Samacheer Kalvi 10th Maths Solutions Chapter 5 Unit Exercise 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 (Unit Exercise 5) |
| Chapter Name | Coordinate Geometry |
TNSCERT Class 10th Maths Pdf | all Unit Exercise Solution
UNIT Exercise – 5
(1) PQRS is a rectangle formed by joining the points P (-1, -1), Q (-1, 4), R (5, 4) and S (5, -1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.
Solution:
Now, A be the mid-point of PQ
∴ A (x1, y1) = (-1-1/2, -1+4/2 = (-1, 3/2)
B be the mid-point of QR
B (x2, y2) = (-1+5/2, 4+4/2) = (2, 4)
C be the mid-point of SR
C (x3, y3) = (5+5/2, 4-1/2) = (5, 3/2)
D be the mid-point of SP
D (x4, y4) = (5-1/2, -1-1/2) = (2, -1)
We know that, two point (x1, y1) and (x2, y2)
Are distance = √(x2 – x1)2 + (y2 – y1)2
Now, The length of AB is
= √(2+1)2 + (4 – 3/2)
= √(3)2 + (3/2)2
= √9+25/4
= √36+25/4
= √61/4
= √61/2
The length of BC is
= √(5-2)2 + (3/2 – 4)2
= √(7)2 + (- 5/2)2
= √9 + 25/4
= √36+25/4
= √61/2
The length of CD is
= √(2-5)2 + (-1-3/2)2
= √(-3)2 + (-5/2)2
= √9+25/4
= √36+25/4
= √61/4
The length of DA is
= √(-1-2)2 + (3/2+1)2
= √(-3)2 + (5/2)2
= √9+25/4
= √36+25/2
= √61/2
Now, AB = BC = CD = DA = √61/2
Now, four sides are equal.
So, ABCD is rhombus
Thus, the ABCD is a rhombus.
(2) The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, -2). The third vertex is (x, y) where y = x + 3. Find the coordinates of the third vertex.
Solution:
Let, (x1, y1) = (2, 1), (x2, y2) = (3, -2) and (x3, y3) = (x, y)
We know that the area of a triangle vertex are (x1, y1), (x2, y2) and (x3, y3) is
= 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)} Sq. units
Now, 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)} = 5
2 (-2-y) + 3 (y – 1) + x (1 + 2) = 10
-4-2y+3y – 3 + 3x = 10
3x + y – 7 = 10
3x + y = 17 —– (i)
Given that, y = x + 3
x – y = -3 —– (ii)
(i) now, (i) + (ii), we get
3x + y = 17
x – y = -3
__________
4x = 14
x = 14/4 = 7/2
From (i), x = 7/2 putting, we get
3x + y = 17
3 × 7/2 + y = 17
y = 17 – 21/2 = 34-21/2
y = 13/2
Thus, the third vertex is (7/2, 13/2)
(3) Find the area of a triangle formed by the lines 3x + y – 2 = 0, 5x + 2y – 3 = 0 and 2x – y – 3 = 0
Solution:
Now, Equation of AB = 3x + y – 2 = 0 —- (i)
Equation of BC = 5x + 2y – 3 = 0 —- (ii)
Equation of CA = 2x – y – 3 = 0 —- (iii)
Now, (i) × 2 – (ii) × 1; We get
6x + 2y – 4 = 0
5x + 2y – 3 = 0
(-) (-) (+)
_____________
x – 1 = 0
x = 1
From (i), x = 1, putting, we get
3x + y – 2 = 0
3.2 + y – 2 = 0
3 – 2 + y = 0
1 + y = 0
y = -1
∴ Now, D = (1, -1) = (x2, y2) (let)
Now, (i) + (ii), we get
3x + y – 2 = 0
2x – y – 3 = 0
_____________
5x – 5 = 0
x = 1
From (iii), x = 1 Putting, we get
2x – y – 3 = 0
2 – y – 3 = 0
y = -1
∴ A = (1, 1) = (x1, y1) (Let)
Now, (ii) × 1 + (iii) × 2, we get
5x + 2y – 3 = 0
4x – 2y – 6 = 0
______________
9x – 9 = 0
9x = 9
x = 1
From (ii), x = 1, Putting, we get
5x + 2y – 3 = 0
5 + 2y – 3 = 0
2 + 2y = 0
y = -1
∴ Now, C = (1, -1) = (x3, y3) (Let)
The area of ∆ABC
= ½ {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)} Sq. units
= ½ {1 (-1+1) + 1 (-1 + 1) + 1 (-1 + 1)} Sq. units
= ½ × 0 Sq. units
= 0 Sq. units
(4) If vertices of a quadrilateral are at A (-5, 7), B (-4, k), C (-1, -6) and D (4, 5) and its area is 72 Sq. units. Find the value of k.
