Revision: Coordinate Geometry >> Coordinate Geometry Maths English Medium Class 10 CBSE

The two mutually perpendicular number lines intersecting each other at their zeroes are called rectangular axes or coordinate axes, or axes of reference. 

The position of a point in a plane is expressed by a pair of numbers, one concerning the x-axis and the other concerning the y-axis. called co-ordinates. 

• x → distance from y-axis (abscissa)

• y → distance from x-axis (ordinate)

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.

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.

Sign Convention

• Right of y-axis → +x

• Left of y-axis → −x

• Above x-axis → +y

• Below x-axis → −y

Standard Line Results

• x = 0 → y-axis

• y = 0 → x-axis

• x = a → line parallel to the y-axis

• y = b → line parallel to the x-axis

Quadrant Reminder