Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.5 Pdf
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TN Samacheer Kalvi 10th Maths Solutions Chapter 5 – Coordinate Geometry
| Board | TNSCERT Class 10th Maths |
| Class | 10 |
| Subject | Maths |
| Chapter | 5 (Exercise 5.5) |
| Chapter Name | Coordinate Geometry |
TNSCERT Class 10th Maths Pdf | all Exercise Solution
Exercise – 5.5
Multiple choice questions:
(1) The area of triangle formed by the points (−5, 0), (0,−5) and (5,0) is
(A) 0 sq. units
(B) 25 sq. units
(C) 5 sq. units
(D) None of these
Solution:
Let, (x1, y1) = (-5, 0), (x1, y2) = (0, -5), (x3, y3) = (5, 0)
The area of the triangle
= 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)} Sq. units
= ½ {-5 (-5 – 0) + 0 (0 – 0) + 5 (0 + 5)} Sq. units
= 1/2 {25 + 25} Sq. units
= 25 sq. units (B)
(2) A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the Y axis. The path travelled by the man is
(A) x = 10
(B) y = 10
(C) x = 0
(D) y = 0
Solution:
AO = wall
OB = distance the man to wall
OB is x – axis.
So, x = OB = 10 [Given that]
∴ x = 10 (A)
(3) The straight line given by the equation x = 11 is
(A) Parallel to X axis
(B) parallel to Y axis
(C) Passing through the origin
(D) Passing through the point (0,11)
Solution:
Given figures, see that parallel + y-axis (B)
(4) If (5,7), (3,p) and (6,6) are collinear, then the value of p is
(A) 3
(B) 6
(C) 9
(D) 12
Solution:
Let, (x1, y1) = (5, 7), (x2, y2) = (3, p), (x3, y3) = (6, 6)
Now three points are collinear
1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)} = 0
x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0
5 (p – 6) + 3 (6 – 7) + 6 (7 – p) = 0
5p – 30 – 3 + 42 – 6p = 0
-p + 9 = 0
p = 9
Thus, the value of p = 9 (C)
(5) The point of intersection of 3x – y = 4 and x + y = 8 is
(A) (5, 3)
(B) (2, 4)
(C) (3, 5)
(D) (4, 4)
Solution:
Given that, x – y = 4 —- (i)
x + y = 8 —– (ii)
Now, (i) + (ii), we get, 3x + x – y + y = 4 + 8
4x = 12
x = 3
From (ii) x = 3 putting, we get
3x – y = 4
9 – y = 4
y = 9 – 4
y = 5
Thus, the intersection point (3, 5) (C)
(6) The slope of the line joining (12, 3), (4, a) is 1/8. The value of ‘a’ is
(A) 1
(B) 4
(C) -5
(D) 2
Solution:
Now, the point slope is
= a-3/4-12
= a-3/-8
Now, a-3/-8 = 1/8
8a – 24 = -8
8a = 16
a = 12
∴ Thus, the value of a is 2 (D)
(7) The slope of the line which is perpendicular to a line joining the points (0, 0 and (-8, 8) is
(A) -1
(B) 1
(C) 1/3
(D) -8
Solution:
(0, 0) and (-8, 8) two points slope is m2 = 8-0/-8-0 = -1
∴ Perpendicular line slope is m1 (let)
Now, m1× m2 = -1
m1 × -1 = -1
m1 = 1 (B)
(8) If slope of the line PQ is 1/√3 then slope of the perpendicular bisector of PQ is
(A) √3
(B) – √3
(C) 1/√3
(D) 0
Solution:
Let, m1 = 1/√3 and perpendicular bisector slope = m2
Now, m1× m2 = -1
1/√3 × m2 = -1
m2 = – √3 (B)
(9) If A is a point on the Y axis whose ordinate is 8 and B is a point on the X axis whose abscissae is 5 then the equation of the line AB is
(A) 8x + 5y = 40
(B) 8x – 5y = 40
(C) x = 8
(D) y = 5
Solution:
Now, x – intercepts = 5
y – intercepts = 8
Now, the given equation is x/5 + y/8 = 1
8x + 5y/40 = 1
8x + 5y = 40 (A)
(10) The equation of a line passing through the origin and perpendicular to the line7x – 3y + 4 = 0 is
(A) 7x – 3y + 4 = 0
(B) 3x – 7y + 4 = 0
(C) 3x + 7y = 0
(D) 7x – 3y = 0
Solution:
7x – 3y + 4 = 0 is perpendicular line equation is -3x – 7y + k = 0
k = 0
Now, the required line equation is
-3x – 7y + 0 = 0
3x + 7y = 0 (C)
(11) Consider four straight lines
(i) l2; 3y = 4x + 5
(ii) l2; 4y = 3x – 1
(iii) l3; 4y + 3x – 1
(iv) l4; 4x + 3y = 2
Which of the following statement is true?
(A) l1 and l2 are perpendicular
(B) l1 and l4 are parallel
(C) l2 and l4 are perpendicular
(D) l2 and l3 are parallel
Solution:
Now, slope of l1 = – (-4)/2 = 4/3
Slope of l2 = -(-3)/4 = 3/4
Slope of l3 = -3/4 = – 3/4
Slope of l11 = -4/3 = – 4/3
Now, slope of l2 × slope of l4
= ¾ × – 4/3
= -1
∴ l2 an l4 are perpendicular (C)
(12) A straight line has equation 8y = 4x + 21. Which of the following is true
(A) The slope is 0.5 and the y intercept is 2.6
(B) The slope is 5 and the y intercept is 1.6
(C) The slope is 0.5 and the y intercept is 1.6
(D) The slope is 5 and the y intercept is 2.6
Solution:
8y = 4x + 21
y = 4/8 x + 21/8
Slope = 4/8 = 1/2
= 0.5
and y intercepts = 21/8
= 2.6
(A)
(13) When proving that a quadrilateral is a trapezium, it is necessary to show
(A) Two sides are parallel.
(B) Two parallel and two non-parallel sides.
(C) Opposite sides are parallel.
(D) All sides are of equal length.
Solution:
(B) Two parallel and two non-parallel sides.
(14) When proving that a quadrilateral is a parallelogram by using slopes you must find
(A) The slopes of two sides
(B) The slopes of two pair of opposite sides
(C) The lengths of all sides
(D) Both the lengths and slopes of two sides
Solution:
Ans:
(B) The slopes of two pair of opposite sides.
(15) (2, 1) is the point of intersection of two lines.
(A) x – y – 3 = 0; 3x – y – 7 = 0
(B) x + y = 3; 3x + y = 7
(C) 3x + y = 3; x + y = 7
(D) x + 3y – 3 = 0; x – y – 7 = 0
Solution:
Now, option (A) x – y – 3 = 0
Putting the point (2, 1)
2 – 1 – 3 = 0
1 – 3 = 0
-2 ≠ 0
Option (B) x + y = 3
Putting the point (2, 1)
2 + 1 = 3
3 = 3
and 3x + y = 7
3.2 + 2 = 7
6 + 1 = 7
7 = 7
Option (C) 3x + y = 3
Putting the point (2, 1)
3.2 + 1 = 3
6+1 = 3
7≠3
Option (D), x + 3y – 3 = 0
Putting the point (2, 1)
2 + 3.1 – 3 = 0
2+3 – 3 = 0
2 ≠ 0
∴ Ans (B)
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