Solution:
Let, A (x1, y1) = (-5, 7), B (x2, y2) = (-4, k)
C (x3, y3) = (-1, -6) and D (x4, y4) = (4, 5)
We know that, the area of the quadrilateral of ABCD is
= ½ {x1y2 + x2y2 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)} Sq. Units
Now, 1/2 {(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)} = 72
{(-5k + 24 – 5 + 28) – (-28 – k – 24 – 25)} = 72×2
(-5k + 47) – (-5 – 77) = 72×2
-5k + 47 + k + 77 = 72×2
-4k + 124 = 72×2
– 4k = 144 – 124
-4k = 20
k = -5
Thus, the value of k = -5.
(5) Without using distance formula, show that the points (-2, – 1), (4, 0), (3, 3) and (-3, 2) are vertices of a parallelogram.
Solution:
Now, Let A (x1, y1) = (-2, -1), B (x2, y2) = (4, 0)
c (x3, y3) = (3, 3) and D (x4, y4) = (-3, 2)
Now, Slope of AB (m1) = y2-y1/x2-x1
= 0+1/4+2 = 1/6
Slope of BC (m2) = y3-y2/x3-x2
= 3+0/3-4 = 3/-1 = -3
Slope of CD (m3) = y4-y3/x4-x3 = 2-3/-3-3 = -1/-6
= 1/6
Slope of DA (m4) = y1-y4/x1-x4 = -1-2/-2+3 = -3
Now, slope of AB = slope of CD
m1 = m3
∴AB and CD are parallel
Now, slope of BC = slope of DA
m2 = m4
∴BC and DA are parallel
Thus, the ABCD be a parallelogram.
(6) Find the equations of the lines, whose sum and product of intercepts are 1 and –6 respectively.
Solution:
Let x intercept = a
∴y intercept = 1 – a
∴Now, x intercept x y intercept = -6
a× (1-a) = -6
a – a2 = -6
-a2 + a + 6 = 0
a2 – a – 6 = 0
a2 – (3-2) a – b = 0
a2 – 3a + 2a – 6 = 0
a (a – 3) + 2 (a – 3) = 0
(a – 3) (a + 2) = 0
a – 3 = 0
a = 3
a + 2 = 0
a = -2
When a = 3
y intercept = 1 – a = 1 – 3 = -2
∴Now, equation of a line is
x/a + y/b = 1
x/3 + y/-2 = 1
x/3 – y/2 = 1
2x-3y/6 = 1
2x – 3y = 6
2x – 3y – 6 = 0
When, a = -2
y intercept = 1+2 = 3
∴Equation of line is
x/-2 + y/3 = 1
-3x + 2y/6
-3x + 2y = 6
3x – 2y = -6
3x – 2y + 6 = 0
Thus, the required equation
2x – 3y – 6 = 0
3x – 2y + 6 = 0
(7) The owner of a milk store finds that, he can sell 980 litres of milk each week at₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could be sell weekly at ₹17/litre?
Solution:
Given that, sell 980 litres of milk each week at ₹14/litre.
