Revision: Co-ordinate Geometry Geometry Maths 2 SSC (English Medium) 10th Standard Maharashtra State Board

The centroid of a triangle is the point of intersection of its medians

The slope m of a line is m = tan⁡θ

where θ is the inclination of the line with the positive x-axis.

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

\[\sqrt{x^2+y^2}\]

\[P\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)\]

\[M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\]

The point of concurrence (centroid) divides the median in the ratio 2:1.

\[G\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)\]

\[m=\frac{y_2-y_1}{x_2-x_1}\]

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

Let A(a, a²), B(b, b²) and C(0, 0) be the coordinates of the given points.

We know that the area of triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is ∣∣`1/2`[x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)]∣∣ square units.

So,

Area of ∆ABC

`= |1/2|a(b^2 - 0) + b(0 - a^2) + 0(a^2 -  b^2)||`

`=|1/2(ab^2 - a^2b)|`

`= 1/2|ab(b -a)|`

`!= 0     (∵ a!= b != 0)`

Since the area of the triangle formed by the points (a, a²), (b, b²) and (0, 0) is not zero, so the given points are not collinear.

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.

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)`

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.

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`. 

Nature of Slope

• m > 0 → rising line

• m < 0 → falling line

• m = 0 → horizontal line

• m = ∞→ vertical line

Parallel Lines 

Two lines are parallel ⇔ , their slopes are equal, m₁ = m₂

​Perpendicular Lines

Two lines are perpendicular ⇔ m₁ × m₂ = −1

​Collinearity of Three Points

Points A, B, and C are collinear

Method 1: Distance method

AB + BC = AC

Method 2: Slope method

Slope of AB = Slope of BC