## Coordinate Geometry | Aster Classes

Tamilnadu, Samacheer, Kalvi, 10th, sslc, Maths, Solutions, Chapter 5, Coordinate Geometry, Unit Exercise 5,

Question 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.

Mid point of a line = ([Math Processing Error])
Mid point of PQ (A) = ([Math Processing Error],[Math Processing Error])
= ([Math Processing Error],[Math Processing Error]) = (-1,[Math Processing Error])
Mid point of QR (B) = ([Math Processing Error],[Math Processing Error]) = ([Math Processing Error],[Math Processing Error]) = (2,4)
Mid point of RS (C) = ([Math Processing Error],[Math Processing Error]) = ([Math Processing Error],[Math Processing Error]) = (5,[Math Processing Error])
Mid point of PS (D) = ([Math Processing Error],[Math Processing Error]) = ([Math Processing Error],[Math Processing Error]) = (2,-1)

img 355
AB = BC = CD = AD = [Math Processing Error]
Since all the four sides are equal,
∴ ABCD is a rhombus.

Question 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.
Let the vertices A(2,1), B(3, – 2) and C(x, y)
Area of a triangle = 5 sq. unit

[Math Processing Error] [x1y2 + x2y3 + x3y1– (x2y1 + x3y2 + x1y3)] = 5
[Math Processing Error] [-4 + 3y + x – (3 – 2x + 2y)] = 5
-4 + 3y + x – 3 + 2x – 2y = 10
3x + y – 7 = 10
3x + y = 17 ……(1)
Given y = x + 3
Substitute the value ofy = x + 3 in (1)
3x + x + 3 = 17
4x = 17 – 3
4x = 14
x = [Math Processing Error] = [Math Processing Error]
Substitute the value of x in y = x + 3
y = [Math Processing Error] + 3 ⇒ y = [Math Processing Error] = [Math Processing Error]
∴ The coordinates of the third vertex is ([Math Processing Error],[Math Processing Error])

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Question 3.
Find the area of a triangle formed by the lines 3x + y – 2 = 0, 5x + 2y – 3 = 0 and 2x – y – 3 = 0
3x + y = 2 ……..(1)
5x + 2y = 3 ………(2)
2x – y = 3 ……….(3)
Solve (1) and (2) to get the vertices Bp

Substitute the value of x = 1 in (1)
3(1) + y = 2
y = 2 – 3 = – 1
The point B is (1, – 1)
Solve (2) and (3) to get the vertices C

Substitute the value of x = 1 in (3)
2(1) – y = 3 ⇒ -y = 3 – 2
– y = 1 ⇒ y = – 1
The point C is (1, – 1)
Solve (1) and (3) to get the vertices A

Substitute the value of x = 1 in (1)
3(1) + y = 2
y = 2 – 3 = -1
The point A is (1, – 1)
The points A (1, – 1), B (1, -1), C(1, -1)
Area of ∆ABC = [Math Processing Error]

[x1y2 + x2y3 + x3y1 – (x2y1 + x3y2 + x1y3)]

Area of the triangle = 0 sq. units.
Note: All the three vertices are equal, all the point lies in a same points.

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Question 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.

Area of the quadrilateral ABCD = 72 sq. units.
[Math Processing Error] [(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)] = 72

-5k + 24 – 5 + 28 – (- 28 – K – 24 – 25) = 144
– 5k + 47 – k – 77 = 144
– 5k + 47 + k + 77 = 144
– 4k + 124 = 144
-4k = 144 – 124
– 4k = 20
k = -5
The value of k = – 5

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Question 5.
Without using distance formula, show that the points (-2,-1), (4,0), (3,3) and (-3,2) are vertices of a parallelogram.
The vertices A(-2, -1), B(4, 0), C(3, 3) and D(- 3, 2)

Slope of a line = [Math Processing Error]
Slope of AB = [Math Processing Error] = [Math Processing Error]
Slope of BC = [Math Processing Error] = [Math Processing Error] = -3
Slope of CD = [Math Processing Error] = [Math Processing Error] = [Math Processing Error]
Slope of AD = [Math Processing Error] = [Math Processing Error] = -3
Slope of AB = Slope of CD = [Math Processing Error]
∴ AB || CD ……(1)
Slope of BC = Slope of AD = -3
From (1) and (2) we get ABCD is a parallelogram.

Question 6.
Find the equations of the lines, whose sum and product of intercepts are 1 and – 6 respectively.
Let the “x” intercept be “a”
y intercept = 1 – a (sum of the intercept is 1)
Product of the intercept = – 6
a (1 – a) = – 6 ⇒ a – a2 = – 6
– a2 + a + 6 = 0 ⇒ a2 – a – 6 = 0
(a – 3) (a + 2) = 0 ⇒ a – 3 = 0 (or) a + 2 = 0
a = 3 (or) a = -2
When a = 3
x – intercept = 3
y – intercept = 1 – 3 = – 2
Equation of a line is
[Math Processing Error] + [Math Processing Error] = 1
[Math Processing Error] + [Math Processing Error] = 1
[Math Processing Error] – [Math Processing Error] = 1
2x – 3y = 6
2x – 3y – 6 = 0

When a =-2
x – intercept = -2
y – intercept = 1 – (- 2) = 1 + 2 = 3
Equation of a line is
[Math Processing Error] + [Math Processing Error] = 1
[Math Processing Error] + [Math Processing Error] = 1
– [Math Processing Error] + [Math Processing Error] = 1
– 3x + 2y = 6
3x – 2y + 6 = 0

Question 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 he sell weekly at ₹ 17/litre?
Let the selling price of a milk be “x”
Let the demand be “y”
We have to find the linear equation connecting them
Two points on the line are (14, 980) and (16,1220)
Slope of the line = [Math Processing Error]
[Math Processing Error] = [Math Processing Error] = 120
Equation of the line is y – y1 = m (x – x1)
y – 980 = 120 (x – 14) ⇒ y – 980 = 120 x – 1680
-120 x + y = -1680 + 980 ⇒ -120 x + y = -700 ⇒ 120 x – y = 700
Given the value of x = 17
120(17) – y = 700
-y = 700 – 2040 ⇒ – y = – 1340
y = 1340
The demand is 1340 liters

