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

Question 1. Find, which of the following points lie on the line x β 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₯β 2π¦ + 5 = 0
Here, π₯ = 1 and π¦ = 3
Put the value of x and y in given equation,
(1)β 2(3) + 5 = 0
1β 6 + 5 = 0
β5 + 5 = 0
0 = 0
Here, πΏπ»π = ππ»π
Hence, the point (1,3) lies on the line π₯β 2π¦ + 5 = 0.
(ii) Given that, the line is π₯β 2π¦ + 5 = 0
Here, π₯ = 0 and π¦ = 5
Put the value of x and y in given equation,
(0)β 2(5) + 5 = 0
0β 10 + 5 = 0
β10 + 5 = 0
β5 = 0
πΏπ»π β  ππ»π
Hence, the point (0,5) does not lies on the line π₯β 2π¦ + 5 = 0.
(iii) Given that, the line is π₯β 2π¦ + 5 = 0
Here, π₯ = β5 and π¦ = 0
Put the value of x and y in given equation,
(β5)β 2(0) + 5 = 0
β5β 0 + 5 = 0
β5 + 5 = 0
0 = 0
πΏπ»π = ππ»π
Hence, the point (-5,0) lies on the line π₯β 2π¦ + 5 = 0.
(iv) Given that, the line is π₯β 2π¦ + 5 = 0
Here, π₯ = 5 and π¦ = 5
Put the value of x and y in given equation,
(5)β 2(5) + 5 = 0
5β 10 + 5 = 0
10 β 10 = 0
0 = 0
πΏπ»π = ππ»π
Hence, the point (5,5) lies on the line π₯β 2π¦ + 5 = 0.
(v) Given that, the line is π₯β 2π¦ + 5 = 0
Here, π₯ = 2 and π¦ = β1.5
Put the value of x and y in given equation,
(2)β 2(β1.5) + 5 = 0
2 + 3 + 5 = 0
10 = 0
πΏπ»π β  ππ»π
Hence, the point (2,-1.5) does not lies on the line π₯β 2π¦ + 5 = 0.
(vi) Given that, the line is π₯β 2π¦ + 5 = 0
Here, π₯ = β2 and π¦ = β1.5
Put the value of x and y in given equation,
(β2)β 2(β1.5) + 5 = 0
β2 + 3 + 5 = 0
6 = 0
πΏπ»π β  ππ»π
Hence, the point (-2,-1.5) does not lies on the line π₯β 2π¦ + 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,

2 + (β2) = 0
0 = 0
Hence, the given statement is true.
(iii) The given line is π¦ β 7 = 0
Here, π₯ = 8 and π¦ = 7
Put the value of x and y in given equation,
7 β 7 = 0
0 = 0
Hence, the given statement is true.
(iv) The given line is π₯ + 3 = 0
Here, π₯ = β3 and π¦ = 0
Put the value of x and y in given equation,
β3 + 3 = 0
0 = 0
Hence, the given statement is true.
(v) The given line is 2π₯ β π¦ = 3
Here, π₯ = 2 and π¦ = π
Put the value of x and y in given equation,
2(2) β π = 3
4 β π = 3
βπ = 3 β 4
βπ = β1
π = 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 β 4k = 3
4k = 27 β 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,

β11π + 10 = β1 Γ 15
β11π + 10 = β15
β11π = β15 β 10
β11π = β25
11π = 25
π = 25/11
π = 2(3/11)
Hence, the value of m is 2(3/11)
.
Question 6. Does the line 3x β 5y = 6 bisect the join of (5, -2) and (-1, 2)?
Solution:

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

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

Question 7. (i) The line y = 3x β 2 bisects the join of (a, 3) and (2, -5), find the value of a.
(ii) The line x β 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 π¦ = 3π₯β 2 bisect the join of (a, 3) and (3, -5). The co-ordinates of the mid-point of AB are

