Question

Consider a line pair $2x^2+3xy-2y^2-10x+15y-28=0$ and another line $L$ passing through origin with gradient 3. The line pair and line $L$ form a triangle whose vertices are $A,B$ and $C$.
Sum of the contangents of the interior angles of the triangle $ABC$ is

• A. $\frac{50}{7}$
• B. 7
• C. $5\sqrt{\frac{2}{7}}$
• D. Not defined

Answer 1

Answer: (A)

The two lines are
$(2x-y+4)(x+2y-7)=0$
$\text{ Line Lis }$
$y=3x$
$\angle B=90^{\circ }$
$\therefore \cot B=0$
$\cot A=\frac{21\times \sqrt{5}}{\sqrt{5}\times 3}=7$

$\cot C=\frac{3\times \sqrt{5}}{\sqrt{5}\times 21}=\frac{1}{7}$
$\therefore \text{ sum of cotangents }=7+\frac{1}{7}=\frac{50}{7}$