Question 8.
Find the image of the point (3,8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Let the image of P(3, 8) and P’ (a, b)
Let the point of intersection be O

Slope of x + 3y = 7 is – [Math Processing Error]
Slope of PP’ = 3 (perpendicular)
Equation of PP’ is
y – y1 = m(x – x1)
y – 8 = 3 (x – 3)
y – 8 = 3x – 9
-8 + 9 = 3x – y
∴ 3x – y = 1 ………(1)
The two line meet at 0

Substitute the value of x = 1 in (1)
3 – y = 1
3 – 1 = y
2 = y
The point O is (1,2)
Mid point of pp’ = ([Math Processing Error])
(1,2) = ([Math Processing Error],[Math Processing Error])
∴ [Math Processing Error] = 1 ⇒ 3 + a = 2
a = 2 – 3 = -1
[Math Processing Error] = 2
8 + b = 4
b = 4 – 8 = – 4
The point P’ is (-1, -4)

Question 9.
Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = O and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Given lines

Substitute the value of y = [Math Processing Error] in (2)
2x – 3 × [Math Processing Error] = -1
2x – [Math Processing Error] = -1
26x – 15 = -13
26x = -13 + 15
26x = 2
x = [Math Processing Error] = [Math Processing Error]
The point of intersection is ([Math Processing Error],[Math Processing Error])

Let the x – intercept and y intercept be “a”
Equation of a line is
[Math Processing Error] + [Math Processing Error] = 1
[Math Processing Error] + [Math Processing Error] = 1 (equal intercepts)
It passes through ([Math Processing Error],[Math Processing Error])
[Math Processing Error] + [Math Processing Error] = 1
[Math Processing Error] = 1
13a = 6
a = [Math Processing Error]
The equation of the line is

Question 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.
Two straight path will intersect at one point.
Solving this equations

2x – 3y + 4 = 0

Substitute the value of x = [Math Processing Error] in (2)

The point of intersection is (-[Math Processing Error],[Math Processing Error])
Any equation perpendicular to 6x – 7y + 8 = 0 is 7x + 6y + k = 0
It passes through (-[Math Processing Error],[Math Processing Error])
7(-[Math Processing Error]) + 6 ([Math Processing Error]) + k = 0
Multiply by 17
-7 + 6 (22) + 17k = 0
-7 + 132 + 17k = 0
17k = -125 ⇒ k = – [Math Processing Error]
The equation of a line is 7x + 6y – [Math Processing Error] = 0
119x + 102y – 125 = 0
∴ Equation of the path is 119x + 102y – 125 = 0

Chapter 5, Coordinate Geometry, Ex 5.5,

Question 1.
The area of triangle formed by the points (-5, 0), (0, – 5) and (5, 0) is …………..
(1) 0 sq.units
(2) 25 sq.units
(3) 5 sq.units
(4) none of these
(2) 25 sq.units Hint.
Hint:
Area of the ∆

Question 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 …………
(1) x = 10
(2) y = 10
(3) x = 0
(4) y = 0
(1) x = 10
Hint:

Question 3.
The straight line given by the equation x = 11 is …………….
(1) parallel to X axis
(2) parallel to Y axis
(3) passing through the origin
(4) passing through the point (0,11)
(2) parallel to Y axis

Question 4.
If (5,7), (3,p) and (6,6) are collinear, then the value of p is ……………
(1) 3
(2) 6
(3) 9
(4) 12
(3) 9
Hint:
Since the three points are collinear. Area of a triangle is 0

5p + 18 + 42 – (21 + 6p + 30) = 0
5p + 60 – (51 + 6p) = 0
5p + 60 – 51 – 6p = 0
-p + 9 = 0
-p = -9
p = 9

Question 5.
The point of intersection of 3x – y = 4 and x + 7 = 8 is ……………
(1) (5,3)
(2) (2,4)
(3) (3,5)
(4) (4, 4)
(3) (3, 5)

Substitute the value of x = 3 in (2)
3 + 7 = 8
y = 8 – 3 = 5
The point of intersection is (3, 5)

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Question 6.
The slope of the line joining (12, 3), (4, a) is 18. The value of ‘a’ is …………….
(1) 1
(2) 4
(3) -5
(4) 2
(4) 2
Hint:
Slope of a line = 18

Question 7.
The slope of the line which is perpendicular to a line joining the points (0, 0) and (- 8, 8) is ………..
(1) -1
(2) 1
(3) 13
(4) -8
(2) 1
Hint:
Slope of a line = y2−y1x2−x1
= 8−0−8−0 = 8−8 = -1
Slope of the Perpendicular = 1

Question 8.
If slope of the line PQ is 13√ then slope of the perpendicular bisector of PQ is …………..
(1) 3–√
(2) –3–√
(3) 13√
(4) 0
(2) –3–√
Hint:
Slope of a line = 13√
Slope of the ⊥r bisector = –3–√

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Question 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 ……………
(1) 8x + 5y = 40
(2) 8x – 5y = 40
(3) x = 8
(4) y = 5
(1) 8x + 5y = 40
Hint:
Let the point A be (0, 8) and B (5, 0)

Question 10.
The equation of a line passing through the origin and perpendicular to the line lx -3y + 4 = 0 is
(1) 7x – 3y + 4 = 0
(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
(3) 3x + 7y = 0
Hint:
Slope of the line of 7x – 3y + 4 = 0
= −7−3 = 73
Slope of its ⊥r = −37
The line passes through (0,0)
Equation of a line is
y – y1 = m(x – x1)
y – 0 = −37 (x – 0)
y = −37 x ⇒ 7y = -3x
3x + 7y = 0

