Question

If the medians $AD$ and $BE$ of the triangle with vertices $A(0,b),B(0,0)$ and $C(a,0)$ are perpendicular, then an equation of the line through $(a,b)$ perpendicular to $AC$ is

• A. $y=\sqrt{2}x+b$
• B. $y=-\sqrt{2}x-b$
• C. $y=\sqrt{2}x-b$
• D. $x=\sqrt{2}y-a$

Answer 1

Answer: (B), (C)

$y=\sqrt{2}x-b$
$y=-\sqrt{2}x-b$

Coordinate of D and E are $(\frac{a}{2},0)$ and $(\frac{a}{2},\frac{b}{2})$ respectively
Slope of $AD$ $\times$ slope of $BE=-1$
$\frac{0-b}{\frac{a}{2}-0}\times \frac{\frac{b}{2}}{\frac{a}{2}}=-1\Rightarrow -\frac{b^2}{2}=-\frac{a^2}{4}\Rightarrow a^2-2b^2=0$......(1)
Slope of $AC$ is $-\frac{b}{a}$
Then slope of required line is $\frac{a}{b}$

Line passes through $(a,b)$

so equation of line is

$y-b=\frac{a}{b}(x-a)$

$by-b^2=ax-a^2$

put $a=\sqrt{2}b$ from 1
Hence equation of required line is
$y=\pm \sqrt{2}x-b$