Definitions [2]
The centroid of a triangle is the point of intersection of its medians
The slope m of a line is m = tanθ
where θ is the inclination of the line with the positive x-axis.
Formulae [5]
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.
\[G\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)\]
\[m=\frac{y_2-y_1}{x_2-x_1}\]
Theorems and Laws [3]
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"(bar"r")` 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.
Key Points
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 ⇔
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
Important Questions [15]
- Show that points A(–1, –1), B(0, 1), C(1, 3) are collinear.
- Determine whether the points are collinear. A(1, −3), B(2, −5), C(−4, 7)
- Determine whether the points are collinear. P(–2, 3), Q(1, 2), R(4, 1)
- From the Given Number Line, Find D(A, B):
- Show that the points (2, 0), (–2, 0), and (0, 2) are the vertices of a triangle. Also, a state with the reason for the type of triangle.
- Find the distance between the points O(0, 0) and P(3, 4).
- ∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that PQLTPQLT=34.
- Find the Co-ordinates of the Centroid of the δ Pqr, Whose Vertices Are P(3, –5), Q(4, 3) and R(11, –4)
- Draw Seg Ab of Length 9.7 Cm. Take a Point P on It Such that A-p-b, Ap = 3.5 Cm. Construct a Line Mn Sag Ab Through Point P.
- Write Down the Equation of a Line Whose Slope is 3/2 and Which Passes Through Point P, Where P Divides the Line Segment AB Joining A(-2, 6) and B(3, -4) in the Ratio 2 : 3.
- ΔRST ~ ΔUAY, In ΔRST, RS = 6 cm, ∠S = 50°, ST = 7.5 cm. The corresponding sides of ΔRST and ΔUAY are in the ratio 5 : 4. Construct ΔUAY.
- Construct the Circumcircle and Incircle of an Equilateral Triangle Abc with Side 6 Cm and Centre O. Find the Ratio of Radii of Circumcircle and Incircle.
- Δ Amt ∼ δAhe. in δ Amt, Ma = 6.3 Cm, ∠Mat = 120°, at = 4.9 Cm, M a H a = 7 5 . Construct δ Ahe.
- Construct the Circumcircle and Incircle of an Equilateral ∆Xyz with Side 6.5 Cm and Centre O. Find the Ratio of the Radii of Incircle and Circumcircle.
- Write the Equation of the Line Passing Through A(–3, 4) and B(4, 5) in the Form of Ax + by + C = 0