Question 11.
Consider four straight lines
(i) l1 : 3y = 4x + 5
(ii) l2 : 4y = 3x – 1
(iii) l3 : 4y + 3x = 7
(iv) l4 : 4x + 3y = 2
Which of the following statement is true?
(1) l1 and l2 are perpendicular
(2) l2 and l4 are parallel
(3) l2 and l4 are perpendicular
(4) l2 and l3 are parallel
(3) l2 and l4 are perpendicular
Hint:
Slope of l1 = 43; Slope of l2 = 34
Slope of l3 = – 34; Slope of l4 = –43
(1) l1 × l2 = 43 × 34 = 1 …….False
(2) l1 = 43; l4 = – 43 not parallel ………False
(3) l2 × l4 = 34 × – 43 = -1 …….True
(4) l2 = 34; l3 = – 34 not parallel ………False

Question 12.
A straight line has equation 87 = 4x + 21. Which of the following is true …………………….
(1) The slope is 0.5 and the y intercept is 2.6
(2) The slope is 5 and the y intercept is 1.6
(3) The slope is 0.5 and they intercept is 1.6
(4) The slope is 5 and the y intercept is 2.6
(1) The slope is 0.5 and they intercept is 2.6
Hint:
8y = 4x + 21
y = 48 x + 218
= 12 x + 218
12 = 0.5
218 = 2.625
Slope = 12 = 0.5
y intercept = 218 = 2.6

Question 13.
When proving that a quadrilateral is a trapezium, it is necessary to show
(1) Two sides are parallel.
(2) Two parallel and two non-parallel sides.
(3) Opposite sides are parallel.
(4) All sides are of equal length.
Solution:
(2) Two parallel and two non-parallel sides.

Question 14.
When proving that a quadrilateral is a parallelogram by using slopes you must find …………………
(1) The slopes of two sides
(2) The slopes of two pair of opposite sides
(3) The lengths of all sides
(4) Both the lengths and slopes of two sides
(2) The slopes of two pair of opposite sides

Question 15.
(2,1) is the point of intersection of two lines.
(1) x – y – 3 = 0; 3x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
(3) 3x + y = 3; x + y = 7
(4) x + 3y – 3 = 0; x – y – 7 = 0
Solution:
(2) x + y = 3; 3x + y = 7

Tamilnadu, Samacheer, Kalvi, 10th, sslc, Maths, Solutions, Chapter 5, Coordinate Geometry, Ex 5.4,

Question 1.
Find the slope of the following straight lines.
(i) 5y – 3 = 0
(ii) 7x – 317 = 0
Solution:
(i) 5y – 3 = 0
5y = 3 ⇒ y = 35
Slope = 0

(ii) 7x – 317 = 0 (Comparing with y = mx + c)
7x = 317
Slope is undefined

Question 2.
Find the slope of the line which is
(i) parallel to y = 0.7x – 11
(ii) perpendicular to the line x = -11
Solution:
(i) y = 0.7x – 11
Slope = 0.7 (Comparing with y = mx + c)
(ii) Perpendicular to the line x = – 11
Slope is undefined (Since the line is intersecting the X-axis)

Question 3.
Check whether the given lines are parallel or perpendicular
(i) x3 + y4 + 17 = 0 and 2×3 + y2 + 110 = 0
(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Solution:
(i) x3 + y4 + 17 = 0 ; 2×3 + y2 + 110 = 0
Slope of the line (m1) = −ab
= – 13 ÷ 14 = –13 × 41 = – 43
Slope of the line (m2) = – 23 ÷ 12 = –23 × 21= – 43
m1 = m2 = – 43
∴ The two lines are parallel.

(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Slope of the line (m1) = −523
Slope of the line (m2) = −23−5 = 235
m1 × m2 = −523 × 235 = -1
∴ The two lines are perpendicular

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Question 4.
If the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’
Solution:
Slope of the first line 12y = -(p + 3)x +12
y = −(p+3)x12+1 (Comparing with y = mx + c)
Slope of the second line (m1) = −(p+3)12
Slope of the second line 12x – 7y = 16
(m2) = −ab = −12−7 = 127
Since the two lines are perpendicular
m1 × m2 = -1
−(p+3)12 × 127 = -1 ⇒ −(p+3)7 = -1
-(p + 3) = -7
– p – 3 = -7 ⇒ -p = -7 + 3
-p = -4 ⇒ p = 4
The value of p = 4

Question 5.
Find the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3, -2) and R(-5,4).
Solution:
Slope of the line = y2−y1x2−x1
Slope of the line QR = 4+2−5−3 = 6−8 = 3−4 ⇒ – 34
Slope of its parallel = – 34
The given point is p(-5, 2)
Equation of the line is y – y1 = m(x – x1)
y – 2 = – 34 (x + 5)
4y – 8 = -3x – 15
3x + 4y – 8 + 15 = 0
3x + 4y + 7 = 0
The equation of the line is 3x + 4y + 7 = 0

Question 6.
Find the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Solution:
Let the vertices A (6, 7), B (2, -3), D (6, -2)

Slope of a line = y2−y1x2−x1
Slope of AB = −3−72−6 = −10−4 = 52
Slope of its perpendicular (CD) = – 25
Equation of the line CD is y – y1 = m(x – x1)
y + 2 = –25 (x – 6)
5(y + 2) = -2 (x – 6)
5y + 10 = -2x + 12
2x + 5y + 10 – 12 = 0
2x + 5y – 2 = 0
The equation of the line is 2x + 5y – 2 = 0

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Question 7.
A(-3,0) B(10, -2) and C(12,3) are the vertices of ∆ABC. Find the equation of the altitude through A and B.
Solution:
To find the equation of the altitude from A.
The vertices of ∆ABC are A(-3, 0), B(10, -2) and C(12, 3)

Slope of BC = y2−y1x2−x1
= 3+212−10 = 52
Slope of the altitude AD is – 25
Equation of the altitude AD is
y – y1 = m (x – x1)
y – 0 = – 25 (x + 3)
5y = -2x -6
2x + 5y + 6 = 0
Equation of the altitude AD is 2x + 5y + 6 = 0
Equation of the altitude from B

Slope of AC = 3−012+3 = 315 = 15
Slope of the altitude AD is -5
Equation of the altitude BD is y – y1= m (x – x1)
7 + 2 = -5 (x – 10)
y + 2 = -5x + 50
5x + 7 + 2 – 50 = 0 ⇒ 5x + 7 – 48 = 0
Equation of the altitude from B is 5x + y – 48 = 0

Question 8.
Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6, -4).
Solution:
“C” is the mid point of AB also CD ⊥ AB.

