Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Selina ICSE Solutions

Question 1. Find, which of the following points lie on the line x – 2y + 5 = 0:
(i) (1, 3)
(ii) (0, 5)
(iii) (-5, 0)
(iv) (5, 5)
(v) (2, -1.5)
(vi) (-2, -1.5)
Solution:
(i) Given that, the line is π‘₯– 2𝑦 + 5 = 0
Here, π‘₯ = 1 and 𝑦 = 3
Put the value of x and y in given equation,
(1)– 2(3) + 5 = 0
1– 6 + 5 = 0
βˆ’5 + 5 = 0
0 = 0
Here, 𝐿𝐻𝑆 = 𝑅𝐻𝑆
Hence, the point (1,3) lies on the line π‘₯– 2𝑦 + 5 = 0.
(ii) Given that, the line is π‘₯– 2𝑦 + 5 = 0
Here, π‘₯ = 0 and 𝑦 = 5
Put the value of x and y in given equation,
(0)– 2(5) + 5 = 0
0– 10 + 5 = 0
βˆ’10 + 5 = 0
βˆ’5 = 0
𝐿𝐻𝑆 β‰  𝑅𝐻𝑆
Hence, the point (0,5) does not lies on the line π‘₯– 2𝑦 + 5 = 0.
(iii) Given that, the line is π‘₯– 2𝑦 + 5 = 0
Here, π‘₯ = βˆ’5 and 𝑦 = 0
Put the value of x and y in given equation,
(βˆ’5)– 2(0) + 5 = 0
βˆ’5– 0 + 5 = 0
βˆ’5 + 5 = 0
0 = 0
𝐿𝐻𝑆 = 𝑅𝐻𝑆
Hence, the point (-5,0) lies on the line π‘₯– 2𝑦 + 5 = 0.
(iv) Given that, the line is π‘₯– 2𝑦 + 5 = 0
Here, π‘₯ = 5 and 𝑦 = 5
Put the value of x and y in given equation,
(5)– 2(5) + 5 = 0
5– 10 + 5 = 0
10 βˆ’ 10 = 0
0 = 0
𝐿𝐻𝑆 = 𝑅𝐻𝑆
Hence, the point (5,5) lies on the line π‘₯– 2𝑦 + 5 = 0.
(v) Given that, the line is π‘₯– 2𝑦 + 5 = 0
Here, π‘₯ = 2 and 𝑦 = βˆ’1.5
Put the value of x and y in given equation,
(2)– 2(βˆ’1.5) + 5 = 0
2 + 3 + 5 = 0
10 = 0
𝐿𝐻𝑆 β‰  𝑅𝐻𝑆
Hence, the point (2,-1.5) does not lies on the line π‘₯– 2𝑦 + 5 = 0.
(vi) Given that, the line is π‘₯– 2𝑦 + 5 = 0
Here, π‘₯ = βˆ’2 and 𝑦 = βˆ’1.5
Put the value of x and y in given equation,
(βˆ’2)– 2(βˆ’1.5) + 5 = 0
βˆ’2 + 3 + 5 = 0
6 = 0
𝐿𝐻𝑆 β‰  𝑅𝐻𝑆
Hence, the point (-2,-1.5) does not lies on the line π‘₯– 2𝑦 + 5 = 0.

Question 2. State, true or false:
(i) The line x/2 + y/3 = 0 passes through the point (2,3).
(ii) The line x/2 + y/3 = 0 passes through the point (4,-6).
(iii) The point (8,7) lies on the line y – 7 = 0.
(iv) The point (-3,0) lies on the line x + 3 = 0.
(v) The point (2,a) lies on the line 2x – y = 3, then a = 5.
Solution:

(i) The given line is x/2 + y/3 = 0
Here, x = 2 and y = 3
Put the value of x and y in given equation,

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

2 + (βˆ’2) = 0
0 = 0
Hence, the given statement is true.
(iii) The given line is 𝑦 βˆ’ 7 = 0
Here, π‘₯ = 8 and 𝑦 = 7
Put the value of x and y in given equation,
7 βˆ’ 7 = 0
0 = 0
Hence, the given statement is true.
(iv) The given line is π‘₯ + 3 = 0
Here, π‘₯ = βˆ’3 and 𝑦 = 0
Put the value of x and y in given equation,
βˆ’3 + 3 = 0
0 = 0
Hence, the given statement is true.
(v) The given line is 2π‘₯ βˆ’ 𝑦 = 3
Here, π‘₯ = 2 and 𝑦 = π‘Ž
Put the value of x and y in given equation,
2(2) βˆ’ π‘Ž = 3
4 βˆ’ π‘Ž = 3
βˆ’π‘Ž = 3 βˆ’ 4
βˆ’π‘Ž = βˆ’1
π‘Ž = 1
Hence, the given statement is false.

Question 3. The line given by the equation 2x – y/3 = 7 passes through the point (k,6); calculate the value of k.
Solution:
It is given that, the equation 2x – y/3 = 7 passes through the point (k,6).
Here, x = k and y = 6
Put the value of x and y in given equation,
2(k) – 6/3 = 7
2(k) – 2 = 7
2k-2 = 7
2k = 7 + 2
2k = 9
k = 9/2
k = 4.5
Hence, the value of k is 4.5.

Question 4. For what value of k will the point (3,-k) lie on the line 9x + 4y = 3?
Solution:

It is given that, the equation of the line is 9x + 4y = 3.
Here, x =3 and y =-k
Put the value of x and y in given equation,
9(3) + 4(-k) =3
27 – 4k = 3
4k = 27 – 3
4k = 24
k = 24/4
k = 6
Hence, the value of k is 6.

