Thread: Triangle Geometry

1. The points A (-4,-2), B (6,4) and C (4,-4) are vertices of a triangle ABC.
I found the equation of the line passing through A and B which is 3x- 5y+2=0.
The question is Point D is the midpoint of AB. Show that CD is perpendicular to AB.

2. Curve C has equation f(x) and point P (2, 10.25) lies on C.
Given that f'(x)= (x^3/2 - 2x^-3/2)^2 +2
Show that f'(x) can be written as Ax^3 + B/x^3 + D

3. x^2 +2kx-5=0
Prove that, for all real values of k, the roots of the equation are real and different.

Thanks so much in advance if you can help me.

to1.
D(1,1)
slope of AB is 3/5
slope of CD is (-4-1)/(4-1)=-5/3
Because (3/5)*(-5/3)=-1 CD is perpendicular to AB

to 2.:
(x^3/2 - 2x^-3/2)^2 +2=(x^(3/2))^2-4+(2/(x^(3/2))^2)+2=x^3+4/(x^3)-2
therefore A=1, B=4, D=-2

to 3.:
x=-k+-sqrt(k^2+5)
k^2+5>=5 for all k in R.

Dear Confused:

Earboth hit all three problems right on the money, and I could not have achieved the indicated results more expeditiously than did he(she?)--on my best day. Nevertheless, I offer the following food for thought which may solidify your understanding some and, subsequent to a bit of trusty reflection on your end, you will soon thereafter be wielding about an enhanced problem solving arsenal the likes of which promises to make you the talk of the town [people will flock from miles too great to count, just hoping to catch an autograph or a perhaps a quick snapshot intended to impress their friends and family. Returning now for a more candid glimpse of reality, I draw your attention to a key objective here, which seeks to provide you with mathematical prowess equal to the task of commanding success upon inevitability finding yourself alone, face-to-face, just you and that nasty mid-term or final examination, with nowhere to turn but within.

Enough.

a) The distance formula applied to the information provided for triangle ABC indicates quickly that ABC is isosceles. Recall from intro-geometry that a segment extending from vertex to midpoint of opposite side in an isosceles t.angle is a perpendicular bisector of that side.

b) Pythagoras applied to side lengths determined as prescribed above indicates ABC a right isosceles. Hence SSS provides supplementary angles BDC, ADC congruent, further implying BDC, ADC right angles.

c) In #2, letting u=x^(1/2)may facilitate visualization. I.e., f'(x)=[u^3 - 2u^-3]^2 +2 indicates coefficients quickly. Squaring the binomial gives A and B without lifting the pencil, and the product of u^3 with its inverse,u^-3, gives 1 which multiplied by the coefficient, -2, and then doubled gives -4 (assuming proficiency with squaring a binomial...if not, make it so). Finally, summing
-4 with constant 2 gives D=-2.

d)The(k^2+5) to which"Earboth" refers in #3 is just the "discriminant" in the q.formula,i.e., b^2-4ac. Recall that b^2-4ac<0 implies both roots complex, b^2-4ac=0 implies exactly one real root(repeated), and b^2-4ac>0 implies two distinct real roots. If not readily obvious by parabolic nature of graph, the fundamental theorem of algebra dictates the number of roots of a polynomial function as equal to the degree of that function, i.e., 2 in the present case.


e) The graph of f(x)=x^2+2kx-5=0 intersects y-axis at(0,-4) which is below the x-axis. But leading coefficient(x^2 term)is positive (i.e., a=1) implying graph opens upward and, consequently, crosses the x-axis at two distinct points [where f(x)=0, i.e., the determining equation for roots, aka zeros in present case).

I hope this helps.

Regards,

Rich B.

Originally Posted by Confuzzled?

1. The points A (-4,-2), B (6,4) and C (4,-4) are vertices of a triangle ABC.
I found the equation of the line passing through A and B which is 3x- 5y+2=0.
The question is Point D is the midpoint of AB. Show that CD is perpendicular to AB.

2. Curve C has equation f(x) and point P (2, 10.25) lies on C.
Given that f'(x)= (x^3/2 - 2x^-3/2)^2 +2
Show that f'(x) can be written as Ax^3 + B/x^3 + D

3. x^2 +2kx-5=0
Prove that, for all real values of k, the roots of the equation are real and different.

Thanks so much in advance if you can help me.