Question

In a $△ABC$, $A\equiv (\alpha ,\beta ),B\equiv (2,3)$ and $C\equiv (1,3)$ and point $A$ lies on line $y=2x+3$ where $\alpha \in I$. Area of $△ABC=\Delta$, is such that $\left[\Delta \right]=5$. Possible coordinates of $A$ are (where $[.]$ represents greatest integer function)

• A. $(2,3)$
• B. $(5,13)$
• C. $(-5,-7)$
• D. $(-3,-5)$

Answer 1

The correct option is B $(5,13)$

Let $A$ be $(a,2a+3)$.

Length of $BC$ is $1$ unit.

Equation of $BC$: $y-3=\frac{3-3}{2-1}\times (x-2)$

$\Rightarrow y=3$ is the perpendicular distance of A from

$BC$ $=\left|\frac{2a+3-3}{1}\right|=|2a|$

Using the formula of perpendicular distance of any point $(x_0,y_0)$ from line $ax+by+c=0$ is $\frac{|a{x}_0+b{y}_0+c|}{\sqrt{a^2+b^2}}$

Area of triangle ABC $=\frac{1}{2}\times$ base $\times$ height $=\frac{1}{2}\times |2a|\times 1=|a|$

Given: $\left[\Delta \right]$ = 5

Thus $5\le |a|<6$

Therefore, co-ordinates are $(5,13)$.