Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line
Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line
Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line Exercise 14(C)
Exercise 14(C) Page No: 197
Question 1. Find the equation of a line whose:
y – intercept = 2 and slope = 3.
Solution:
Given,
y – intercept = c = 2 and slope = m = 3.
The line equation is given by: y = mx + c
On substituting the values of c and m, we get
y = 3x + 2
The above is the required line equation.
Question 2. Find the equation of a line whose:
y – intercept = -1 and inclination = 45o.
Solution:
Given,
y – intercept = c = -1 and inclination = 45o.
So, slope = m = tan 45o = 1
Hence, on substituting the values of c and m in the line equation y = mx + c, we get
y = x – 1
The above is the required line equation.
Question 3. Find the equation of the line whose slope is -4/3 and which passes through (-3, 4).
Solution:
Given, slope = -4/3
The line passes through the point (-3, 4) = (x1, y1)
Now, on substituting the values in y – y1 = m(x – x1), we have
y – 4 = -4/3 (x + 3)
3y – 12 = -4x – 12
4x + 3y = 0
Hence, the above is the required line equation.
Question 4. Find the equation of a line which passes through (5, 4) and makes an angle of 60o with the positive direction of the x-axis.
Solution:
The slope of the line, m = tan 60o = √3
And, the line passes through the point (5, 4) = (x1, y1)
Hence, on substituting the values in y – y1 = m(x – x1), we have
y – 4 = √3 (x – 5)
y – 4 = √3x – 5√3
y = √3x + 4 – 5√3, which is the required line equation.
Question 5. Find the equation of the line passing through:
(i) (0, 1) and (1, 2) (ii) (-1, -4) and (3, 0)
Solution:
(i) Let (0, 1) = (x1, y1) and (1, 2) = (x2, y2)
So,
Slope of the line = (2 – 1)/ (1 – 0) = 1
Now,
The required line equation is given by,
y – y1 = m(x – x1)
y – 1 = 1(x – 0)
y – 1 = x
y = x + 1
(ii) Let (-1, -4) = (x1, y1) and (3, 0) = (x2, y2)
So,
Slope of the line = (0 + 4)/ (3 + 1) = 4/4 = 1
The required line equation is given by,
y – y1 = m(x – x1)
y + 4 = 1(x + 1)
y + 4 = x + 1
y = x – 3
Question 6. The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find:
(i) the gradient of PQ;
(ii) the equation of PQ;
(iii) the co-ordinates of the point where PQ intersects the x-axis.
Solution:
Given,
The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively.
(i) Gradient of PQ = (5 – 6)/ (-3 – 2) = -1/-5 = 1/5
(ii) The line equation of PQ is given by
y – y1 = m(x – x1)
y – 6 = 1/5 (x – 2)
5y – 30 = x – 2
5y = x + 28
(iii) Let the line PQ intersect the x-axis at point A (x, 0).
So, on putting y = 0 in the line equation of PQ, we get
0 = x + 28
x = -28
Hence, the co-ordinates of the point where PQ intersects the x-axis are A (-28, 0).
Question 7. 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 point where the line AB intersects the y-axis.
Solution:
(i) Given, co-ordinates of two points A and B are (-3, 4) and (2, -1).
Slope = (-1 – 4)/ (2 + 3) = -5/5 = -1
The equation of the line AB is given by:
y – y1 = m(x – x1)
y + 1 = -1(x – 2)
y + 1 = -x + 2
x + y = 1
(ii) Let’s consider the line AB intersect the y-axis at point (0, y).
On putting x = 0 in the line equation, we get
0 + y = 1
y = 1
Thus, the co-ordinates of the point where the line AB intersects the y-axis are (0, 1).
Question 8. The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equation of AB and CD.
Solution:
The slope of line AB = tan 45o = 1
And, the line AB passes through P (3, 4).
Hence, the equation of the line AB is given by
y – y1 = m(x – x1)
y – 4 = 1(x – 3)
y – 4 = x – 3
y = x + 1
The slope of line CD = tan 60o = √3
And, the line CD passes through P (3, 4).
Hence, the equation of the line CD is given by
y – y1 = m(x – x1)
y – 4 = √3 (x – 3)
y – 4 = √3x – 3√3
y = √3x + 4 – 3√3
Question 9. In ΔABC, A = (3, 5), B = (7, 8) and C = (1, -10). Find the equation of the median through A.
Solution:
Given,
Vertices of ΔABC, A = (3, 5), B = (7, 8) and C = (1, -10).
Coordinates of the mid-point D of BC = (x1 + x2) / 2, (y1 + y2) / 2
The slope of AD = (y2 – y1)/ (x2 – x1)
= (-1 – 5)/ (4 – 3)
= -6/1
= -6
Hence, the equation of the median AD through A is given by
y – y1 = m (x – x1)
y – 5 = -6(x – 3)
y – 5 = -6x + 18
6x + y = 23
Question 10. The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, ∠A = 60o and vertex C = (7, 5). Find the equations of BC and CD.
Solution:
Given, ∠A = 60o and vertex C = (7, 5)
As, ABCD is a parallelogram, we have
∠A + ∠B = 180o [corresponding angles]
∠B = 180o – 60o = 120o
Slope of BC = tan 120o = tan (90o + 30o) = cot 30o = √3
So, the equation of line BC is given by
y – y1 = m(x – x1)
y – 5 = √3 (x – 7)
y – 5 = √3x – 7√3
y = √3x + 5 – 7√3
As, CD || AB and AB || x-axis
Slope of CD = Slope of AB = 0 [As slope of x-axis is zero]
So, the equation of the line CD is given by
y – y1 = m(x – x1)
y – 5 = 0(x – 7)
y = 5
Question 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:
The given line equations are:
x + 2y = 7 ….(1)
x – y = 4 ….(2)
On solving the above line equations, we can find the point of intersection of the two lines.
So, subtracting (2) from (1), we get
3y = 3
y = 1
Now,
x = 4 + y = 4 + 1 = 5 [From (2)]
It’s given that,
The required line passes through (0, 0) and (5, 1).
The slope of the line = (1 – 0)/ (5 – 0) = 1/5
Hence, the required equation of the line is given by
y – y1 = m(x – x1)
y – 0 = 1/5(x – 0)
5y = x
x – 5y = 0
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