0)$ $\Rightarrow 2x+y=0$ Hence, the correctoption is 1 .