Question 5. The line 3x/5-2y/3 + 1 = 0 contains the point (m,2m-1); calculate the value of m.
Solution:

It is given that, the equation is 3x/5 – 2y/3+1 = 0.
Here, x=m and y = 2m – 1
Put the value of x and y in given equation,

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

βˆ’11π‘š + 10 = βˆ’1 Γ— 15
βˆ’11π‘š + 10 = βˆ’15
βˆ’11π‘š = βˆ’15 βˆ’ 10
βˆ’11π‘š = βˆ’25
11π‘š = 25
π‘š = 25/11
π‘š = 2(3/11)
Hence, the value of m is 2(3/11)
.
Question 6. Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2)?
Solution:

It is given that, the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2). The co-ordinates of the mid-point of AB are

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Put the value of x and y in given equation,
3x – 5y = 6
3(2) – 5(0) = 6
6 – 0 = 6
6 = 6
Hence, the line 3π‘₯– 5𝑦 = 6 bisect the join of (5,-2) and (-1,2).

Question 7. (i) The line y = 3x – 2 bisects the join of (a, 3) and (2, -5), find the value of a.
(ii) The line x – 6y + 11 = 0 bisects the join of (8, -1) and (0, k). Find the value of k.
Solution:

(i) It is given that, the line 𝑦 = 3π‘₯– 2 bisect the join of (a, 3) and (3, -5). The co-ordinates of the mid-point of AB are

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 8. (i) The point (-3, 2) lies on the line ax + 3y + 6 = 0, calculate the value of a.
(ii) The line y = mx + 8 contains the point (-4, 4), calculate the value of m.
Solution:

(i) It is given that, the point (-3, 2) lies on the line ax + 3y + 6 = 0.
Here, π‘₯ = βˆ’3 and 𝑦 = 2
Put the value of x and y in given equation,
π‘Ž(βˆ’3) + 3(2) + 6 = 0
βˆ’3π‘Ž + 12 = 0
3π‘Ž = 12
π‘Ž = 4
Hence, the value of a is 4.
(ii) It is given that, the line y = mx + 8 contains the point (-4, 4).
Here, π‘₯ = βˆ’4 and 𝑦 = 4
Put the value of x and y in given equation,
4 = βˆ’4π‘š + 8
4π‘š = 4
π‘š = 1
Hence, the value of m is 1.

Question 9. The point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3. Does P lie on the line x – 5y + 15 = 0?
Solution:

It is given that, point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3. By section formula,

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

(π‘₯, 𝑦) = 0,3
Here, π‘₯ = 0 and 𝑦 = 3
Put the value of x and y in given equation,
π‘₯– 5𝑦 + 15 = 0
0– 5(3) + 15 = 0
0– 15 + 15 = 0
– 15 + 15 = 0
0 = 0
Hence, the point P lies on the line π‘₯– 5𝑦 + 15 = 0.

Question 10. The line segment joining the points (5,-4) and (2,2) is divided by the point Q in the ratio 1: 2. Does the line x – 2y = 0 contain Q?
Solution:
It is given that, point P divides the join of (5, -4) and (2, 2) in the ratio 1: 2. By section formula,

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

(x,y)=4,-2
Here, x=4 and y=-2
Put the value of x and y in given equation,
x – 2y = 0
4 – 2(-2) = 0
4 + 4 = 0
8 β‰  0
Hence, the point Q does not lie on the line x – 2y = 0.

Question 11. Find the point of intersection of the lines:
4x + 3y = 1 and 3x – y + 9 = 0. If this point lies on the line (2k – 1)x – 2y = 4; find the value of k.

Solution:
It is given that,
4x + 3y = 1…..….(1)
3x – y + 9 = 0 ………(2)
From equation (1) we get, the value of x,
4x + 3y = 1

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

π‘₯ = βˆ’2
Here, π‘₯ = βˆ’2 and 𝑦 = 3
Put the value of x and y in given equation,
(2π‘˜β€“ 1)π‘₯– 2𝑦 = 4.
(2π‘˜β€“ 1)(βˆ’2)– 2(3) = 4
βˆ’4π‘˜ + 2– 6 = 4
βˆ’4π‘˜ = 8
π‘˜ = βˆ’2
Hence, the value of k is -2.

Question 12. Show that the lines 2x + 5y = 1,x – 3y = 6 and x + 5y + 2 = 0 are concurrent.
Solution:
When two or more lines meet at a single location, they are said to be concurrent.
The point of intersection of the first two lines,
2π‘₯ + 5𝑦 = 1 … . … . (1)
π‘₯– 3𝑦 = 6 … … . . (2)
Form equation (2) we get the value of (2),
π‘₯– 3𝑦 = 6
π‘₯ = 6 + 3𝑦 … … … . (3)
Put the value of (3) in equation (1)
2π‘₯ + 5𝑦 = 1
2(6 + 3𝑦) + 5𝑦 = 1
12 + 6𝑦 + 5𝑦 = 1
11𝑦 = 1 βˆ’ 12
11𝑦 = βˆ’11
𝑦 = βˆ’1
Put the value of y in equation (3)
π‘₯ = 6 + 3𝑦
π‘₯ = 6 + 3(βˆ’1)
π‘₯ = 6 βˆ’ 3
π‘₯ = 3
Hence, the point of intersection is (3, -1).
x + 5y + 2 = 0, then the given lines will be concurrent.
Here, x = 3 and y = -1
Put the value of x and y in equation π‘₯ + 5𝑦 + 2 = 0
π‘₯ + 5𝑦 + 2 = 0
3 + 5(βˆ’1) + 2 = 0
5– 5 = 0
0 = 0
(3, -1) also lie on the third line.
Hence, the given lines are concurrent

Exercise 14B

Question 1. Find the slope of the line whose inclination is:
(i) 0Β°
(ii) 30Β°
(iii) 72Β°30’
(iv) 46Β°
Solution:

(i) tan 0Β°=0
(ii) tan 30°=1/√3
(iii) tan 72Β°30’=3.1716
(iv) tan 46Β°=1.0355

Question 2. Find the inclination of the line whose slope is:
(i) 0
(ii) √3
(iii) 0.7646
(iv) 1.0875

Solution:
(i) ta n⁑0° = 0
Ɵ = 0°
(ii) tan Ɵ = √3
Ɵ = 60°
(iii) tan Ɵ = 0.7646
Ɵ = 37Β° 24′
(iv) tan Ɵ =1.0875
Ɵ = 47Β° 24′

Question 3. 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:

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 4. Find the slope of the line parallel to AB if:
(i) A = (-2, 4) and B = (0, 6)
(ii) A = (0, -3) and B = (-2, 5)
Solution:

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 5. 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) A = (0, -5) and B = (-2, 4)

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slop of the line perpendicular to AB = -1/Slope of AB
Slop of the line perpendicular to AB = -1/-1
Slop of the line perpendicular to AB = -1/-1
Slop of the line perpendicular to AB = 1

Question 6. The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.
Solution:

It is given that, the line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a).

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

According to question, lines are parallel,
1 = π‘Žβˆ’5/5
1 Γ— 5 = π‘Ž βˆ’ 5
5 = π‘Ž βˆ’ 5
5 + 5 = π‘Ž
π‘Ž = 10
Hence, the value of a is 10.

Question 7. The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.
Solution:

It is given that, the line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1)

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

By cross-multiplication,
-1(6) = 6(2 – a)
-6 =12 – 6a
6a = 12 + 6
a =18/6
a =3
Hence, the value of a is 3.

Question 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:
It is given that, the points are A(4,-2), B(-4,4) and C(10,6) are the vertices of a right-angled triangle.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line
Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 9. 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:
It is given that, the points A(4, 5), B(1, 2), C(4, 3) and D(7, 6) are the vertices of a parallelogram.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line
Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 10. (-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.
Solution:

It is given that, A(-2, 4), B(4, 8), C(10, 7) and D(11, -5) are the vertices of a quadrilateral.
Let us assumed that, P, Q, R and S be the mid-points of AB, BC, CD and DA .

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line
Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line
Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 11. Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.
Solution:

It is given that, the points P (a, b + c), Q (b, c + a) and R (c, a + b).

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 12. Find x, if the slop of the line joining (x,2) and (8,-11) is -3/4.
Solution:

It is given that, the slop of the line joining (x,2) and (8,-11) is (-3)/4.
Let us assumed that, A = (x,2) and B= (8,-11)

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 13. The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slope of all its sides.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of AB = 0. As the slope of any line parallel to x-axis is 0.
Ξ”ABC is an equilateral triangle ∠A = 60Β°
Slope of AC = tan 60° = √3. As the slope of any line parallel to x-axis is 0.
Slope of BC = -tan 60Β° = βˆ’βˆš3. As the slope of any line parallel to x-axis is 0.

Question 14. The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides. Also find:
(i) the slope of the diagonal AC,
(ii) the slope of the diagonal BD.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Solution:
It is given that, the side AB of a square ABCD is parallel to the x-axis.
The slop of any line parallel to x-axis is 0.
Slope of AB = 0
According to figure,
AB | | CD
Slope of AB = Slope of CD = 0
BC is perpendicular to AB,
Slope of BC = βˆ’1/π‘†π‘™π‘œπ‘π‘’ π‘œπ‘“ 𝐴𝐡
Slope of BC = βˆ’1/0
Slope of BC = not defined
AD is perpendicular to AB,
Slope of AD = βˆ’1/π‘†π‘™π‘œπ‘π‘’ π‘œπ‘“ 𝐴𝐡
Slope of AD = βˆ’1/0
Slope of AD = not defined
(i) Diagonal of a square is 45Β°, Diagonal AC makes an angle of 45Β° with the positive direction of x-axis.
Slope of AC = tan45Β°
Slope of AC = 1
(ii) Diagonal of a square is 45Β°, Diagonal AC makes an angle ofβˆ’45Β° with the positive direction of x-axis.
Slope of AC = tan (βˆ’45Β°)
Slope of AC = βˆ’1

Question 15. A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find:
(i) the slope of the altitude of AB,
(ii) the slope of the median AD, and
(iii) the slope of the line parallel to AC.
Solution:

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line
Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 16. The slope of the side BC of a rectangle ABCD is 2/3. Find:
(i) the slope of the side AB.
(i) the slope of the side AD.

Solution:
(i) It is given that, the slope of the side BC of a rectangle ABCD is 2/3.
Here, BC is perpendicular to AB.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Q17. Find the slope and the inclination of the line AB if:
(i) A = (-3,-2) and B = (1,2)
(ii) A = (0,-√3) and B = (3,0)
(ii) A = (-1,2√3) and B = (-2,√3)

Solution:
(i) Co-ordinates of A (-3,-2) and B (1,2)

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line
Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 18. The points (-3, 2), (2, -1) and (a, 4) are collinear. Find a.
Solution:
It is given that, points A(-3, 2), B(2, -1) and C(a, 4) are collinear.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 19. The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K.
Solution:
It is given that, the points A(K, 3), B(2, -4) and C(-K+1, -2) are collinear.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 20. Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C. Which segment appears to have the steeper slope, AB or AC? Justify your conclusion by calculating the slopes of AB and AC.
Solution:

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

The formula of slope of the line = 𝑦2βˆ’π‘¦1/π‘₯2βˆ’π‘₯1
Co-ordinates of A(1, 1) and B(4, 7)
Slope of AB = 7βˆ’1/4βˆ’1
Slope of AB = 6/3
Slope of AB = 2
Co-ordinates of A(1, 1) and C(4, 10)
Slope of AC = 10βˆ’1/4βˆ’1
Slope of AC = 9/3
Slope of AC = 3
We can say that, the line with greater slope is steeper.
Hence, AC has steeper slope.