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₯ = β3 and π¦ = 2
Put the value of x and y in given equation,
π(β3) + 3(2) + 6 = 0
β3π + 12 = 0
3π = 12
π = 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₯ = β4 and π¦ = 4
Put the value of x and y in given equation,
4 = β4π + 8
4π = 4
π = 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 β 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,

(π₯, π¦) = 0,3
Here, π₯ = 0 and π¦ = 3
Put the value of x and y in given equation,
π₯β 5π¦ + 15 = 0
0β 5(3) + 15 = 0
0β 15 + 15 = 0
β 15 + 15 = 0
0 = 0
Hence, the point P lies on the line π₯β 5π¦ + 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 β 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,

(x,y)=4,-2
Here, x=4 and y=-2
Put the value of x and y in given equation,
x β 2y = 0
4 β 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 β y + 9 = 0. If this point lies on the line (2k β 1)x β 2y = 4; find the value of k.

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

π₯ = β2
Here, π₯ = β2 and π¦ = 3
Put the value of x and y in given equation,
(2πβ 1)π₯β 2π¦ = 4.
(2πβ 1)(β2)β 2(3) = 4
β4π + 2β 6 = 4
β4π = 8
π = β2
Hence, the value of k is -2.

Question 12. Show that the lines 2x + 5y = 1,x β 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₯ + 5π¦ = 1 β¦ . β¦ . (1)
π₯β 3π¦ = 6 β¦ β¦ . . (2)
Form equation (2) we get the value of (2),
π₯β 3π¦ = 6
π₯ = 6 + 3π¦ β¦ β¦ β¦ . (3)
Put the value of (3) in equation (1)
2π₯ + 5π¦ = 1
2(6 + 3π¦) + 5π¦ = 1
12 + 6π¦ + 5π¦ = 1
11π¦ = 1 β 12
11π¦ = β11
π¦ = β1
Put the value of y in equation (3)
π₯ = 6 + 3π¦
π₯ = 6 + 3(β1)
π₯ = 6 β 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₯ + 5π¦ + 2 = 0
π₯ + 5π¦ + 2 = 0
3 + 5(β1) + 2 = 0
5β 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°30β
(iv) 46Β°
Solution:

(i) tan 0Β°=0
(ii) tan 30Β°=1/β3
(iii) tan 72Β°30β=3.1716
(iv) tan 46Β°=1.0355

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

Solution:
(i) ta nβ‘0Β° = 0
Ζ = 0Β°
(ii) tan Ζ = β3
Ζ = 60Β°
(iii) tan Ζ = 0.7646
Ζ = 37Β° 24′
(iv) tan Ζ =1.0875
Ζ = 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:

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:

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)

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).

According to question, lines are parallel,
1 = πβ5/5
1 Γ 5 = π β 5
5 = π β 5
5 + 5 = π
π = 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)

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.

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.

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 .

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).

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)

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

Slope of AB = 0. As the slope of any line parallel to x-axis is 0.
ΞABC is an equilateral triangle β A = 60Β°
Slope of AC = tan 60Β° = β3. As the slope of any line parallel to x-axis is 0.
Slope of BC = -tan 60Β° = ββ3. 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.

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 = β1/πππππ ππ π΄π΅
Slope of BC = β1/0
Slope of BC = not defined
Slope of AD = β1/πππππ ππ π΄π΅
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 ofβ45Β° with the positive direction of x-axis.
Slope of AC = tan (β45Β°)
Slope of AC = β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:

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.

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

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

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.

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.

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:

The formula of slope of the line = π¦2βπ¦1/π₯2βπ₯1
Co-ordinates of A(1, 1) and B(4, 7)
Slope of AB = 7β1/4β1
Slope of AB = 6/3
Slope of AB = 2
Co-ordinates of A(1, 1) and C(4, 10)
Slope of AC = 10β1/4β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 π¦ = ππ₯ + π
Hence, the required equation is π¦ = 3π₯ + 2

Question 2. Find the equation of a line whose: y-intercept = -1 and inclination ππ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 π¦ = ππ₯ + π
Hence, the required equation is π¦ = π₯ β 1.