Let selling price = x and demand = y let (x1, y1) = (14, 980)
Given that, sell 1220 litres of mill each week at ₹16/litre.
Let (x2, y2) = (16, 1220)
Slope of the line = y2 – y1/x2 – x1
= 1220-980/16-14
= 240/2
= 120
∴Equation of the line is
(y – y1) = m (x – x1)
(y – 980) = 120 (x – 14)
y – 980 = 120x – 1680
120x – y = 1680 – 980
120x – y = 700
Now, given that x = 17
120×17 – y = 700
y = 2040 – 700
y = 1340
Thus, the demand is 1340 litres
(8) Find the image of the point (3, 8) with respect to the line z + 3y = 7 assuming the line to be a plane mirror.
Solution:
Let, B the mirror point of A is (x1, y1)
x + 3y = 7 —- (i)
This equation slop is (m1) = coefficient of x/coefficient of y
= -1/3
= – 1/3
Let, AB line slope = m2
Now, x + 3y = 7 and AB is perpendicular.
m1× m2 =-1
– 1/3 × m2 = -1
m2 = 3
Throw, the equation of AB is y – y1 = m (x – x1)
(y – 8) = 3 (x – 3)
3x – 9 = y – 8
3x – y = -8+9
3x – y = 1 — (ii)
Now, (ii) × 1 + (i) × 3, we get
x + 3y = 7
9x – 3y = 3
__________
10x = 10
x = 1
From (ii) x = 1 putting, we get
3x – y = 1
3 – y = 1
y = 3 – 1 = 2
y = 2
∴Point 0 = (1, 2)
Now, 0 be the mid-point of AB.
Now, (3+x1/2, 8+y1/2) = (1, 2)
3 + x1 = 2
x1 = 2 – 3
x1 = -1
8+y1/2 = 2
8 + y1 = 4
y1 = 4-8
y1 = -4
Thus, the mirror point B is (-1, -4)
(9) Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Solution:
Now, 4x + 7y = 3 —- (i)
2x – 3y = -1 —– (ii)
Now, (i) × 3 + (ii) × 7, we get
12x + 21y = 9
14x – 21y = -7
_____________
26x = 2
x = 2/26
x = 1/13
From (i) putting x = 1/13, we get
4x + 7y = 3
4 × 1/13 + 7y = 3
7y = 3 – 4/13
7y = 39 – 4/13
7y = 35/13
y = 5/13
The intersection point is = (1/13, 5/13)
Let, x intercepts = y intercepts = a
∴Equation of line is x/a + y/a = 1
(1/13, 5/13) point passing through this line
1/13a + 5/13a = 1
1+5/13a = 1
13a = 6
a = 6/13
∴The required equation the line is
x/(6/13) + y/(6/13) = 1
13x/6 + 13y/6 = 1
13x + 13y = 6
13x + 13y – 6 = 0
Thus, the required equation is 13x + 13y – 6 = 0
(10) A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.
Solution:
Given that 2x – 3y = -4 —- (i)
3x + 4y = 5 —- (ii)
Now, (i) × 4 + (ii) × 3, we get
8x – 12y = -16
9x + 12y = 15
____________
17x = -1
x = – 1/17
From (i) putting x = – 1/17, we get
2x – 3y = -4
2 × – 1/17 – 3y = -4
3y = 4 – 2/17 = 68-2/17 = 66/17
y = 22/17
Now, two straight paths intersection point is (- 1/17, 22/17)
Now, 6x – 7y + 8 = 0 is perpendicular equation is – 7x – 6y + k = 0 —- (i)
Now, (- 1/17, 22/17) point passing through the equation (i)
– 7 × – 1/17 – 6 × 22/7 + k = 0
7/17 – 132/17 + k = 0
7-132/17 + k = 0
k = 125/17
The required equation is – 7k – 6y + 135/17 = 0
-119x – 102y + 125 = 0
119x + 102y – 125 = 0
Thus, the required equation is 119x + 102y – 125 = 0.
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