Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14B Solutions
Exercise – 14B
(1) Find the slope of the line whose inclination is:
(i) 0°
(ii) 30°
(iii) 72°
(iv) 46°
Solution:-
(i) Θ = 0°
Slope of x – axis parallel to x – axis
∴ m = tan Θ = tan 0
= 0
(ii) Θ = 30°
Slope of inclination of 30°
Tan 30° = 1/√3
(iii) Θ = 72° 30’
Slope of inclination of 72° 30’
Tan 72° 30’ = 3.1716
(iv) Θ = 46°
Slope of inclination of 46°
Tan 46° = 1.0355
(Q2) Find the inclination of the line whose slope is:
(i) 0
(ii) √3
(iii) 0.7646
(iv) 1.0875
Solution:
Given that
Slope = 0°
We have to find inclination that is Θ
We have to use formula of slope-
Slope (m) = tan Θ
∴ tan Θ = o
Θ = 0°
(ii) Given that,
Slope = √3
We have to find inclination that is ‘Θ’
We have to use formula slope (m) = tan Θ
∴ tan Θ = √3
Tan Θ = 60°
∴ Θ = 60°
(iii) Given that,
Slope = 0.7646
We have to find inclination that is ‘Θ’
We have to use formula –
Slope (m) = tan Θ
∴ tan Θ = 0.7646
∴ Θ = 37° 24’
(iv) Given that,
Slope = 1.0875
We have to find inclination that is ‘Θ’
We have to use formula –
Slope (m) = tan Θ
∴ tan Θ = 1.0875
∴ Θ = 47° 24’
(Q3) Find the slope of the line passing through the following pairs of points:
(i) (-2, -3) and (1, 2)
(ii) (-4, 0) and origin
(iii) (a, -b) and (b, -a)
Solution:
(i) (-2, -3) = (x1, y1) and (1, 2) = (x2, y2)
We have to find slope so, we have to use formula-
Slope = (y2 – y1)/(x2 – x1)
= (2-(-3))/(1–(-2))
= (2+3)/(1+2)
Slope = 5/3
(ii) (-4, 0) = (x1, y1) and (0, 0) = (x2 , y2) (origin).
We have to find slope so, we have to use formula –
Slope = (y2 – y1)/(x2 – x1)
Slope = (0-0)/(0 –(-4))
= 0/4
Slope = 0
(iii) (a1 – b) = (x1, y1) and (b1 – a) = (x2, y2)
We have to find slope so, we have to use formula-
Slope = (y₂ – y₁)/x₂ – x₁)
= (-a-(-b))/(b – a)
= (-a+b)/(b – a)
= (b-a)/(b-a)
Slope = 1
(Q4) Find the slope of the line parallel to AB if:
(i) A = (-2, 4) and B= (0, 6)
(ii) A = (01 -3) and B = (-2, 5)
Solution:
(i) Given that,
A= (-2, 4) = (x1, y1) and B = (0, 6) = (x2, y2)
Now, we have to find slope of AB –
∴ Slope of AB = (y2 – y1)/(x2 – x1)
= (6-4)/(0-(-2))
= 2/2
Slope of AB = 1
(ii) Given that,
A = (0, -3) = (x1, y1) and B = (-2, 5) = (x2, y2)
Now, we have to find slope of AB –
∴ Slope of AB = (y2 – y1)/(x2 – x1)
= (5–(-3))/(-2 -0)
= (5+3)/-2
= 8/-2
∴ Slope of AB = – 4
(Q5) Find the slope of the line perpendicular to AB if:
(i) A= (0, -5) and B = (-2, 4)
(ii) A = (3, -2) and B = (-1, 2)
Solution:-
(i) Given that,
A = (0, -5) = (x1, y1)
And B = (-2, 4) = (x2, y2)
First we have to find slope
So, we have to use formula –
Slope (m1) = y2–y1/x2–x1
= 4-(-5)/-2 -0
= 4+5/-2
m1 = 9/-2
Now,
Line (l) perpendicular to AB –
∴ m2 × m1 = – 1
m2 × 9/-2 = – 1
m2 = -1 × -2/9
m2 = 2/9
(ii) Given that,
A = (3, -2) and B = (-1, 2)
A = (3, -2) = x1, y1) and B = (-1, 2) = (x2, y2)
First we have to find slope –
So, we have to use formula –
Slope (m1) = y2 – y1/x2 – x1
= 2 – (-2)/- 1 – 3
= 2+2/-4
m = 4/-4
m1 = -1
Now, Line (l) perpendicular to line AB –
∴ m2 × m1 = -1
m2 × (-1) = -1
m2 = -1/-1
m2 = 1
∴ Slope of the line perpendicular to AB is y
(Q6) The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a) find a.
