10 Dec 2020 3:39 pm

**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.

m_{1} × m_{2} = -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

Slope of AD = y−2x−2

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)