Exercise 14C

Question 1. Find the equation of a line whose: y-intercept = 2 and slope = 3.
Solution:

It is given that, y-intercept c = 2 and slope m = 3
Substituting the values of c and m in the equation 𝑦 = π‘šπ‘₯ + 𝑐
Hence, the required equation is 𝑦 = 3π‘₯ + 2

Question 2. Find the equation of a line whose: y-intercept = -1 and inclination πŸ’πŸ“Β°
Solution:
It is given that, y-intercept c = -1 and inclination = 45Β°
The slope of m is tan 45Β° = 1
Substituting the values of c and m in the equation 𝑦 = π‘šπ‘₯ + 𝑐
Hence, the required equation is 𝑦 = π‘₯ βˆ’ 1.

Question 3. Find the equation of the line whose slope isΒ βˆ’4/3Β and which passes through (-3, 4).
Solution:
It is given that, the slope isΒ βˆ’4/3.
The equation passes through (-3,4) = (π‘₯1, 𝑦1)
Substitution the values in 𝑦 βˆ’ 𝑦1Β = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 4 = βˆ’4
3 (π‘₯ βˆ’ (βˆ’3))
𝑦 βˆ’ 4 = βˆ’4
3 (π‘₯ + 3)
3(𝑦 βˆ’ 4) = βˆ’4(π‘₯ + 3)
3𝑦 βˆ’ 12 = βˆ’4π‘₯ βˆ’ 12
4π‘₯ + 3𝑦 = 0
Hence, the required equation is 4π‘₯ + 3𝑦 = 0

Question 4. Find the equation of a line which passes through (5,4) and makes an angle of πŸ”πŸŽΒ° with the positiveΒ direction of the x-axis.
Solution:
The slope of the line tan60° is √3.
The equation passes through (5,4) = (π‘₯1, 𝑦1)
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 4 = √3(π‘₯ βˆ’ 5)
𝑦 βˆ’ 4 = √3π‘₯ βˆ’ 5√3
𝑦 = √3π‘₯ βˆ’ 5√3 + 4
Hence, the required equation is 𝑦 = √3π‘₯ βˆ’ 5√3 + 4.

Question 5. Find the equation of the line passing through:
(i) (0, 1) and (1, 2)
(ii) (-1, -4) and (3, 0).
Solution:
(i) The given co-ordinates of are (0, 1) and (1, 2)
The formula of slope of the line = 𝑦2βˆ’π‘¦1/π‘₯2βˆ’π‘₯1
Slope of the line =Β 2βˆ’1/1βˆ’0
Slope of the line =Β 1/1
Slope of the line = 1
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 1 = 1(π‘₯ βˆ’ 0)
𝑦 βˆ’ 1 = π‘₯
𝑦 = π‘₯ + 1
Hence, the required equation is 𝑦 = π‘₯ + 1.
(ii) The given co-ordinates of are (-1, -4) and (3, 0)

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ (βˆ’4) = 1(π‘₯ βˆ’ (βˆ’1))
𝑦 + 4 = 1(π‘₯ + 1)
𝑦 + 4 = π‘₯ + 1
𝑦 = π‘₯ βˆ’ 3
Hence, the required equation is 𝑦 = π‘₯ βˆ’ 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:

(i) It is given that, the co-ordinates of two points P and Q are (2, 6) and (-3, 5)

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

(ii) The equation of PQ,
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 6 = 1
5 (π‘₯ βˆ’ 2)
5(𝑦 βˆ’ 6) = π‘₯ βˆ’ 2
5𝑦 βˆ’ 30 = π‘₯ βˆ’ 2
5𝑦 βˆ’ π‘₯ = βˆ’2 + 30
5𝑦 βˆ’ π‘₯ = 28
5𝑦 = 28 + π‘₯
Hence, the required equation is 5𝑦 = 28 + π‘₯.
(iii) Let us assumed that, the co-ordinates of the point where PQ intersects the x-axis at point 𝐴(π‘₯, 0).
Put the value of 𝑦 = 0 in the above equation.
5𝑦 = 28 + π‘₯
5(0) = 28 + π‘₯
0 = 28 + π‘₯
π‘₯ = βˆ’28
Hence, the value of x is -28.

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) It is given that, co-ordinates of two points A and B are (-3,4) and (2,-1).
The formula of slope of the line = 𝑦2βˆ’π‘¦1/π‘₯2βˆ’π‘₯1
Slope of the line =Β βˆ’1βˆ’4/2+3
Slope of the line =Β βˆ’5/5
Slope of the line = βˆ’1
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ (βˆ’1) = βˆ’1(π‘₯ βˆ’ 2)
𝑦 + 1 = βˆ’1(π‘₯ βˆ’ 2)
𝑦 + 1 = βˆ’π‘₯ + 2
𝑦 + 1 = βˆ’π‘₯ + 2
π‘₯ + 𝑦 = 2 βˆ’ 1
π‘₯ + 𝑦 = 1
(ii) Let us assumed that, the co-ordinates of the point where AB intersects the y-axis at point 𝐴(0, 𝑦).
Put the value of π‘₯ = 0 in the above equation.
π‘₯ + 𝑦 = 1
0 + 𝑦 = 1
𝑦 = 1
Hence, the value of y is 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.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Solution:
It is given that, the angle is 45Β°
Slope of line AB = π‘‘π‘Žπ‘› 45Β°
Slope of line AB = 1
The line AB passes through P(3,4).
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 4 = 1(π‘₯ βˆ’ 3)
𝑦 βˆ’ 4 = π‘₯ βˆ’ 3
𝑦 = π‘₯ βˆ’ 3 + 4
𝑦 = π‘₯ + 1
Again,
The angle is 60Β°
Slope of line CD = π‘‘π‘Žπ‘› 60Β°
Slope of line CD = √3
The line AB passes through P(3,4).
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 4 = √3(π‘₯ βˆ’ 3)
𝑦 βˆ’ 4 = π‘₯√3 βˆ’ 3√3
𝑦 = π‘₯√3 βˆ’ 3√3 + 4