Question 3. Find the equation of the line whose slope isΒ β4/3Β and which passes through (-3, 4).
Solution:
It is given that, the slope isΒ β4/3.
The equation passes through (-3,4) = (π₯1, π¦1)
Substitution the values in π¦ β π¦1Β = π(π₯ β π₯1)
π¦ β 4 = β4
3 (π₯ β (β3))
π¦ β 4 = β4
3 (π₯ + 3)
3(π¦ β 4) = β4(π₯ + 3)
3π¦ β 12 = β4π₯ β 12
4π₯ + 3π¦ = 0
Hence, the required equation is 4π₯ + 3π¦ = 0

Question 4. Find the equation of a line which passes through (5,4) and makes an angle of ππΒ° with the positiveΒ direction of the x-axis.
Solution:
The slope of the line tan60Β° is β3.
The equation passes through (5,4) = (π₯1, π¦1)
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 4 = β3(π₯ β 5)
π¦ β 4 = β3π₯ β 5β3
π¦ = β3π₯ β 5β3 + 4
Hence, the required equation is π¦ = β3π₯ β 5β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 =Β π¦2βπ¦1/π₯2βπ₯1
Slope of the line =Β 2β1/1β0
Slope of the line =Β 1/1
Slope of the line = 1
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 1 = 1(π₯ β 0)
π¦ β 1 = π₯
π¦ = π₯ + 1
Hence, the required equation is π¦ = π₯ + 1.
(ii) The given co-ordinates of are (-1, -4) and (3, 0)

Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β (β4) = 1(π₯ β (β1))
π¦ + 4 = 1(π₯ + 1)
π¦ + 4 = π₯ + 1
π¦ = π₯ β 3
Hence, the required equation is π¦ = π₯ β 3

Question 6. The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find:
(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)

(ii) The equation of PQ,
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 6 = 1
5 (π₯ β 2)
5(π¦ β 6) = π₯ β 2
5π¦ β 30 = π₯ β 2
5π¦ β π₯ = β2 + 30
5π¦ β π₯ = 28
5π¦ = 28 + π₯
Hence, the required equation is 5π¦ = 28 + π₯.
(iii) Let us assumed that, the co-ordinates of the point where PQ intersects the x-axis at point π΄(π₯, 0).
Put the value of π¦ = 0 in the above equation.
5π¦ = 28 + π₯
5(0) = 28 + π₯
0 = 28 + π₯
π₯ = β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 =Β π¦2βπ¦1/π₯2βπ₯1
Slope of the line =Β β1β4/2+3
Slope of the line =Β β5/5
Slope of the line = β1
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β (β1) = β1(π₯ β 2)
π¦ + 1 = β1(π₯ β 2)
π¦ + 1 = βπ₯ + 2
π¦ + 1 = βπ₯ + 2
π₯ + π¦ = 2 β 1
π₯ + π¦ = 1
(ii) Let us assumed that, the co-ordinates of the point where AB intersects the y-axis at point π΄(0, π¦).
Put the value of π₯ = 0 in the above equation.
π₯ + π¦ = 1
0 + π¦ = 1
π¦ = 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.

Solution:
It is given that, the angle is 45Β°
Slope of line AB = π‘ππ 45Β°
Slope of line AB = 1
The line AB passes through P(3,4).
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 4 = 1(π₯ β 3)
π¦ β 4 = π₯ β 3
π¦ = π₯ β 3 + 4
π¦ = π₯ + 1
Again,
The angle is 60Β°
Slope of line CD = π‘ππ 60Β°
Slope of line CD = β3
The line AB passes through P(3,4).
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 4 = β3(π₯ β 3)
π¦ β 4 = π₯β3 β 3β3
π¦ = π₯β3 β 3β3 + 4

Question 9. In ΞABC,Β AΒ = (π, π),Β BΒ = (7,Β 8) and πͺ = (1, β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, βΒ 10).

