Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.5

Question 1.
Find the centroid of the triangle whose vertices are
(i) (2, -4), (-3, -7) and (7, 2)
Solution:
Let the vertices of a triangle be A (2, -4), B (-3, -7) and C (7, 2) Centroid
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 1
Centroid is (2, -3)

(ii) (-5, -5), (1, -4) and (-4, -2)
Solution:
Let the vertices of a triangle be A (-5, -5), B (1, -4) and C (-4, -2)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 2

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 2.
If the centroid of a triangle is at (4, -2) and two of its vertices are (3, -2) and (5, 2) then find the third vertex of the triangle.
Solution:
Let the vertices of a triangle be A (3, -2), B (5, 2) and C (x3, y3)
Centroid of a triangle is (4, -2)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 3
∴ \(\frac{8+x_{3}}{3}\) = 4
8 + x3 = 12
x3 = 12 – 8
= 4
and
\(\frac{y_{3}}{3}\) = -2
y3 = -6
∴ The third vertex is (4, -6)

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 3.
Find the length of median through A of a triangle whose vertices are A(-1, 3), B (1, -1) and C (5, 1).
Solution:
AD is the median of the ΔABC
D is the mid-point of BC
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 4
Length of the median AD is 5 units.

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 4.
The vertices of a triangle are (1, 2), (h, -3) and (-4, k). If the centroid of the triangle is at the point (5, -1) then find the value of \(\sqrt{(h+k)^{2}+(h+3k)^{2}}\)
Solution:
Let the vertices A (1, 2), B (h, -3) and C (-4, k)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 5
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 6
\(\frac{-3+h}{3}\) = 5
-3 + h = 15
h = 15 + 3 = 18
and
\(\frac{-1+k}{3}\) = -1
-1 + k = -3
k = -3 + 1
k = -2
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 7

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 5.
Orthocentre and centroid of a triangle are A(-3, 5) and B(3, 3) respectively. If C is the circumcentre and AC is the diameter of this circle, then find the radius of the circle.
Solution:
Let PQR be any triangle orthocentre, centroid and circumcentre.
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 8
A orthocentre is (-3, 5)
B centroid is (3, 3)
C orthocentre is (a, 6)
Also \(\frac{AB}{BC}\) = \(\frac{2}{1}\)
B divides AC in the ratio 2 : 1
A line divides internally in the ratio point P is (\(\frac{mx_{2}+nx_{1}}{m+n}\), \(\frac{my_{2}+ny_{1}}{m+n}\))
m = 2
x1 = 3
y1 = 5
amd
n = 1
x2 = a
y2 = b
∴ The point B (\(\frac{2a-3}{2+1}\), \(\frac{2b+5}{2+1}\))
(3, 3) = (\(\frac{2a-3}{3}\), \(\frac{2b+5}{3}\))
\(\frac{2a-3}{3}\) = 3
2a – 3 = 9
2a = 9 + 3
2a = 12
a = \(\frac{12}{2}\) = 6
and
\(\frac{2b+5}{3}\)
2b + 5 = 9
2b = 9 – 5
2b = 4
b = \(\frac{4}{2}\) = 2
∴ Orthocentre C is (6, 2)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 9

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 6.
ABC is a triangle whose vertices are A (3, 4), B (-2, -1) and C (5, 3). If G is the centroid and BDCG is a parallelogram then find the coordinates of the vertex D.
Solution:
The vertices of a triangle are A (3, 4), B (-2, -1) and C (5, 3)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 10
The point G is (2, 2)
Let the vertices D be (a, b)
Since BDCG is a parallelogram
Mid-point of BC = Mid-point of DG
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 11
\(\frac{2+a}{2}\) = \(\frac{3}{2}\)
4 + 2a = 6
2a = 6 – 4
2a = 2
a = 1
and
\(\frac{2+b}{2}\) = 1
2 + b = 2
b = 2 – 2 = 0
The vertices D is (1, 0).

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

Question 7.
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 12
Solution:
In ΔABC, Let A (x1, y1), B (x2, y1) and C (x3, y3)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 13
x1 + x2 = 3 → (1)
y1 + y2 = 10 → (2)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 14
x2 + x3 = 14 → (3)
y2 + y3 = 10 → (4)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 15
x1 + x3 = 3 → (5)
y1 + y3 = 10 → (6)
Add (1) + (3) + (5) We get
2x1 + 2x2 + 2x3 = 30
2(x1 + x2 + x3) = 30
x1 + x2 + x3 = 15
From (1), x1 + x2 = 3
∴ x3 = 12
From (3), x2 + x3 = 14
∴ x1 = 1
From (5), x1 + x3 = 13
∴ x2 = 2
Add (2), (4) and (6) we get
2y1 + 2y2 + 2y3 = -12
2(y1 + y2 + y3) = – 12
∴ y1 + y2 + y3 = -6
From (2), y1 + y2 = 10
∴ y3 = -16
From (4) y2 + y3 = -9
∴ y1 = 3
From (6) y1 + y3 = -13
∴ y2 = 7
The vertices of the A are A (1, 3), B (2, 7) and C (12, -16)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5 16
The Centroid of Δ is (5, -2)

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.5

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