Question 9. In Ξ”ABC,Β AΒ = (πŸ‘, πŸ“),Β BΒ = (7,Β 8) and π‘ͺ = (1, βˆ’10). Find the equation of the median through A.
Solution:

It is given that, the vertices of Ξ”ABC,Β AΒ = (3,Β 5),Β BΒ = (7,Β 8) andΒ CΒ = (1, βˆ’Β 10).

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of the line = βˆ’6
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 5 = βˆ’6(π‘₯ βˆ’ 3)
𝑦 βˆ’ 5 = βˆ’6π‘₯ + 18
6π‘₯ + 𝑦 = 18 + 5
6π‘₯ + 𝑦 = 18 + 5
6π‘₯ + 𝑦 = 23

Question 10. 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.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Solution:
It is given that, ABCD is a parallelogram.
∠A + ∠B = 180°
60° + ∠B = 180°
∠B = 180Β° βˆ’ 60Β°
∠B = 120°
Slope of line CD = tan 60Β°
Slope of line CD = tan (90Β° + 30Β°)
Slope of line CD = cot 30Β°
Slope of line CD = √3
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 5 = √3(π‘₯ βˆ’ 7)
𝑦 βˆ’ 5 = √3π‘₯ βˆ’ 7√3
𝑦 = √3π‘₯ βˆ’ 7√3 + 5
It is given that, AB is parallel to the x-axis.
CD||AB
Slope of line AB = 0
Slope of line AB = Slope of line CD
Substitution the values in 𝑦 βˆ’ 𝑦1 = m(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 5 = 0(π‘₯ βˆ’ 7)
𝑦 βˆ’ 5 = 0
𝑦 = 5

Question 11. Find the equation of the straight line passing through origin and the point of intersection of the lines XΒ + 2Y =Β 7 and X– Y = 4.
Solution:

It is given that, the point of intersection of the lines π‘₯ + 2𝑦 = 7 and π‘₯– 𝑦 = 4.
π‘₯ + 2𝑦 = 7 … … … . (𝑖)
π‘₯– 𝑦 = 4 … … … . . (𝑖𝑖)
From equation (ii) get the value of x,
π‘₯ = 4 + 𝑦 … … … . . (𝑖𝑖𝑖)
Put the value of x in equation (i)
(4 + 𝑦) + 2𝑦 = 7
4 + 𝑦 + 2𝑦 = 7
3𝑦 = 3
𝑦 = 3/3
𝑦 = 1
Put the value of y in equation (iii)
π‘₯ = 4 + 1
π‘₯ = 5
The required line passes through (0,0) and (5,1)
The formula of slope of the line = 𝑦2βˆ’π‘¦1/π‘₯2βˆ’π‘₯1
Slope of the line =Β 1βˆ’0/5βˆ’0
Slope of the line =Β 1/5
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 0 = 1/5 (π‘₯ βˆ’ 0)
5𝑦 = π‘₯
5𝑦 = π‘₯
0 = π‘₯ βˆ’ 5𝑦
π‘₯ βˆ’ 5𝑦 = 0

Question 12. In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A. Also, find the equation of the line through vertex B and parallel to AC.
Solution:
It is given that, the co-ordinated of Δ𝐴𝐡𝐢 are (4,7), (-2,3) and (0,1).
Let us assumed that, AD is the median through vertex A

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of the line AD = 1
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 2 = 1(π‘₯ βˆ’ (βˆ’1))
𝑦 βˆ’ 2 = 1(π‘₯ + 1)
𝑦 βˆ’ 2 = π‘₯ + 1
𝑦 βˆ’ π‘₯ = 1 + 2
𝑦 = 3 + π‘₯
Slope of the line AC = 1βˆ’7/0βˆ’4
Slope of the line AC = βˆ’6/βˆ’4
Slope of the line AC = 3/2
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 3 = 3/2 (π‘₯ βˆ’ (βˆ’2))
2(𝑦 βˆ’ 3) = 3(π‘₯ + 2)
2𝑦 βˆ’ 6 = 3π‘₯ + 6
2𝑦 βˆ’ 3π‘₯ = 6 + 6
2𝑦 = 3π‘₯ + 12

Question 13. A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.
Solution:

It is given that, The co-ordinates A(0, 3), B(4, 4) and C(8, 0).

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of line perpendicular to BC = 1
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 3 = 1(π‘₯ βˆ’ 0)
𝑦 = π‘₯ + 3
Hence, the required equation is 𝑦 = π‘₯ + 3

Question 14. Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1,4) and (2, 3).
Solution:

Let us assumed that, the co-ordinates are A(1, 4), B(2,3) and C(-1,2)

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of line perpendicular to BC = 1
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 2 = 1(π‘₯ βˆ’ (βˆ’1))
𝑦 βˆ’ 2 = (π‘₯ + 1)
𝑦 βˆ’ 2 = π‘₯ + 1
𝑦 = π‘₯ + 1 + 2
𝑦 = π‘₯ + 3
Hence, the required equation is 𝑦 = π‘₯ + 3

Question 15. Find the equation of the line, whose:
(i) x-intercept = 5 and y-intercept = 3
(ii) x-intercept = -4 and y-intercept = 6
(iii) x-intercept = -8 and y-intercept = -4
Solution:

(i) x-intercept = 5 and y-intercept = 3
x-intercept = 5 means corresponding point on x-axis is (5,0)
y-intercept = 3 means corresponding point on y-axis is (0,3)

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

5𝑦 = βˆ’3(π‘₯ βˆ’ 5)
5𝑦 = βˆ’3π‘₯ + 15
5𝑦 + 3π‘₯ = 15
Hence, the required equation is 5𝑦 + 3π‘₯ = 15
(ii) x-intercept = -4 and y-intercept = 6
x-intercept = -4 means corresponding point on x-axis is (-4,0)
y-intercept = 6 means corresponding point on y-axis is (0,6)

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of the line = 6/4
Slope of the line = 3/2
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 0 = 3
2 (π‘₯ βˆ’ (βˆ’4))
𝑦 = 3
2 (π‘₯ + 4)
2𝑦 = 3π‘₯ + 12
Hence, the required equation is 2𝑦 = 3π‘₯ + 12
(iii) x-intercept = -8 and y-intercept = -4
x-intercept = -8 means corresponding point on x-axis is (-8,0)
y-intercept = -4 means corresponding point on y-axis is (0,-4)

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 16. Find the equation of the line whose slope is βˆ’5/6 and x-intercept is 6.
Solution:

It is given that, x-intercept = 6 means corresponding point on x-axis is (6, 0).
Slope of the line = βˆ’5/6
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 0 = βˆ’5
6 (π‘₯ βˆ’ 6)
6𝑦 = βˆ’5(π‘₯ βˆ’ 6)
6𝑦 = βˆ’5π‘₯ + 30
6𝑦 + 5π‘₯ = 30

Question 17. Find the equation of the line with x-intercept 5 and a point on it (-3, 2).
Solution:

It is given that, x-intercept = 5 means corresponding point on x-axis is (-3, 2).

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of the line = 2/βˆ’8
Slope of the line = βˆ’1/4
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 0 = βˆ’1
4 (π‘₯ βˆ’ 5)
4𝑦 = βˆ’1(π‘₯ βˆ’ 5)
4𝑦 = βˆ’π‘₯ + 5
4𝑦 + π‘₯ = 5
Hence, the required equation is 4𝑦 + π‘₯ = 5

Question 18. Find the equation of the line through (1, 3) and making an intercept of 5 on the y-axis.
Solution:

It is given that, y-intercept = 5 means corresponding point on y-axis is (0, 5).

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of the line = βˆ’2/1
Slope of the line = βˆ’2
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 5 = βˆ’2(π‘₯ βˆ’ 0)
𝑦 βˆ’ 5 = βˆ’2(π‘₯)
𝑦 βˆ’ 5 = βˆ’2π‘₯
𝑦 + 2π‘₯ = 5
Hence, the required equation is 𝑦 + 2π‘₯ = 5

Question 19. Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axis.
Solution:
Let us assumed that, AB and CD be two equally inclined lines.
Slope of line AB,
π‘‘π‘Žπ‘› 45Β° = 1
(π‘₯1, 𝑦1) = (βˆ’2,0)
Substitution the values in 𝑦 βˆ’ 𝑦1Β = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 0 = 1(π‘₯ βˆ’ (βˆ’2))
𝑦 = 1(π‘₯ + 2)
𝑦 = π‘₯ + 2
Let us assumed that, AB and CD be two equally inclined lines.
Slope of line AB,
π‘‘π‘Žπ‘› βˆ’ 45Β° = βˆ’1
(π‘₯1, 𝑦1) = (βˆ’2,0)
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 0 = βˆ’1(π‘₯ + 2)
𝑦 = βˆ’1(π‘₯ + 2)
𝑦 = βˆ’π‘₯ βˆ’ 2
𝑦 + π‘₯ + 2 = 0

Question 20. The line through P(5,3) intersects y-axis at Q.
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co-ordinates of Q.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Solution:
(i) The equation of the y-axis is x = 0
It is given that the required line through P(5,3) intersects the y-axis at Q and the angle of inclination is 45˚.
Hence, the slope of line PQ = π‘‘π‘Žπ‘› 45Β° = 1
(ii) The equation of a line passing through the point A(π‘₯1, 𝑦1) with slope β€˜m’ is
The equation of the line passing through the point P(5,3) with slope 1 is
(π‘₯1, 𝑦1) = (5,3)
Substitution the values in 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 3 = 1(π‘₯ βˆ’ 5)
𝑦 βˆ’ 3 = π‘₯ βˆ’ 5
𝑦 βˆ’ π‘₯ = βˆ’5 + 3
𝑦 βˆ’ π‘₯ = βˆ’2
π‘₯ βˆ’ 𝑦 = 2
(iii) The equation of the line PQ is π‘₯ βˆ’ 𝑦 = 2
It is given that, the line intersects with y-axis π‘₯ = 0.
Substituting π‘₯ = 0 in the equation π‘₯ βˆ’ 𝑦 = 2
Put the value of π‘₯
0 βˆ’ 𝑦 = 2
𝑦 = βˆ’2
Hence, the co-ordinates points of intersection Q are (0.-2).

