Question

A(1,1) B(7,3) C(3,6) are the vertices of the triangle ABC.D is midpoint of BC and AL perpendicular to BC . Find the slope of AD and AL

Answer 1

Answer: The answer is $\dfrac{7}{8},~~\dfrac{4}{3}.$

Step-by-step explanation: As given in the question and shown in the attached figure, the vertices of ΔABC are A(1,1), B(7,3) and C(3,6). AD is drawn where 'D' is the mid-point of BC and AL is drawn perpendicular to BC. We need to find the slope of AD and AL.

Since D is the mid-point of BC, so the coordinates of D are

$\left(\dfrac{7+3}{2},\dfrac{3+6}{2}\right)=(5,\dfrac{9}{2}).$

Therefore, slope of AD is given by

$m_1=\dfrac{\frac{9}{2}-1}{5-1}=\dfrac{7}{2\times 4}=\dfrac{7}{8}.$

Now, the slope of BC is

$m_2=\dfrac{6-3}{3-7}=-\dfrac{3}{4}.$

Also, since AD ⊥ BC, so slope of AD will be

$m_2^\prime=-\dfrac{1}{m_2}=\dfrac{4}{3}.$

Thus, the slopes of AD and AL are

$\dfrac{7}{8}~~\textup{and}~~\dfrac{4}{3}~\textup{respectively}.$

Answer 2

Answer:

Step-by-step explanation: