Calculate a for Triangle ABC & 45° Angle

[ferry2]

Points $$A (-3, -1), B (5, a), C (3, 13)$$ are vertices of a triangle. Find the values ​​of parameter $$a$$ for which the angle between $$AB$$ and the median, passing through $$A$$ is equal to $$45$$ degrees.

 

[tiny-tim]

hi ferry2!

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[ferry2]

Let the median is $$AM$$. I started with calculating the coordinates of the point $$M$$ which is $$M(\frac{5+3}{2}, \frac{a+13}{2})\Rightarrow M(4, \frac{a+13}{2})$$. Then I calculate the dot product of $$\vec{AM} \vec{AB}$$. To be an angle $$\angle (\vec{AM}, \vec{AB})=45$$ degrees should $$\vec{AM} \vec{AB}=|\vec{AM}||\vec{AB}|*\frac{\sqrt{2}}{2}$$. And here comes my difficulty. I get very complicated equation. What wrong?

 

[tiny-tim]

… I get very complicated equation. What wrong?

looks ok …

what equation did you get? ​

(btw, you could have used (8, a + 13) instead of (4, (a + 13)/2) )

 

[ferry2]

Well $$\vec{AB}(8, a+1), \vec{AM}(7, \frac{15+a}{2}) \Rightarrow \vec{AM}\vec{AB}=56+\frac{1}{2}(a+1)(15+a)$$

$$|\vec{AM}||\vec{AB}|=\sqrt{64+(a+1)^2}\sqrt{16+\frac{(15+a)^2}{4}}$$

Then $$\vec{AM}\vec{AB}=|\vec{AM}||\vec{AB}|\frac{\sqrt{2}}{2} \Leftrightarrow 56+\frac{1}{2}(a+1)(15+a)=\sqrt{64+(a+1)^2}\sqrt{16+\frac{(15+a)^2}{4}}\frac{\sqrt{2}}{2}$$

How to solve this?

 

[tiny-tim]

(shouldn't that 16 be 49?)

you could simplify that last equation by squaring it,

also you could make it more symmetrical by substituting b = a + 8

 

[ferry2]

Thanks a lot. I solved by squaring and then I apply Horner's method. The real root's are -1 and -21.

 

[tiny-tim]

ooooh … what's Horner's method ?

 

[ferry2]

http://math.fullerton.edu/mathews/n2003/hornermod.html is the Horner's method.