Slope of AB = y2−y1x2−x1
= −4−26+4 = −610 = – 35
Slope of the ⊥r AB is 53
Mid point of AB = (x1+x22,y1+y22)
= (−4+62,2−42) = (22,−22) = (1,-1)
Equation of the perpendicular bisector of CD is
y – y1 = m(x – x1)
y + 1 = 53 (x – 1)
5(x – 1) = 3(y + 1)
5x – 5 = 3y + 3
5x – 3y – 5 – 3 = 0
5x – 3y – 8 = 0
Equation of the perpendicular bisector is 5x – 3y – 8 = 0

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Question 9.
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0.
Solution:
Given lines are.

x = 4343 = 1
Substitute the value of x = 1 in (1)
7(1) + 3y = 10 ⇒ 3y = 10 – 7
y = 33 = 1
The point of intersection is (1,1)
Equation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0
This line passes through (1,1)
13 (1) + 5 (1) + k = 0
13 + 5 + k = 0 ⇒ 18 + k = 0
k = -18
∴ The equation of the line is 13x + 5y – 18 = 0

Question 10.
Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0.
Solution:
Given lines are.

Substitute the value of x = 167 in (2)
3 × 167 + 2y = 10 ⇒ 2y = 10 – 487
2y = 70−487 ⇒ 2y = 227
y = 222×7 = 117
The point of intersect is (167,117)
Equation of the line perpendicular to 4x – 7y + 13 = 0 is 7x + 4y + k = 0
This line passes through (167,117)
7 (167) + 4 (117) + k = 0 ⇒ 16 + 447 + k = 0
112+447 + k = 0 ⇒ 1567 + k = 0
k = – 1567
Equation of the line is 7x + 4y – 1567 = 0
49x + 28y – 156 = 0

Question 11.
Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y -4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3.
Solution:
The given lines are.

Substitute the value of x = 0 in (1)
3 (0) + y = -2
y = -2
The point of intersection is (0, -2).
The given equation is

Substitute the value of y = 911 in (6)
– x + 2 (911) = 3 ⇒ -x + 1811 = 3
-x = 3 – 1811 = 33−1811 = 1511
x = – 1511
The point of intersection is (-1511,911)
Equation of the line joining the points (0, -2) and (-1511,911) is

31 × (- 11x) = 11 × 15 (y + 2) = 165 (y + 2)
– 341 x = 165 y + 330
– 341 x – 165 y – 330 = 0
341 x + 165 y + 330 = 0
(÷ by 11) ⇒ 31 x + 15 y + 30 = 0
The required equation is 31 x + 15 y + 30 = 0

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Question 12.
Find the equation of a straight line through the point of intersection of the lines 8JC + 3j> = 18, 4JC + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7,6).
Solution:
Given lines are.
8x + 3y = 18 …..(1)
4x + 5y = 9 …..(2)

x = 6328 = 94
Substitute the value of x = 94 in (2)
4 (94) + 5y = 9
9 + 5y = 9 ⇒ 5y = 9 – 9
5y = 0 ⇒ y = 0
The point of intersection is (94,0)
Mid point of the points (5, -4) and (-7, 6)

Equation of the line joining the points (94,0) and (-1,1)

-13y = 4x – 9
-4x – 13y + 9 = 0 ⇒ 4x + 13y – 9 = 0
The equation of the line is 4x + 13y – 9 = 0

Tamilnadu, Samacheer, Kalvi, 10th, sslc, Maths, Solutions, Chapter 5, Coordinate Geometry, Ex 5.3,

Question 1.
Find the equation of a straight line passing through the mid-point of a line segment joining the points (1, -5), (4, 2) and parallel to (i) X axis (ii) Y axis
Solution:
Mid point of the line joining to points (1, -5), (4, 2)
Mid point of the line = (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})
= (\frac { 1+4 }{ 2 } ,\frac { -5+2 }{ 2 } ) = (\frac { 5 }{ 2 } ,\frac { -3 }{ 2 } )

(i) Any line parallel to X-axis. Slope of a line is 0.
Equation of a line is y – y1 = m (x – x1)
y + 32 = 0 (x – 52)
y + 32 = 0 ⇒ 2y+32 = 0
2y + 3 = 0

(ii) Equation of a line parallel to Y-axis is
x = 52 ⇒ 2x = 5
2x – 5 = 0

Question 2.
The equation of a straight line is 2(x – y) + 5 = 0. Find its slope, inclination and intercept on the Y axis.
Solution:
Equation of a line is
2 (x – y) + 5 = 0
2 x – 2y + 5 = 0
-2y = -2x – 5
2y = 2x + 5
y = 2×2 + 52
y = x + 52
Slope of line = 1
Y intercept = 52
tan θ = 1
tan θ = tan 45°
∴ angle of inclination = 45°

Question 3.
Find the equation of a line whose inclination is 30° and making an intercept -3 on the Y axis.
Solution:
Angle of inclination = 30°
Slope of a line = tan 30°
(m) = 13√
y intercept (c) = -3
Equation of a line is y = mx + c
y = 13√ x – 3
3–√ y = x – 3 3–√
∴ x – 3–√ y – 3 3–√ = 0

Question 4.
Find the slope and y intercept of 3x−−√ + (1 – 3–√)y = 3.
Solution:
The equation of a line is 3–√x + (1 – 3–√)y = 3
(1 – 3–√)y = 3–√ x + 3

Question 5.
Find the value of ‘a’, if the line through (-2,3) and (8,5) is perpendicular to y = ax + 2
Solution:
Given points are (-2, 3) and (8, 5)
Slope of a line = y2−y1x2−x1
= 5−38+2 = 210 = 15
Slope of a line y = ax + 2 is “a”
Since two lines are ⊥r
m1 × m2 = -1
15 × a = -1 ⇒ a5 = -1 ⇒ a = -5
∴ The value of a = -5

Question 6.
The hill in the form of a right triangle has its foot at (19,3). The inclination of the hill to the ground is 45°. Find the equation of the hill joining the foot and top.
Solution:
Slope of AB (m) = tan 45°
= 1
Equation of the hill joining the foot and the top is
y – y1 = m(x – x1)
y – 3 = 1 (x – 19)

y – 3 = x – 19
– x + y – 3 + 19 = 0
– x + y + 16 = 0
x – y – 16 = 0
The required equation is x – y – 16 = 0

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Question 7.
Find the equation of a line through the given pair of points.
(i) (2,23) and (−12,-2)
(ii) (2,3) and (-7,-1)
Solution:
(i) Equation of the line passing through the point (2,23) and (−12,-2)

-5(3y – 2) = -8 × 2 (x – 2)
-15y + 10= -16 (x – 2)
-15y + 10= -16x + 32
16x – 15y + 10 – 32 = 0
16x – 15y – 22 = 0
The required equation is 16x – 15y – 22 = 0

(ii) Equation of the line joining the point (2, 3) and (-7, -1) is

-9 (y – 3) = -4 (x – 2)
-9y + 27 = – 4x + 8
4x – 9y + 27 – 8 = 0
4x – 9y + 19 = 0
The required equation is 4x – 9y + 19 = 0

Question 8.
A cat is located at the point(-6, -4) in xy plane. A bottle of milk is kept at (5,11). The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take its milk.
Solution:

Equation of the line joining the point is
y−y1y2−y1=x−x1x2−x1
y+411+4 = x+65+6
y+415 = x+611
15(x + 6) = 11(y + 4)
15x + 90 = 11y + 44
15x – 11y + 90 – 44 = 0
15x – 11y + 46 = 0
The equation of the path is 15x – 11y + 46 = 0

Question 9.
Find the equation of the median and altitude of AABC through A where the vertices are A(6,2), B(-5, -1) and C(1,9).
Solution:
(i) To find median

Equation of the median AD is

2(x – 6) = -8 (y – 2)
2x – 12= -8y + 16
2x + 8y – 28 = 0
(÷ by 2) x + 4y – 14 = 0
∴ Equation of the median is x + 4y – 14 = 0
Equation of the altitude is 3x + 5y – 28 = 0

(ii) To find the equation of the altitude

Slope of BC = y2−y1x2−x1
= 9+11+5
= 106 = 53
Slope of the altitude = – 35
Equation of the altitude AD is
y – y1 = m (x – x1)
y – 2 = – 35 (x – 6)
-3 (x – 6) = 5 (y – 2)
-3x + 18 = 5y – 10
-3x – 5y + 18 + 10 = 0
-3x – 5y + 28 = 0
3x + 5y – 28 = 0

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Question 10.
Find the equation of a straight line which has slope −54 and passing through the point (-1,2).
Solution:
Slope of a line (m) = −54
The given point (x1, y1) = (-1, 2)
Equation of a line is y – y1 = m (x – x1)
y – 2= −54 (x + 1)
5(x + 1) = -4(y – 2)
5x +5 = -4y + 8
5x + 4y + 5 – 8 = 0
5x + 4y – 3 = 0
The equation of a line is 5x + 4y – 3 = 0

Question 11.
You are downloading a song. The percenty (in decimal form) of mega bytes remaining to get downloaded in x seconds is given by y = -0.1x + 1.
(i) graph the equation.
(ii) find the total MB of the song.
(iii) after how many seconds will 75% of the song gets downloaded?
Solution:
(i) y = – 0. 1x + 1

(ii) y = -0.1x + 1
x → seconds
y → Mega byte of the song.
The total mega byte of the song is at the beginning that is when x = 0
y = 1 mega byte

In other words y = 25%
y = 0.25
By using the equation we get
0.25 = -0.1x + 1
0.1 x = 0.75
x = 0.750.1 = 7.5 seconds

i.e y = 0
0 = -0.1 x + 1
0.1 x = 1 second
x = 10.1 = 11 × 10
x = 10 seconds

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Question 12.
Find the equation of a line whose intercepts on the x and y axes are given below.
(i) 4, -6
(ii) -5, – 4
Solution:
(i) x intercept (a) = 4; y intercept (b) = – 6
Equation of a line is xa + yb = 1
x4 + y−6 = 1 ⇒ x4 – y6 = 1
(LCM of 4 and 6 is 12)
3x – 2y = 12
3x – 2y – 12 = 0
The equation of a line is 3x – 2y – 12 = 0

(ii) x intercept (a) = -5; y intercept (b) = 34
Equation of a line is xa + yb = 1 ⇒ x−5+y34=1
x−5 + 4y3 = 1
(LCM of 5 and 3 is 15)
– 3x + 20y = 15
– 3x + 20y – 15 = 0
3x – 20y + 15 = 0
∴ Equation of a line is 3x – 20y + 15 = 0

