Question

A(1.1), B(7,3) and C(3,6) are the vertices of a $\Delta ABC$. If D is the midpoint of BC and $AL\bot BC$A, find the slopes of [i] AD [ii] AL.

Answer 1

Hint: Find the coordinates of D using the property that the coordinates of the midpoint of $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ are given by $C\equiv \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$. Hence find the slope of AD using the fact that the slope of the line joining $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. Use the fact that if the slopes of two perpendicular lines are ${{m}_{1}}$ and ${{m}_{2}}$ then ${{m}_{1}}{{m}_{2}}=-1$ to find the slope of AL.
Complete step-by-step answer:

Finding the coordinates of D:
We have $B\equiv \left( 7,3 \right)$ and $C\equiv \left( 3,6 \right)$
We know that the coordinates of the midpoint of $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ are given by $C\equiv \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$.
Here ${{x}_{1}}=7,{{x}_{2}}=3,{{y}_{1}}=3$ and ${{y}_{2}}=6$
Hence the coordinates of D are given by $\left( \dfrac{7+3}{2},\dfrac{6+3}{2} \right)=\left( 5,\dfrac{9}{2} \right)$
Finding the slope of AD:
We have $A\equiv \left( 1,1 \right)$ and $D\equiv \left( 5,\dfrac{9}{2} \right)$
We know that the slope of the line joining $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Here ${{x}_{1}}=1,{{x}_{2}}=5,{{y}_{1}}=1$ and ${{y}_{2}}=\dfrac{9}{2}$
Hence the slope of the line is given by $m=\dfrac{\dfrac{9}{2}-1}{5-1}=\dfrac{7}{2\times 4}=\dfrac{7}{8}$
Hence the slope of AD is $\dfrac{7}{8}$.
Finding the slope of BC:
We have $B\equiv \left( 7,3 \right)$ and $C\equiv \left( 3,6 \right)$
We know that the slope of the line joining $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Here ${{x}_{1}}=7,{{x}_{2}}=3,{{y}_{1}}=3$ and ${{y}_{2}}=6$
Hence the slope of the line is given by $m=\dfrac{6-3}{3-7}=-\dfrac{3}{4}$
Hence the slope of BC is $-\dfrac{3}{4}$.
Finding the slope of AL:
Let the slope of AL be m.
We know that if the slopes of two perpendicular lines are ${{m}_{1}}$ and ${{m}_{2}}$ then ${{m}_{1}}{{m}_{2}}=-1$ to find the slope of AL.
Since AL is perpendicular to BC, we have
$m\times \left( -\dfrac{3}{4} \right)=-1$
Multiplying both sides by $-\dfrac{4}{3}$, we get
$m=\dfrac{4}{3}$
Hence the slope of AL is $\dfrac{4}{3}$.
Note: Alternatively, we can find the slope of the line BC and AD by assuming that their equations are y = mx+c.
Now the points through which these lines pass must satisfy the equation of these lines. Hence form a linear equation system in two variables. Solve for m. The value of m gives the slope of the line.