Question

An equilateral triangle ABC has coordinates O(0, 0), B(a, 0) and C(c, d).

a Find c and d in terms of a by using the fact that OB = BC = CO.

b Find the equation of the lines OB, BC and CO.

In any triangle PQR prove that PQ2 + PR2 = 2(PS2 + SR2)

Where S is the midpoint of QR.

Prove that the midpoints of a parallelogram bisect each other using coordinate geometry.

Solution

Answer 1

1.

a)  

$OB = BC = CO$

$$OB^2 = BC^2 = CO^2$$

$$a^2=(a-c)^2+(-d)^2=c^2+d^2$$

$$a^2-2ac+c^2=c^2$$

$$c=\dfrac{a}{2}, a\not=0$$

$$a^2=\dfrac{a^2}{4}+d^2$$

$$|d|=\dfrac{\sqrt{3}}{2}|a|$$

Let $a>0, d>0.$ Then

$$c=\dfrac{a}{2}, d=\dfrac{\sqrt{3}}{2}a$$

b)

$$OB: y=0$$

$$BC: y=-\sqrt{3}x+\sqrt{3}a$$

$$CO:y=\sqrt{3}x$$

2.

Let $P(0,0), Q(2a, 2b), R(2c, 2d).$ Then $S(a+c, b+d).$

$$PQ^2=(2a)^2+(2b)^2$$

$$PR^2=(2c)^2+(2d)^2$$

$$PS^2=(a+c)^2+(b+d)^2$$

$$SR^2=(2c-(a+c))^2+(2d-(b+d))^2$$

Substitute

$$2(PS^2+SR^2)=2PS^2+2SR^2$$

$$=2((a+c)^2+(b+d)^2)+2((c-a)^2+(d-b)^2)$$

$$=2(a^2+2ac+c^2+b^2+2bd+d^2)$$

$$+2(c^2-2ac+a^2+d^2-2bd+d^2)$$

$$=4a^2+4b^2+4c^2+4d^2$$

$$=(2a)^2+(2b)^2+(2c)^2+(2d)^2$$

$$=PQ^2+PR^2$$

Therefore

$$PQ^2+PR^2=2(PS^2+SR^2)$$

3. Let $ABCD$ be a parallelogram

$$A(0, 0), B(b, h), C(b+d, h), D(d, 0)$$

$$BC\parallel AD, AD=BC$$

Then

$$AC: y=\dfrac{h}{b+d}x$$

$$BD: y=\dfrac{h}{b-d}x-\dfrac{hd}{b-d}$$

Intersection

$$\dfrac{h}{b+d}x=\dfrac{h}{b-d}x-\dfrac{hd}{b-d}$$

$$x(bh-dh)=x(bh+dh)-bdh-d^2h$$

$$2dhx=bdh+d^2h$$

$$x=\dfrac{b+d}{2}$$

$$y=\dfrac{h(b+d)}{2(b+d)}=\dfrac{h}{2}$$

Point $E(\dfrac{b+d}{2}, \dfrac{hd}{2})$ is the midpoint of the diagonal $AC.$

Point $E(\dfrac{b+d}{2}, \dfrac{hd}{2})$ is the midpoint of the diagonal $BD.$

Therefore the diagonal of a parallelogram bisect each other.