Question

Let $A(1,1),B(7,3)$ and $C(3,6)$ are the vertices of a $\Delta ABC$. If $D$ is the mid-point of $BC$ and $AL$ is the altitude of the triangle along $BC,$ then the product of slopes of $AD$ and $AL$ is

• A. $\frac{6}{7}$
• B. $-\frac{6}{7}$
• C. $\frac{7}{6}$
• D. $-\frac{7}{6}$

Answer 1

The correct option is C $\frac{7}{6}$

The coordinate of $D$
$=(\frac{7+3}{2},\frac{3+6}{2})=(5,\frac{9}{2})$

Slope of $AD=\frac{\frac{9}{2}-1}{5-1}=\frac{7}{8}$
Slope of $BC=\frac{6-3}{3-7}=-\frac{3}{4}$

As $AL\perp BC$, so the slope of $AL$
$=\frac{4}{3}(\because m_1{m}_2=-1)$

Product of slope of $AD$ and $AL$
$=\frac{7}{8}\times \frac{4}{3}=\frac{7}{6}$