Slope of the line = β6
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 5 = β6(π₯ β 3)
π¦ β 5 = β6π₯ + 18
6π₯ + π¦ = 18 + 5
6π₯ + π¦ = 18 + 5
6π₯ + π¦ = 23

Question 10. The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, β A = 60Β° and vertex C = (7,5). Find the equations of BC and CD.

Solution:
It is given that, ABCD is a parallelogram.
β A + β B = 180Β°
60Β° + β B = 180Β°
β B = 180Β° β 60Β°
β B = 120Β°
Slope of line CD = tan 60Β°
Slope of line CD = tan (90Β° + 30Β°)
Slope of line CD = cot 30Β°
Slope of line CD = β3
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 5 = β3(π₯ β 7)
π¦ β 5 = β3π₯ β 7β3
π¦ = β3π₯ β 7β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 π¦ β π¦1 = m(π₯ β π₯1)
π¦ β 5 = 0(π₯ β 7)
π¦ β 5 = 0
π¦ = 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 XβΒ Y = 4.
Solution:

It is given that, the point of intersection of the lines π₯ + 2π¦ = 7 and π₯β π¦ = 4.
π₯ + 2π¦ = 7 β¦ β¦ β¦ . (π)
π₯β π¦ = 4 β¦ β¦ β¦ . . (ππ)
From equation (ii) get the value of x,
π₯ = 4 + π¦ β¦ β¦ β¦ . . (πππ)
Put the value of x in equation (i)
(4 + π¦) + 2π¦ = 7
4 + π¦ + 2π¦ = 7
3π¦ = 3
π¦ = 3/3
π¦ = 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 =Β π¦2βπ¦1/π₯2βπ₯1
Slope of the line =Β 1β0/5β0
Slope of the line =Β 1/5
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 0 = 1/5 (π₯ β 0)
5π¦ = π₯
5π¦ = π₯
0 = π₯ β 5π¦
π₯ β 5π¦ = 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 Ξπ΄π΅πΆ are (4,7), (-2,3) and (0,1).
Let us assumed that, AD is the median through vertex A

Slope of the line AD = 1
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 2 = 1(π₯ β (β1))
π¦ β 2 = 1(π₯ + 1)
π¦ β 2 = π₯ + 1
π¦ β π₯ = 1 + 2
π¦ = 3 + π₯
Slope of the line AC = 1β7/0β4
Slope of the line AC = β6/β4
Slope of the line AC = 3/2
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 3 = 3/2 (π₯ β (β2))
2(π¦ β 3) = 3(π₯ + 2)
2π¦ β 6 = 3π₯ + 6
2π¦ β 3π₯ = 6 + 6
2π¦ = 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).

Slope of line perpendicular to BC = 1
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 3 = 1(π₯ β 0)
π¦ = π₯ + 3
Hence, the required equation is π¦ = 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)

Slope of line perpendicular to BC = 1
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 2 = 1(π₯ β (β1))
π¦ β 2 = (π₯ + 1)
π¦ β 2 = π₯ + 1
π¦ = π₯ + 1 + 2
π¦ = π₯ + 3
Hence, the required equation is π¦ = 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)

5π¦ = β3(π₯ β 5)
5π¦ = β3π₯ + 15
5π¦ + 3π₯ = 15
Hence, the required equation is 5π¦ + 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)

Slope of the line = 6/4
Slope of the line = 3/2
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 0 = 3
2 (π₯ β (β4))
π¦ = 3
2 (π₯ + 4)
2π¦ = 3π₯ + 12
Hence, the required equation is 2π¦ = 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)

Question 16. Find the equation of the line whose slope is β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 = β5/6
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 0 = β5
6 (π₯ β 6)
6π¦ = β5(π₯ β 6)
6π¦ = β5π₯ + 30
6π¦ + 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).

Slope of the line = 2/β8
Slope of the line = β1/4
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 0 = β1
4 (π₯ β 5)
4π¦ = β1(π₯ β 5)
4π¦ = βπ₯ + 5
4π¦ + π₯ = 5
Hence, the required equation is 4π¦ + 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).