Solution:-
Given that, the line passing through (0, 2) and (-3, -1)
First we have to find slope of line l1 –
Slope (m1) = y2 – y1/ x2 – x1
(0, 2) = (x1, y1) and (-3, -1) = (x2, y2)
Slope (m1) = -1-2/-3-0
= -3/-3
Slope (m1) = 1
Now, we have to find slope of line l2 –
Slope (m2) = y2 – y1/x2 – x1
(-1, 5) = (x1, y1) and (4, a) = (x2, y2)
Slope (m2) = a-5/4 – (-1)
= a-5/4+1
m2 = a – 5/5
If the two lines are parallel the slope of these two lines are same or equal
m1 =m2
1 = a-5/5
5 = a – 5
5 + 5 = a
∴ a = 10
(Q7) The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a
Solution:
Given that, the line passing through (-4, -2) and (2, -3)
First we have to find the slope of line l1 –
Slope (m1) = y2 – y1/x2 – x1
(-4, -2) = (x1, y1) and (2, -3) = (x2, y2)
Slope (m1) = -3 – (-2)/2 – (-4)
= -3 +2/2+4
m1 = -1/6
Now, we have to find slope of line l2–
Slope (m2) = y2 – y1/x2 – x1
(a, 5) = (x1, y1) and (2, -1) = (x2, y2)
Slope = (m2) = -1-5/2-a
= -6/2-a
But these two lines l1 and l2 are perpendicular-
∴ m1 × m2 = -1
-1/6 × -6/2-a = -1
-1/6 = -1 × (2-a/-6)
-1/6 = +2–a/6
-6/6 = 2-a
-1 = 2-a
-1 -2 = -a
-3 = -a
Multiply by (-) on both side –
a = 3
(8) Without using the distance formula, show that the points A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right angled triangle.
Solution:-
Given that,
The points
A (4, -2),
B (-4, 4) and C (10, 6)
First we have to find the slope of AB –
A (4, -2) = (x1, y1) and B (-4, 4) = (x2, y2)
Slope (m1) = y2 – y1/x2 – x1
= 4 – (-2)/- 4 – 4
= 4+2/-8
Slope (m1) = 6/-8
= 3/-4
Now, we have to find slope of BC –
B (-4, 4) = (x1, y1) and C (10, 6) = (x2, y2)
Slope (m2) = y2 – y1/x2 – x1
= 6-4/ 10-(-4)
= 2/10+4
= 2/14
m2 = 1/7
And now, we have to find slope of AC –
A (4, -2) = (x1, y1) and C (10, 6) = (x2, y2)
Slope (m3) = y2 – y1/x2 – x1
= 6- (-2)/10-4
= 6+2/6
= 8/6
m3 = 4/3
We have to seen that, slope m1 and slope m3 is Slope of AB = -1/slope of AC
∴ That is AB⊥AC.
∴ AB perpendicular AC
Means the given points A, B and C are the vertices of right angled triangle.
(Q9) Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7,6) are the vertices of a parallelogram.
Solution:
Given points are A (4, 5), B (1, 2), C (4, 3) and D (7, 6)
We have to show that the points A, B, C and D are the vertices of a parallelogram.
So, now we have to find slope of AB and CD –
Slope of AB = y2 – y1/x2 –x1
A (4, 5) = (x1, y1) and B = (1, 2) = (x2, y2)
Slope of AB = 2-5/1-4
= -3/-3
= 1
Slope of CD = y2 – y1/x2 – x1
C (4, 3) = (x1, y1) and D (7, 6) = (x2, y2)
Slope of CD = 6-3/7-4
= 3/3
= 1
We have seen that slope of AB = slope of CD
So, AB parallel to CD
AB || CD
Now, we have to find slope of BC and DA
Slope of BC = y2 – y1/x2 – x1
B = (1, 2) = (x1, y1) and C = (4, 3) = (x2, y2)
Slope of BC = 3-2/4-1
= 1/3
Slope of DA = y2 – y1/x2 – x1
D = (7, 6) = (x1, y1) and A = (4, 5) = (x2, y2)
Slope of DA = 5-6/4-7
= -1/-3
= 1/3
We have seen that slope BC = slope of DA
So, BC parallel DA
BC || DA
∴ ABCD is a parallelogram
(10) Show that the points P (a,b + c) Q ( b, c + a) are collinear
Solution:
If P, Q, R are collinear then the slope of PQ and QR are same.
Now, we have to find the slope of PQ –
Slope of PQ = y2 – y1/x2 – x1
P (a, b + c) = (x1, y1) and Q (b, c + a) = (x2, y2)
Slope of PQ = c + a – (b + c)/b – a
= c+a–b-c/b-a
= a-b/ b-a
= -1
Slope of QR = y2 – y1/x2 – x1
Q (b, c + a) = (x1, y1) and
R (c, a + b) = (x2, y2)
Slope of QR = a + b – (c + a)/c – b
= a + b – c – a/c – b
= b-c/c-b
Slope of QR = -1
∴ We can seen that,
The slope of PQ and slope of QR are same.
∴ the points P, Q and R are collinear.
Here is your solution of Selina Concise Class 10 Math Chapter 14 Equation of a line
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