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Question 13.
Find the intercepts made by the following lines on the coordinate axes.
(i) 3x – 2y – 6 = 0
(ii) 4x + 3y + 12 = 0
Solution:
(i) 3x – 2y – 6 =0.
x intercept when y = 0
⇒ 3x – 6 = 0
⇒ x = 2
y intercept when x = 0
⇒ 0 – 2y – 6 = 0
⇒ y = -3
(ii) 4x + 3y + 12 = 0. x intercept when y = 0
⇒ 4x + 0 + 12 = 0
⇒ x = -3
y intercept when x = 0
⇒ 0 + 3y + 12 = 0
⇒ y = -4

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Question 14.
Find the equation of a straight line.
(i) passing through (1,-4) and has intercepts which are in the ratio 2 : 5
(ii) passing through (-8, 4) and making equal intercepts on the coordinate axes.
Solution:
(i) Let the x-intercept be 2a and the y intercept 5 a
The equation of a line is
xa + ya = 1 ⇒ x2a + y5a = 1
The line passes through the point (1, -4)
12a + (−4)5a = 1 ⇒ 12a – 45a = 1
Multiply by 10a
(L.C.M of 2a and 5a is 10a)
5 – 8 = 10a ⇒ -3 = 10a
a = −310
The equation of the line is

Multiply by 3
-5x – 2y = 3 ⇒ -5x – 2y – 3 = 0
5x + 2y + 3 = 0
The equation of a line is 5x + 2y + 3 = 0

(ii) Let the x-intercept andy intercept “a”
The equation of a line is
xa + ya = 1
The line passes through the point (-8, 4)
−8a + 4a = 1
−8+4a = 1
-4 = a
The equation of a line is
x−4 + y−4 = 1
Multiply by -4
x + y = -4
x + y + 4 = 0
The equation of the line is x + y + 4 = 0

Question 1.
What is the slope of a line whose inclination with positive direction of x -axis is
(i) 90°
(ii) 0°
Solution:
Here θ = 90°
Slope (m) = tan θ
Slope = tan 90°
= undefined.

(ii) Here θ = 0°
Slope (m) = tan θ
Slope = tan 0°
= 0

Question 2.
What is the inclination of a line whose slope is
(i) 0
(ii) 1
Solution:
(i) m = 0
tan θ = 0 ⇒ θ = 0°
(ii) m = 1 ⇒ tan θ = tan 45° ⇒ 0 = 45°

Question 3.
Find the slope of a line joining the points
(i) (5,5–√) with the origin
(ii) (sin θ, -cos θ) and (-sin θ, cos θ)
Solution:
(i) The given points is (5,5–√) and (0, 0)
Slope of a line = y2−y1x2−x1 = 0−5√0−5
= 5√5=15√

(ii) The given points is (sin θ, -cos θ) and (-sin θ, cos θ)
Slope of a line = y2−y1x2−x1=cosθ+cosθ−sinθ−sinθ
= 2cosθ−2sinθ = – cot θ

Question 4.
What is the slope of a line perpendicular to the line joining A(5,1) and P where P is the mid-point of the segment joining (4,2) and (-6,4).
Solution:
Mid point of XY = (x1+x22,y1+y22) = (4−62,2+42)
= (−22,62) = (-1, 3)

Slope of a line = y2−y1x2−x1 = (3−1−1−5)
= 2−6 = – 13

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Question 5.
Show that the given points are collinear: (-3, -4), (7,2) and (12, 5)
Solution:
The vertices are A(-3, -4), B(7, 2) and C(12, 5)
Slope of a line = y2−y1x2−x1
Slope of AB = 2+47+3 = 610 = 35
Slope of BC = 5−212−7 = 35
Slope of AB = Slope of BC = 35
∴ The three points A,B,C are collinear.

Question 6.
If the three points (3, -1), (a, 3) and (1, -3) are collinear, find the value of a.
Solution:
The vertices are A(3, -1), B(a, 3) and C(1, -3)
Slope of a line = y2−y1x2−x1
Slope of AB = 3+1a−3 = 4a−3
Slope of BC = 3+3a−1 = 6a−1
Since the three points are collinear.
Slope of AB = Slope BC
4a−3 = 6a−1
6 (a – 3) = 4 (a – 1)
6a – 18 = 4a – 4
6a – 4a = -4 + 18
2a = 14 ⇒ a = 142 = 7
The value of a = 7

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Question 7.
The line through the points (-2, a) and (9,3) has slope –12 Find the value of a.
Solution:
The given points are (-2, a) and (9, 3)
Slope of a line = y2−y1x2−x1
– 12 = 3−a9+2 ⇒ – 12 = 3−a11
2(3 – a) = -11 ⇒ 6 – 2a = -11
-2a = -11 – 6 ⇒ -2a = -17 ⇒ a = – 172
∴ The value of a = 172

Question 8.
The line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8,12) and (x, 24). Find the value of x.
Solution:
Find the slope of the line joining the point (-2, 6) and (4, 8)
Slope of line (m1) = y2−y1x2−x1
= 8−64+2 = 26 = 13
Find the slope of the line joining the points (8, 12) and (x, 24)
Slope of a line (m2) = 24−12x−8 = 12x−8
Since the two lines are perpendicular.
m1 × m2 = -1
13 × 12x−8 = -1 ⇒ 123(x−8)=−1
-1 × 3 (x – 8) = 12
-3x + 24 = 12 ⇒ – 3x = 12 -24
-3x = -12 ⇒ x = 123 = 4
∴ The value of x = 4

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Question 9.
Show that the given points form a right angled triangle and check whether they satisfies Pythagoras theorem.
(i) A(1, -4) , B(2, -3) and C(4, -7)
(ii) L(0, 5), M(9,12) and N(3,14)
Solution:
(i) The vertices are A(1, -4), B(2, -3) and C(4, -7)
Slope of a line = y2−y1x2−x1
Slope of AB = −3+42−1 = 11 = 1
Slope of BC = −7+34−2 = −42 = -2
Slope of AC = −7+44−1 = – 33 = -1
Slope of AB × Slope of AC = 1 × -1 = -1