Slope of the line = β2/1
Slope of the line = β2
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 5 = β2(π₯ β 0)
π¦ β 5 = β2(π₯)
π¦ β 5 = β2π₯
π¦ + 2π₯ = 5
Hence, the required equation is π¦ + 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π 45Β° = 1
(π₯1, π¦1) = (β2,0)
Substitution the values in π¦ β π¦1Β = π(π₯ β π₯1)
π¦ β 0 = 1(π₯ β (β2))
π¦ = 1(π₯ + 2)
π¦ = π₯ + 2
Let us assumed that, AB and CD be two equally inclined lines.
Slope of line AB,
π‘ππ β 45Β° = β1
(π₯1, π¦1) = (β2,0)
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 0 = β1(π₯ + 2)
π¦ = β1(π₯ + 2)
π¦ = βπ₯ β 2
π¦ + 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.

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 45Λ.
Hence, the slope of line PQ = π‘ππ 45Β° = 1
(ii) The equation of a line passing through the point A(π₯1, π¦1) with slope βmβ is
The equation of the line passing through the point P(5,3) with slope 1 is
(π₯1, π¦1) = (5,3)
Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 3 = 1(π₯ β 5)
π¦ β 3 = π₯ β 5
π¦ β π₯ = β5 + 3
π¦ β π₯ = β2
π₯ β π¦ = 2
(iii) The equation of the line PQ is π₯ β π¦ = 2
It is given that, the line intersects with y-axis π₯ = 0.
Substituting π₯ = 0 in the equation π₯ β π¦ = 2
Put the value of π₯
0 β π¦ = 2
π¦ = β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  βby = 0
(iii) 3xΒ –Β 4y = 5
Solution:
(i) π¦ = 4
Substituting this equation with π¦ = ππ₯ + π
Slope of line = 0
Put the value of y in the above equation,
4 = 0π₯ + π
4 = π
π = 4
y-intercept = c = 4
(ii) ππ₯β ππ¦ = 0
ππ₯ = ππ¦
π/π π₯ = π¦
π¦ = (π/π)π₯
Substituting this equation with π¦ = ππ₯ + π
Slope of line =Β π/π

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

It is given that, the equation of a line is π₯ β π¦ = 4
βπ¦ = 4 β π₯
π¦ = π₯ β 4
Substituting this equation with π¦ = ππ₯ + π
Slope of line = 1
Slope of line = 1
y-intercept = c = -4
Let us assumed that, the inclination be Ζ
Slope of line = 1
π‘ππΖ = π‘ππ45Λ
Ζ = 45Λ
Hence, the inclination us 45Λ.

Question 3. (i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x β 21y + 50 = 0?
(ii) Is the line x β 3y = 4 perpendicular to the line 3x β 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₯ + 4π¦ + 7 = 0
3π₯ + 4π¦ + 7 = 0
4π¦ = β7 β 3π₯

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₯ β 3π¦ = 4
β3π¦ = β4 β π₯
β3π¦ = β(4 + π₯)
3π¦ = π₯ + 4
π¦ = 1/3 π₯ + 4/3
Slope of the line is 1/3
3π₯β π¦ = 7
β π¦ = β3π₯ β 7
π¦ = 3π₯ + 7
Slope of the line = 3
Product of slope of the two lines 11 β  β1
Hence, the lines are not perpendicular to each other.

Slope of the line = β 1/2
Product of slope of the two lines 3 β  β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.