Exercise 14D

Question 1. Find the slope and y-intercept of the line:
(i) y = 4
(ii) axΒ  –by = 0
(iii) 3xΒ –Β 4y = 5
Solution:
(i) 𝑦 = 4
Substituting this equation with 𝑦 = π‘šπ‘₯ + 𝑐
Slope of line = 0
Put the value of y in the above equation,
4 = 0π‘₯ + 𝑐
4 = 𝑐
𝑐 = 4
y-intercept = c = 4
(ii) π‘Žπ‘₯– 𝑏𝑦 = 0
π‘Žπ‘₯ = 𝑏𝑦
π‘Ž/𝑏 π‘₯ = 𝑦
𝑦 = (π‘Ž/𝑏)π‘₯
Substituting this equation with 𝑦 = π‘šπ‘₯ + 𝑐
Slope of line =Β π‘Ž/𝑏

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 2. The equation of a line x – y = 4. Find its slope and y-intercept. Also, find its inclination.
Solution:

It is given that, the equation of a line is π‘₯ βˆ’ 𝑦 = 4
βˆ’π‘¦ = 4 βˆ’ π‘₯
𝑦 = π‘₯ βˆ’ 4
Substituting this equation with 𝑦 = π‘šπ‘₯ + 𝑐
Slope of line = 1
Slope of line = 1
y-intercept = c = -4
Let us assumed that, the inclination be Ɵ
Slope of line = 1
π‘‘π‘Žπ‘›ΖŸ = π‘‘π‘Žπ‘›45˚
Ɵ = 45˚
Hence, the inclination us 45˚.

Question 3. (i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0?
(ii) Is the line x – 3y = 4 perpendicular to the line 3x – y = 7?
(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1?
(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.
Solution:

(i) It is given that, 3π‘₯ + 4𝑦 + 7 = 0
3π‘₯ + 4𝑦 + 7 = 0
4𝑦 = βˆ’7 βˆ’ 3π‘₯

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of the line is 4/3
Hence, the product of the slope of the two lines is -1, the line are perpendicular to each other.
(ii) It is given that, π‘₯ βˆ’ 3𝑦 = 4
βˆ’3𝑦 = βˆ’4 βˆ’ π‘₯
βˆ’3𝑦 = βˆ’(4 + π‘₯)
3𝑦 = π‘₯ + 4
𝑦 = 1/3 π‘₯ + 4/3
Slope of the line is 1/3
3π‘₯– 𝑦 = 7
– 𝑦 = βˆ’3π‘₯ βˆ’ 7
𝑦 = 3π‘₯ + 7
Slope of the line = 3
Product of slope of the two lines 11 β‰  βˆ’1
Hence, the lines are not perpendicular to each other.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of the line = βˆ’ 1/2
Product of slope of the two lines 3 β‰  βˆ’1
Hence, the lines are not perpendicular to each other.
(iv) It is given that, the slope of the line by (1,4) and (x,2) is 2.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

By cross-multiplication,
2(π‘₯ βˆ’ 1) = βˆ’2
π‘₯ βˆ’ 1 = βˆ’1
π‘₯ = βˆ’1 + 1
π‘₯ = 0

Question 4. Find the slope of the line which is parallel to:
(i) x + 2y + 3 = 0

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Solution.
(i) x + 2y + 3 = 0
2y = -x – 3

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 5. Find the slope of the line which is perpendicular to:

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Solution:

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

By cross-multiplication,
y = 2(x + 3)
y = 2x + 6
Slope of the line = 2
Slope of the line which is perpendicular to given line = -1/Slope of the given line
Slope of the line which is perpendicular to given line = -1/2

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 6. (i) Lines 2x – by + 3 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
(ii) Lines mx + 3y + 7 = 0 and 5x – ny – 3 = 0 are perpendicular to each other. Find the relation connecting m and n.
Solution:
(i) The given equation is 2x-by+3=0
2x + 3 = by
by = 2x + 3

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line
Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 7. Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3p = 4 are perpendicular to each other.
Solution:

The given equation is 2π‘₯ βˆ’ 𝑦 + 5 = 0
βˆ’π‘¦ = βˆ’7 βˆ’ 2π‘₯
𝑦 = 2π‘₯ + 7
Slope of the line = 2
𝑝π‘₯ + 3𝑦 = 4
3𝑦 = 4 βˆ’ 𝑝π‘₯
3𝑦 = βˆ’π‘π‘₯ + 4

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 8. The equation of a line AB is 2x – 2y + 3 = 0.
(i) Find the slope of the line AB.
(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.
Solution:

(i) The given equation is 2π‘₯– 2𝑦 + 3 = 0
– 2𝑦 = βˆ’2π‘₯ βˆ’ 3
2𝑦 = 2π‘₯ + 3
𝑦 = π‘₯ + 3/2
Slope of the line = 1
(ii) Let the required angle be Ɵ
Slope of tan Ɵ = 1
tan Ɵ = tan 45˚
Ɵ = 45˚
Hence, the value of Ɵ is 45˚

Question 9. The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.
Solution:

The given equation is 4π‘₯ + 3𝑦 = 9
3𝑦 = βˆ’4π‘₯ + 9

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 10. If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
Solution:

The given equation is 𝑦 = 3π‘₯ + 7
Slope of the line = 3
again,
2𝑦 + 𝑝π‘₯ = 3
2𝑦 = βˆ’π‘π‘₯ + 3

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 11. The line through A(-2,3) and B(4,b) is perpendicular to the line 2x – 4y = 5. Find the value of b.
Solution:
It is given that, the line through A(-2,3) and B(4,b) is perpendicular to the line 2x – 4y = 5

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 12. Find the equation of the line through (-5, 7) and parallel to:
(i) x-axis
(ii) y-axis
Solution:
(i) The slope of the line parallel to x-axis is 0.
(π‘₯1, 𝑦1) = (βˆ’5, 7)
Substituting this equation with 𝑦 βˆ’ 𝑦1Β = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 7 = π‘š(π‘₯ βˆ’ (βˆ’5))
𝑦 βˆ’ 7 = π‘š(π‘₯ + 5)
𝑦 βˆ’ 7 = 0(π‘₯ + 5)
𝑦 βˆ’ 7 = 0
𝑦 = 7
Hence, the value of y is 7.
(ii) Here, the slope of the line parallel to y-axis in not defined. The slope of the line is tan90˚ and hence the given
line is parallel to y-axis.
(π‘₯1, 𝑦1) = (βˆ’5, 7)
Required equation of the line is
π‘₯ βˆ’ π‘₯1Β = 0
π‘₯ βˆ’ (βˆ’5) = 0
π‘₯ + 5 = 0

