Formulae [3]
\[P\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)\]
The distance between P(x1, y1) and Q(x2, y2) is
\[\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]
The distance of a point P(x, y) from the origin is
\[\sqrt{x^2+y^2}\]
\[M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\]
The point of concurrence (centroid) divides the median in the ratio 2:1.
Theorems and Laws [5]
Let `A(bara)` and `B(barb)` be any two points in the space and `R(barr)` be the third point on the line AB dividing the segment AB externally in the ratio m : n, then prove that `barr = (mbarb - nbara)/(m - n)`.
As the point R divides the line segment AB externally, we have either A-B-R or R-A-B.
Assume that A-B-R and `bar(AR) : bar(BR)` = m : n
∴ `(AR)/(BR) = m/n` so n(AR) = m(BR)
As `n(bar(AR))` and `m(bar(BR))` have same magnitude and direction,
∴ `n(bar(AR)) = m(bar(BR))`
∴ `n(barr - bara) = m(barr - barb)`
∴ `nbarr - nbara = mbarr - mbarb`
∴ `mbarr - nbarr = mbarb - nbara`
∴ `(m - n)barr = mbarb - nbara`
∴ `barr = (mbarb - nbara)/(m - n)`
Hence proved.
Let `A(bara)` and `B(barb)` are any two points in the space and `R(barr)` be a point on the line segment AB dividing it internally in the ratio m : n, then prove that `barr = (mbarb + nbara)/(m + n)`.
R is a point on the line segment AB(A – R – B) and `bar(AR)` and `bar(RB)` are in the same direction.
Point R divides AB internally in the ratio m : n
∴ `(AR)/(RB) = m/n`
∴ n(AR) = m(RB)
As `n(bar(AR))` and `m(bar(RB))` have same direction and magnitude,
`n(bar(AR)) = m(bar(RB))`
∴ `n(bar(OR) - bar(OA)) = m(bar(OB) - bar(OR))`
∴ `n(vecr - veca) = m(vecb - vecr)`
∴ `nvecr - nveca = mvecb - mvecr`
∴ `mvecr + nvecr = mvecb + nveca`
∴ `(m + n)vecr = mvecb + nveca`
∴ `vecr = (mvecb + nveca)/(m + n)`
By vector method prove that the medians of a triangle are concurrent.
Let A, B and C be vertices of a triangle.
Let D, E and F be the mid-points of the sides BC, AC and AB respectively.
Let `bara, barb, barc, bard, bare` and `barf` be position vectors of points A, B, C, D, E and F respectively.
Therefore, by mid-point formula,
∴ `bard = (barb + barc)/2, bare = (bara + barc)/2` and `barf = (bara + barb)/2`
∴ `2bard = barb + barc, 2bare = bara + barc` and `2barf = bara + barb`
∴ `2bard + bara = bara + barb + barc`, similarly `2bare + barb = 2barf + barc = bara + barb + barc`
∴ `(2bard + bara)/3 = (2bare + barb)/3 = (2barf + barc)/3 = (bara + barb + barc)/3 = barg` ...(Say)
Then we have `barg = (bara + barb + barc)/3 = ((2)bard + (1)bara)/(2 + 1) = ((2)bare + (1)barb)/(2 + 1) = ((2)barf + (1)barc)/(2 + 1)`
If G is the point whose position vector is `barg`, then from the above equation it is clear that the point G lies on the medians AD, BE, CF and it divides each of the medians AD, BE, CF internally in the ratio 2 : 1.
Therefore, three medians are concurrent.
If D, E, F are the midpoints of the sides BC, CA, AB of a triangle ABC, prove that `bar(AD) + bar(BE) + bar(CF) = bar0`.
Let `bara, barb, barc, bard, bare, barf` be the position vectors of the points A, B, C, D, E, F respectively.
Since D, E, F are the midpoints of BC, CA, AB respectively, by the midpoint formula
`bard = (barb + barc)/2, bare = (barc + bara)/2, barf = (bara + barb)/2`
∴ `bar(AD) + bar(BE) + bar(CF) = (bard - bara) + (bare - barb) + (barf - barc)`
= `((barb + barc)/2 - bara) + ((barc + bara)/2 - barb) + ((bara + barb)/2 - barc)`
= `1/2barb + 1/2barc - bara + 1/2barc + 1/2bara - barb + 1/2bara + 1/2barb - barc`
= `1/2(barb + barc - 2bara + bar c + bara - 2barb + bara + barb - 2barc)`
= `(bara + barb + barc) - (bara + barb + barc) = bar0`.
If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.
Let P(x, y), Q(a + b, b – a) and R (a – b, a + b) be the given points. Then
PQ = PR ...(Given)
⇒ `sqrt({x - (a + b)}^2 + {y - (b - a)}^2) = sqrt({x - (a - b)}^2 + {y - (a + b)}^2`
⇒ `{x - (a + b)}^2 + {y - (b - a)}^2 = {x - (a - b)}^2 + {y - (a + b)}^2`
⇒ x2 – 2x(a + b) + (a + b)2 + y2 – 2y(b – a) + (b – a)2 = x2 + (a – b)2 – 2x(a – b) + y2 – 2(a + b) + (a + b)2
⇒ –2x(a + b) – 2y(b – a) = –2x(a – b) – 2y(a + b)
⇒ ax + bx + by – ay = ax – bx + ay + by
⇒ 2bx = 2ay
⇒ bx = ay
