In a triangle $ABC$ with $A=(1,3) ,B =(q,0), C =(p,-4)$

Question

Let $A=(1,3),B =(q,0), C =(p,-4)$, with $p>0$, the slope of $AB$ is $+45^\circ$ and $AC= \sqrt{50}$.

1. Determine the gradient of $AB$
2. Calculate the equation of the line $AB$
3. Calculate the value of $q$
4. Calculate $p$
5. Calculate midpoint of $AC$
6. Determine the equation of the line parallel to $AB$ passing through $C$

Answer 1

1. Gradient of $\overline{AB}$ is $\tan45^\circ = 1$.

2. Equation of $\overleftrightarrow{AB}$ is $y = mx + c$. Putting $m = 1, y = 3, x = 1$, we get $c = 2$. So equation of $\overleftrightarrow{AB}$ is $y = x + 2$

3. To find $q$, $\frac{q-3}{0-1}=1$. We get $q=2$.

4. To find $p$, using distance formula, $\sqrt{(p-1)^2 + (-4-3)^2} = \sqrt{50}$. You get $p=2$.

5. Midpoint of $\overline{AC}$ is $(\frac{2+1}{2}, \frac{3-4}{2}) = (\frac{3}{2},\frac{-1}{2})$.

6. Line with $m=1$ and passing through $(2,-4)$ is $y = x+c$. Put $y=-4$ and $x=2$ to get $c=-6$. So the required line is $y=x-6$.