∴ AB is ⊥r to AC
∠A = 90°
∴ ABC is a right angle triangle
Verification:

20 = 2 + 18
20 = 20 ⇒ Pythagoras theorem verified

(ii) The vertices are L(0, 5), M(9, 12) and N(3, 14)
Slope of a line = y2−y1x2−x1
Slope of LM = 12−59−0 = 79

Slope of MN = 14−123−9 = 2−6 = – 13
Slope of LN = 14−53−0 = 93 = 3
Slope of MN × Slope of LN = – 13 × 3 = -1
∴ MN ⊥ LN
∠N = 90°
∴ LMN is a right angle triangle
Verification:

130 = 90 + 40
130 = 130 ⇒ Pythagoras theorem is verified

Question 10.
Show that the given points form a parallelogram:
A (2.5,3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5).
Solution:
Let A(2.5, 3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5) are the vertices of a parallelogram.

Slope of AB = Slope of CD = -1
∴ AB is Parallel to CD ……(1)

Slope of BC = Slope of AD
∴ BC is parallel to AD
From (1) and (2) we get ABCD is a parallelogram.

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Question 11.
If the points A(2, 2), B(-2, -3), C(1, -3) and D(x, y) form a parallelogram then find the value of x and y.
Solution:
Let A(2, 2), B(-2, -3), C(1, -3) and D(x, y) are the vertices of a parallelogram.

Slope of a line = y2−y1x2−x1
Slope of AB = −3−2−2−2 = −5−4 = 54
Slope of BC = −3+3−2−1 = 0−3 = 0
Slope of CD = y+3x−1
Since ABCD is a parallelogram
Slope of AB = Slope of CD
54 = y+3x−1
5(x – 1) = 4 (y + 3)
5x – 5 = 4y + 12
5x – 4y = 12 + 5
5x – 4y = 17 ……(1)
Slope of BC = Slope of AD
0 = y−2x−2
y – 2 = 0
y = 2
Substitute the value of y = 2 in (1)
5x – 4(2) = 17
5x -8 = 17 ⇒ 5x = 17 + 18
5x = 25 ⇒ x = 255 = 5
The value of x = 5 and y = 2.

Question 12.
Let A(3, -4), B(9, -4) , C(5, -7) and D(7, -7). Show that ABCD is a trapezium.
Solution:
Let A(3, -4), B(9, -4), C(5, -7) and D(7, -7) are the vertices of a quadrilateral.

Slope of a line = y2−y1x2−x1
Slope of AB = −4+49−3 = 06 = 0
Slope of BC = −7+45−9 = −3−4 = 34
Slope of CD = −7+77−5 = 02 = 0
Slope of AD = −7+47−3 = −34 = – 34
The slope of AB and CD are equal.
∴ AB is parallel to CD. Similarly the slope of AD and BC are not equal.
∴ AD and BC are not parallel.
∴ The Quadrilateral ABCD is a trapezium.

Question 13.
A quadrilateral has vertices at A(-4, -2), B(5, -1) , C(6, 5) and D(-7, 6). Show that the mid-points of its sides form a parallelogram.
Solution:
Let A(-4, -2), B(5, -1), C(6, 5) and D(-7, 6) are the vertices of a quadrilateral.

Slope of EF = Slope of GH = 710
∴ EF || GH …….(1)
Slope of FG= Slope of EH = – 712
∴ FG || EH ……(2)
From (1) and (2) we get EFGH is a parallelogram.
The mid point of the sides of the Quadrilateral ABCD is a Parallelogram.

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Question 14.
PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS = 2PR. If the coordinates of S and M are (1, 1) and (2, -1) respectively, find the coordinates of P.
Solution:
Slope of a line = y2−y1x2−x1
Slope of SM = 1+11−2 = 2−1 = -2
Slope of PM = 12 (Since SM and PM are ⊥r)
Let the point p be (a,b)
Slope of PM = 12
b+1a−2 = 12 ⇒ a – 2 = 2b + 2

a – 2b = 4
a = 4 + 2b ……(1)
Given QS = 2PR
QS2 = PR
∴ SM = PR
SM = 2PM (PR = 2PM)

Squaring on both sides

∴ (b + 1)2 = 14 ⇒ b + 1 = ± 12
b = 12 – 1 (or) b = – 12 – 1
= – 12 – 1 (or) b = –12 – 1
= – 12 (or) – 32
a = 4 + 2b
a = 4 + 2 (−12)
a = 3
a = 4 + 2 (−32)
a = 4 – 3
a = 1
The point of p is (3,−12) (or) (1,−32)

Tamilnadu, Samacheer, Kalvi, 10th, sslc, Maths, Solutions, Chapter 5, Coordinate Geometry, Ex 5.1,

Question 1.
Find the area of the triangle formed by the points
(i) (1,-1), (-4, 6) and (-3, -5)
(ii) (-10, -4), (-8, -1) and (-3, -5)
Solution:
(i) Let the vertices A (1, -1), B (-4, 6) and C (-3, -5)

= 12 [(6 + 20 + 3) – (4 – 18 – 5)] = 12 [29 – (-19)] = 12 [29 + 19]
= 12 × 48 = 24 sq. units.
Area of ∆ABC = 24 sq. units

(ii) Let the vertices be A(-10, -4), B(-8 -1) and C(-3, -5)

Area of ∆ABC = 12[(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)]
= 12 [(50 + 3 + 32) – (12 + 40 + 10)]

= 12 [85 – (62)] = 12 [23] = 11.5
Area of ∆ACB = 11.5 sq.units

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Question 2.
Determine whether the sets of points are collinear?
(i) (-12,3)
(ii) (a,b + c), (b,c + a) and (c,a + b)
Solution:
(i) Let the points be A (-12,3), B (-5, 6) and C(-8, 8)
Area of ∆ABC = 12 [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y3 + x1y3)]
= 12 [(- 3 – 40 – 24) – (-15 – 48 – 4)]

= 12 [-67 + 67] = 12 × 0 = 0
Area of a ∆ is 0.
∴ The three points are collinear.

