First slide

Equation of line in Straight lines

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

Moderate

A

$y = \sqrt{2} x + b$
B

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

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

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

Solution

Slope of AD $= \frac{b - 0}{0 - a / 2} = - \frac{2 b}{a}$
Slope of $B E = \frac{b / 2}{a / 2} = \frac{b}{a}$
Since AD and BE are perpendicular
$\frac{- 2 b}{a} \cdot \frac{b}{a} = - 1 \Rightarrow a^{2} = 2 b^{2} (i)$
Slope of AC is – b/a
$∴$ Equation of the line through (a, b) perpendicular to AC is
$\begin{array}{l} y - b = \frac{c}{b} (x - a) \\ \Rightarrow a x - b y = a^{2} - b^{2} \end{array}$
$\Rightarrow \pm \sqrt{2} b x - b y = b^{2}$ from (i)
$\pm \sqrt{2} x - y = b \Rightarrow y = \pm \sqrt{2} x - b$
So an equation of required line is $y = \sqrt{2} x - b$

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