Question 13. (i) Find the equation of the line passing through (5, -3) and parallel toΒ x – 3yΒ =Β 4.
(ii) Find the equation of the line parallel to the lineΒ 3x + 2y = 8 and passing through the point (0, 1).
Solution:

(i) π‘₯ βˆ’ 3𝑦 = 4
βˆ’3𝑦 = 4 βˆ’ π‘₯
βˆ’3𝑦 = βˆ’(βˆ’4 + π‘₯)
3𝑦 = π‘₯ βˆ’ 4

Slope of the line = 1/3
Required equation of the line passing through (5,-3)
Substituting this equation with 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)

3(𝑦 + 3) = π‘₯ βˆ’ 5
3𝑦 + 9 = π‘₯ βˆ’ 5
3𝑦 βˆ’ π‘₯ = βˆ’5 βˆ’ 9
3𝑦 βˆ’ π‘₯ = βˆ’14
βˆ’(π‘₯ βˆ’ 3𝑦) = βˆ’14
π‘₯ βˆ’ 3𝑦 = 14
π‘₯ βˆ’ 3𝑦 βˆ’ 14 = 0
(ii) 3π‘₯ + 2𝑦 = 8
2𝑦 = 8 βˆ’ 3π‘₯
2𝑦 = βˆ’3π‘₯ + 8

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 14. Find the equation of the line passing through (-2, 1) and perpendicular to πŸ’π’™ + πŸ“π’š = πŸ”.
Solution:

The given equation is 4π‘₯ + 5𝑦 = 6
4π‘₯ + 5𝑦 = 6
5𝑦 = βˆ’4π‘₯ + 6

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

4(𝑦 βˆ’ 1) = 5(π‘₯ + 2)
4𝑦 βˆ’ 4 = 5π‘₯ + 10
4𝑦 βˆ’ 5π‘₯ = 14
5π‘₯ βˆ’ 4𝑦 = βˆ’14
5π‘₯ βˆ’ 4𝑦 + 14 = 0

Question 15. Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).
Solution:

Let the A(6,3) and B(0,3)
The perpendicular bisector of a line is perpendicular to the line and it bisects the line, that it, it passes through the mid-point of the line.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 16. In the following diagram, write down:
(i) the co-ordinates of the points A, B and C.
(ii) the equation of the line through A and parallel to BC.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Solution:
(i) The co-ordinates of point A, B and C are (2,3), (-1,2) and (3,0) respectively.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

The equation of the line is 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 3 = βˆ’1/2 (π‘₯ βˆ’ 2)
2(𝑦 βˆ’ 3) = βˆ’1(π‘₯ βˆ’ 2)
2𝑦 βˆ’ 6 = βˆ’π‘₯ + 2
π‘₯ + 2𝑦 = 2 + 6
π‘₯ + 2𝑦 = 8
Hence, the required line is π‘₯ + 2𝑦 = 8.

Question 17. B(-5, 6) and D(1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.
Solution:

We know that, in a rhombus diagonal bisect each other at right angle. Let O be the point of intersection of the diagonals AC and BD.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

The equation of the line is 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ 5 = 3(π‘₯ βˆ’ (βˆ’2))
𝑦 βˆ’ 5 = 3(π‘₯ + 2)
𝑦 βˆ’ 5 = 3π‘₯ + 6
𝑦 = 3π‘₯ + 6 + 5
𝑦 = 3π‘₯ + 11

Question 18. A = (7, -2) and C = (-1, -6) are the vertices of square ABCD. Find the equations of diagonal BD and of diagonal AC.
Solution:

In a square, diagonals bisect each other at right angle. Let O be the point of intersection of the diagonals AC and BD.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of BD = βˆ’2
The equation of the line is 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)
𝑦 βˆ’ (βˆ’4) = βˆ’2(π‘₯ βˆ’ 3)
𝑦 + 4 = βˆ’2(π‘₯ βˆ’ 3)
𝑦 + 4 = βˆ’2π‘₯ + 6
𝑦 + 2π‘₯ = 6 βˆ’ 4
2π‘₯ + 𝑦 = 2

Question 19. A (1, -5), B (2, 2) and C (-2, 4) are the vertices of triangle ABC, find the equation of:
(i) the median of the triangle through A.
(ii) the altitude of the triangle through B.
(iii) the line through C and parallel to AB.
Solution:

(i) The median through A will pass through the mid-point of BC. Let AD be the median through A.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line
Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line
Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Slope of AB = βˆ’7
Slope of the line parallel to AB = Slope of AB = 7
The equation of the line is 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ x1)
𝑦 βˆ’ 4 = 7(π‘₯ βˆ’ (βˆ’2))
𝑦 βˆ’ 4 = 7(π‘₯ + 2)
𝑦 βˆ’ 4 = 7π‘₯ + 14
𝑦 βˆ’ 7π‘₯ = 14 + 4
𝑦 βˆ’ 7π‘₯ = 18
Hence, the required equation is 𝑦 βˆ’ 7π‘₯ = 18.

Question 20. (i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.
(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin.
Solution:

(i) The given equation is 2𝑦 = 3π‘₯ + 5

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line
Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line

Question 21. Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5. Find its coordinates of point P. Also, find the equation of the line through P and parallel to 3x + 5y = 7.
Solution:

It is given that, Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5.

Selina ICSE Class 10 Maths Solutions Chapter 14 Equation Of A Line