(ii) Let the points be A (a, b + c), B (b, c + a) and C (c, a + b)
Area of the triangle = 12 [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)]

Since the area of a triangle is 0.
∴ The given points are collinear.

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Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’

Solution:
(i) Let the vertices be A (0,0) B (p, 8), c (6, 2)
Area of a triangle = 20 sq. units
12 [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)] = 20

12 [(0 + 2p + 0) – (0 + 48 + 0)] = 20
12 [2p – 48] = 20
2p – 48 = 40 ⇒ 2p = 40 + 48
p = 882 = 44
The value of p = 44

(ii) Let the vertices be A (p, p), B (5, 6) and C (5, -2)
Area of a triangle = 32 sq. units
12 [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)] = 32

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Question 4.
In each of the following, find the value of ‘a’ for which the given points are collinear.
(i) (2,3), (4, a) and (6, -3)
(ii) (a, 2 – 2a), (-a + 1, 2a) and (-4 -a, 6 – 2a).
Solution:
(i) Let the points be A (2, 3), B(4, a) and C(6, -3).
Since the given points are collinear.
Area of a triangle = 0
12 [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)] = 0)

(ii) Let the points be A (a, 2 – 2a), B (-a + 1, 2a) C (-4 -a, 6 – 2a).
Since the given points are collinear.
Area of a ∆ = 0

6a2 – 2a – 2 – (-2a2 – 6a + 2) = 0
6a2 – 2a – 2 + 2a2 + 6a – 2 = 0
8a2 + 4a – 4 = 0 (Divided by 4)
2a2 + a – 1 = 0
2a2 + 2a – a – 1 = 0
2a (a + 1) – 1 (a + 1) = 0

(a + 1) (2a – 1) = 0
a + 1 = 0 (or) 2a – 1 = 0
a = -1 (or) 2a = 1 ⇒ a = 12
The value of a = -1 (or) 12

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Question 5.
Find the area of the quadrilateral whose vertices are at
(i) (-9, -2), (-8, -4), (2, 2) and (1, -3)
(ii) (-9, 0), (-8,6), (-1, -2) and (-6, -3)
Solution:
(i) Let the vertices A (-9, -2), B(-8, -4), C(2, 2) and D(1, -3).
Plot the vertices in a graph.

[Note: Consider the points in counter clock wise order]

Area of the Quadrilateral ABDC = 12 [36 + 24 + 2 – 4 – (16 – 4 – 6 – 18)]
= 12 [58 – (-12)] – 12[58 + 12]
= 12 × 70 = 35 sq. units 2
Area of the Quadrilateral = 35 sq. units

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(ii) Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.

= 12 [33 + 35] = 12 × 68 = 34 sq. units
Area of the Quadrilateral = 34 sq. units

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Question 6.
Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (-4, -2), (-3, k), (3, -2) and (2, 3)
Solution:
Let the vertices A (-A, -2), B (-3, k), C (3, -2) and D (2, 3)
Area of the Quadrilateral = 28 sq. units
12 [(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2+ x4y3 + x1y4)] = 28

-7k + 21 = 56
-7k = 56 – 21
-7k = 35 ⇒ 7k = – 35
k = – 357 = -5
The value of k = -5

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Question 7.
If the points A(-3, 9), B(a, b) and C(4, -5) are collinear and if a + b = 1, then find a and b.
Solution:
Since the three points are collinear
Area of a ∆ = 0

-3b – 5a + 36 – 9a – 4b – 15 = 0
-7b – 14a + 21 = 0
(÷ by 7) – b – 2a + 3 = 0
2a + b – 3 = 0

Substitute the value of a = 2 in (2) ⇒ 2 + b = 1
b = 1 – 2 = -1
The value of a = 2 and b = -1

Question 8.
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
Solution:
Let the vertices of the ∆ABC be A(x1,y1), B(x2,y2), C(x3,y3)

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Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.

Solution:

= 12 [(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)]
= 12 [(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)]
= 12 [212 – (-212)]

= 12 [212 + 212] = 12 [424] = 212 sq. units

= 12 [90 – (-90)]

= 12 [90 + 90]
= 12 × 180 = 90 sq. units
Area of the patio = Area of the Quadrilateral ABCD – Area of the Quadrilateral EFGH
= (212 – 90) sq. units
Area of the patio = 122 sq. units

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Question 10.
A triangular shaped glass with vertices at A(-5, -4), B(l, 6) and C(7, -4) has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
Solution:
Given the vertices of the triangular glass is A (-5, -4), B (1, 6), and C (7, -4)

= 12 [(20 + 42 – 4) – (-28 – 4 – 30)]
= 12 [58 – (-62)]
= 12 [58 + 62]
= 12 × 120 = 60 sq. feet
Number of cans to paint 6 square feet = 1
∴ Number of cans = 606 = 10 ⇒ Number of cans = 10

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Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED

Solution:
Area of a triangle = 12 [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)]
(i) Area of ∆AGF = 12 [(-2.5 – 13.5 – 6) – (-13.5 – 1 – 15)]
= 12 [-22 – (-29.5)]

= 12 [-22 + 29.5]
= 12 × 7.5 = 3.75 sq.units

(ii) Area of ∆FED = 12 [(-2 + 4.5 + 3) – (4.5 + 1 – 6)]
= 12 [5.5 – (-0.5)]

= 12 [5.5 + 0.5] = 12 × 6 = 3 sq.units

(iii)

= 12 [(4 + 2 + 0.75 + 9) – (-4 -1.5 – 4.5 -2)]
= 12 [15.75 + 12]
= 12 [27.75] = 13.875
= 13.88 sq. units

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