By cross-multiplication,
2(π₯ β 1) = β2
π₯ β 1 = β1
π₯ = β1 + 1
π₯ = 0

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

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

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

Solution:

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

Question 6. (i) Lines 2x β 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 β ny β 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

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

The given equation is 2π₯ β π¦ + 5 = 0
βπ¦ = β7 β 2π₯
π¦ = 2π₯ + 7
Slope of the line = 2
ππ₯ + 3π¦ = 4
3π¦ = 4 β ππ₯
3π¦ = βππ₯ + 4

Question 8. The equation of a line AB is 2x β 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₯β 2π¦ + 3 = 0
β 2π¦ = β2π₯ β 3
2π¦ = 2π₯ + 3
π¦ = π₯ + 3/2
Slope of the line = 1
(ii) Let the required angle be Ζ
Slope of tan Ζ = 1
tan Ζ = tan 45Λ
Ζ = 45Λ
Hence, the value of Ζ is 45Λ

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

The given equation is 4π₯ + 3π¦ = 9
3π¦ = β4π₯ + 9

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 π¦ = 3π₯ + 7
Slope of the line = 3
again,
2π¦ + ππ₯ = 3
2π¦ = βππ₯ + 3

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

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, π¦1) = (β5, 7)
Substituting this equation with π¦ β π¦1Β = π(π₯ β π₯1)
π¦ β 7 = π(π₯ β (β5))
π¦ β 7 = π(π₯ + 5)
π¦ β 7 = 0(π₯ + 5)
π¦ β 7 = 0
π¦ = 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 tan90Λ and hence the given
line is parallel to y-axis.
(π₯1, π¦1) = (β5, 7)
Required equation of the line is
π₯ β π₯1Β = 0
π₯ β (β5) = 0
π₯ + 5 = 0

Question 13. (i) Find the equation of the line passing through (5, -3) and parallel toΒ xΒ βΒ 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₯ β 3π¦ = 4
β3π¦ = 4 β π₯
β3π¦ = β(β4 + π₯)
3π¦ = π₯ β 4

Slope of the line = 1/3
Required equation of the line passing through (5,-3)
Substituting this equation with π¦ β π¦1 = π(π₯ β π₯1)

3(π¦ + 3) = π₯ β 5
3π¦ + 9 = π₯ β 5
3π¦ β π₯ = β5 β 9
3π¦ β π₯ = β14
β(π₯ β 3π¦) = β14
π₯ β 3π¦ = 14
π₯ β 3π¦ β 14 = 0
(ii) 3π₯ + 2π¦ = 8
2π¦ = 8 β 3π₯
2π¦ = β3π₯ + 8

Question 14. Find the equation of the line passing through (-2, 1) and perpendicular to ππ + ππ = π.
Solution:

The given equation is 4π₯ + 5π¦ = 6
4π₯ + 5π¦ = 6
5π¦ = β4π₯ + 6

4(π¦ β 1) = 5(π₯ + 2)
4π¦ β 4 = 5π₯ + 10
4π¦ β 5π₯ = 14
5π₯ β 4π¦ = β14
5π₯ β 4π¦ + 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.

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.

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

The equation of the line is π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 3 = β1/2 (π₯ β 2)
2(π¦ β 3) = β1(π₯ β 2)
2π¦ β 6 = βπ₯ + 2
π₯ + 2π¦ = 2 + 6
π₯ + 2π¦ = 8
Hence, the required line is π₯ + 2π¦ = 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.

The equation of the line is π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 5 = 3(π₯ β (β2))
π¦ β 5 = 3(π₯ + 2)
π¦ β 5 = 3π₯ + 6
π¦ = 3π₯ + 6 + 5
π¦ = 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.

Slope of BD = β2
The equation of the line is π¦ β π¦1 = π(π₯ β π₯1)
π¦ β (β4) = β2(π₯ β 3)
π¦ + 4 = β2(π₯ β 3)
π¦ + 4 = β2π₯ + 6
π¦ + 2π₯ = 6 β 4
2π₯ + π¦ = 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.

Slope of AB = β7
Slope of the line parallel to AB = Slope of AB = 7
The equation of the line is π¦ β π¦1 = π(π₯ β x1)
π¦ β 4 = 7(π₯ β (β2))
π¦ β 4 = 7(π₯ + 2)
π¦ β 4 = 7π₯ + 14
π¦ β 7π₯ = 14 + 4
π¦ β 7π₯ = 18
Hence, the required equation is π¦ β 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 2π¦ = 3π₯ + 5

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.