Question 11
The two vertices of ∆ABC are given by A(-3, 0) and B(-8, 5) and its centroid is (-2, 1).What will be the coordinates of the third vertex C?
Solution
The two vertices of triangle are A (-3, 0) and B (-8, 5). Its centroid is (-2, 1).
We know, x(centroid) = (x1+x2+x3)/3 and y(centroid) = (y1+y2+y3)/3
Now, x(centroid) = (−3−8+x3)/3
X(centroid) = -2
-2 = −3–8+x33
-6 = -3 – 8 + x3
5 = x3
Now, y(centroid) = (0+5+y3)/3
Y(centroid) = 1
1 = 0+5+y33
3 = 5 + y3
-2 = y3
The third coordinate is (5, -2).
Question 12
The coordinates of the vertices of ABC are A(14, 4), B(18, 20) and C(2, 8).If E and F are the midpoints of AB and AC respectively. Prove that EF=-1/2(BC).
Solution
Given the coordinates of the vertices of a triangle ABC are
A(7,2), B(9,10) and C(1,4)
And E and F are the midpoint of AB and AC respectively
Hence the coordinates of E = (7+9/2 , 2+10/2 )
= (16/2 , 12/2 ) = (8,6)
And the coordinates of F =( 7+1/2 , 2+4/2 )
=(8/2 , 6/2) = (4,3)
Now, by the distance formula,
The length of BC = 10 unit
Length of EF = 5 unit
Hence EF = 5 = ½ x10= (½)BC
Question 13
Find the ratio in which the segment joining the points (2, -6) and (8,10) is divided by X-axis? Also find the coordinates of this point on Xaxis.
Solution
Let C(x,0) divides the line segment A(1, -3) and B(4,5) in
k:1 ratio
by section formula ,
(x,0) = (4k+1/k+1 , 5k-3/ k+1)
Now 5k-3/k+1 = 0
5k – 3 = 0
K=3/5
And x = 4k + 1/ k+1 = 17/8
therefore the ratio in which C divides A and B is k:1
i.e. 3/5: 1 and the coordinate at C is (17/8 , 0)
Question 14
Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2, -5)
and (6, 3). Find the coordinates of the point of intersection.
Solution
Let the given points be:
A(-2, -5) = (x1, y1)
B(6, 3) = (x2, y2)
The line x – 3y = 0 divides the line segment joining the points A and B in the ratio k :
1.Using section formula,
Point of division P(x, y) = [(kx2 + x1)/(k + 1), (ky2 + y1)/(k + 1)]
x = (6k – 2)/(k + 1) and y = (3k – 5)/(k + 1)
Here, the point of division lies on the line x – 3y = 0.
Thus,
[(6k – 2)/(k + 1)] – 3[(3k – 5)/(k + 1)] = 0
6k – 2 – 3(3k – 5) = 0
6k – 2 – 9k + 15 = 0
-3k + 13 = 0
-3k = -13
k = 13/3
Thus, the ratio in which the line x – 3y = 0 divides the line segment AB is 13 : 3.
Therefore, x = [6(13/3) – 2]/ [(13/3) + 1]
= (78 – 6)/(13 + 3)
= 72/16
= 9/2
And
y = [3(13/3) – 5]/ [(13/3) + 1]
= (39 – 15)/(13 + 3)
= 24/16
= 3/2
Therefore, the coordinates of the point of intersection = (9/2, 3/2)
Question 15
“Two polygons having the same number of sides are similar, if their
corresponding angles are equal and their corresponding sides are
proportional.” In a triangle ABC, where A (0, 4), B (-4, 0) and C (4, 8).Point P (-1,
3) and Q (1, 5) lies on the side AB and AC respectively.
I) P divides AB in the ratio:
a) 1:3 c) 2:3
b) 3:1 d) 3:2
II) Q divides AC in the ratio:
a) 3:1 c) 3:2
b) 1:3 d) 2:3
III) What have you observed?
a) 𝐴𝑃/PB = AQ/QC b)PQ||BC
c) Both (a) and (b)
V) Which theorem have you applied?
a) Thales theorem.
b) Converse of midpoint theorem.
c) Converse of Basic Proportional theorem.
d) None of the above.
Solution
I) A
II) B
III) C
IV) C
SELF ASSESSMENT
Q1. The x- coordinate of a point P is twice it’s y- coordinate. If P is equidistant from Q(2,-5) and
R(-3,6), find the co-ordinates of P.
Q2. Show that △ABC, where A(−2,0), B(2,0),C(0,2) and △DEF where D(−4,0), E(4,0), F(0,4) are
similar triangles.
Q3.(1, -1),(0,4) and (-5,3) are vertices of a triangle. Check whether it is a scalene triangle, isosceles triangle or an equilateral triangle. Also, find the length of its median joining the vertex , (1,-1) the mid-point of the opposite side.
Q4. . Find the ratio is which the line segment joining the points , A(3, -3) and , B(-2, 7) is divided by x-axis. Also find the co-ordinates of the point of division.
Q5.Determine the ratio in which the straight line x-y -2 =0 divides the line segment joining , (3,-1) and (8,9)
