Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14C Solutions
Exercise- 14C
(1) Find the equation of line whose y – intercept = 2 and slope = 3
Solution:
Given that,
y – Intercept = 2 x = 0
Slope = 3
m = 3
∴ We have to use –
Equation of line = y – y1/x – x1
(0, 2) = (x1, y1)
Equation of line = y-y1/x – x1
3 = y – 2/x – 0
3x = y – 2
Y = 3x + 2
∴ The required equation of line is y = 3x + 2
(Q2) Find the equation of a line whose y – intercept = -1 and inclination = 45°
Solution:
Given that,
Y-Intercept = -1 x = 0
Inclination (θ) = 45°
We know that,
Slope (m) = tan θ
m = tan 45°
m = y (∵ tan 45° = 1)
We have to use –
Equation of line = y – y1/x – x1
Slope = y – y1 /x – x1
1 = y – y1/x – x1
(0, -1) = (x1, y1)
1 = y – (-1)/x – 0
1 = y+1/x
x = y + 1
y = x – 1
∴ the required equation of line is y – x – 1
(Q3) Find the equation of the line whose slope is -4/-3 and which passes through (-3, 4).
Solution:
Given that,
The line passes through the points (-3, 4) and
Slope (m) = -4/3
∴ Equation of line = y – y1/x – x1
Slope = y – y1/x – x1
(-3, 4) = (x1, y1)
∴ Slope = y – 4/x – (-3)
-4/3 = y – 4/x+3
-4 × (x + 3) = 3(y-4)
-4x – 12 = 3y – 12
-4x – 3y = -12 +12
-4x – 3y = 0
4x + 3y = 0
∴ The required equation of line is 4x + 3y = 0
(Q4) Find the equation of a line which passes through (5, 4) and makes an angle of 60° with the positive direction of the x – axis.
Solution:
Given that, the line passes through the point (5, 4) and Also given that the angle or inclination (θ) = 60°
We know that,
Slope (m) = tan θ
= tan 60°
Slope (m) = √3 (∵ tan 60° = √3)
(5, 4) = (x1, y1)
∴ Equation of line = y – y1 /x – x1
Slope = y – y1/x – x1
√3 = y-4/x-5
√3 (x-5) = y-4
√3x – 5√3 = y – 4
Y = √3x + 4 -5√3
∴ the required equation of line is y = √3x + 4 – 5√3
(Q5) Find the equation of the line passing through
(i) (0, 1) and (1, 2)
(ii) (-1, -4) and (3, 0)
Solution:
Given that, the equation of line passing through the point (0, 1) = (x1, y1) and (1, 2) = (x2, y2)
First we have to find slope of line –
Slope = y2 – y1/x2 – x1
= 2-1/1-0
= 1/1
= 1
∴ m = y – y1/x – x1
y = y-1/x-0
1 = y-1/x
x = y-1
y = x + 1
∴ this is the required equation of line.
(ii) Given that the equation of line passing through the points (-1, -4) = (x1, y1) and (3, 0) = (x2, y2)
First we have to find slope of line-
Slope = y2 – y1/x2 – x1
= 0 – (-4)/3-(-1)
= 0+4/3+1
= 4/4
= 1
Now, m = y – y1/x – x1
1 = y – (-4)/x – (-1)
1 = y+4/x+1
X + 1 = y + 4
y = x – 3
∴ This is the required equation of line.
(Q6) The co-ordinates of two points A and B are (-3, 4) and (2, -1) Find
(i) The equation of AB
(ii) The co-ordinates of the points where the line AB intersects the y – axis.
Solution:
Given that the co- ordinates of two points are A and B are
(-3, 4) = (x1, y1) and (2, -1) = (x2, y2)
First we have to find slope –
Slope = y2 – y1/x2 – x1
= -1-4/2-(-3)
= -5/2+3
= -5/5
Slope = -1
Now we have find equation of line AB –
We know that,
m = y – y1/x – x1
Slope (m) = y-4/x–(-3)
= y-4/x+3
– 1 = y-4/x+3
-1 × (x+3) = y-4
-x – 3 = y – 4
– x – 3 + 4 = y
– x + 1 = y
y = – x + 1
x + y = 1
(iii) According to question the line AB intersect the y – axis at point (0, y).
So, we have to put x = 0 in equation of line –
x + y = 1
0 + y = 1
y = 1
(Q9) In △ABC, A = (3, 5), B = (7, 8) and C =(1, 10). Find the equation of the median through A .
Solution:
First we have to find the co-ordinates of mid – point of P of BC.
P divides the line BC.
We know that,
The mid-point formula –
P = (x1 + x2/2, y1+y2/2)
B = (7, 8) = (x1, y1) and C = (1, -10) = (x2, y2)
P = (7+1/2, 8+(-10)/2)
= (8/2, 8-10/2)
= (4, -2/2)
P = (4, -1)
Now, we have to find the slope of the line AP –
Slope of line AP = y2 – y1/x2 – x1
A = (3, 5) = (x1, y1) and P = (4, -1) = (x2, y2)
AP = -1-5/4-3
= -6/1
AP = -6
Now, we have to find equation of line AP –
m = y – y1/ x – x1
– 6 = y-5/x -3
– 6 × (x-3) = y-5
– 6x + 18 = y-5
– 6x + 18 + 5 = y
– 6x + 23 = y
y + 6x = 23
(Q10) the following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, ∠A = 60° and vertex C = (7, 5). Find the equations of BC and CD.
Solution:
Given that ∠A = 60° and vertex C = (7, 5)
To find the equation of line BC and CD
BC =?
CD =?
For CD,
Θ = 0° {∵ AB || CD || x- axis}
Slope (m) = tan Θ
= tan 0°
= 0
The line passing through the point (7, 5 )
We know that,
m = y – y1/x-x1
(7, 5) = (x1, y1)
m = y-5/x-7
0 = y -5/x-7
0 × (x-7) = y-5
0 = y-5
y = 5
For BC,
The line passing through the point (7, 5)
LA and LB are corresponding angle,
So, ∠A = ∠B
∴ ∠A = 60° (∵ given)
∴ ∠B = 60°
∴ Θ = 60°
Slope (m) = tan Θ
= tan 60°
= √3
We know that,
m = y – y1/x –x1
(y, 5) = (x1, y1)
√3 = y-5/x-7
√3 (x-7) = y-5
√3x – √3 7 = y-5
√3x – 7√3 = y-5
√3x + 5 – 7√3 = y
y = √3x + 5 – 7√3
(11) Find the equation of the straight line passing through origin and the point of intersection of the lines x+2y = 7 and x – y = 4.
Solution:
Intersection points are common
X + 2y = 7…… (1)
X – y = 4 …….. (2) (∵ given)
Subtracting equation (1) and (2) ,
y = 1
Put y = 1 in equation (2)
x – y = 4
x – 1= 4
x = 4 + 1
x = 5
We know that,
y2 – y1/x2 – x1 = y – y1/x – x1
1-0/5-0 = y -0/x -0
1/5 = y/x
X = 5y
∴ x – 5y = 0
Here is your solution of Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14C
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