ML Aggarwal Equation of Straight Line Exe-12.1 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-12.1 Questions for Equation of Straight Line as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Equation of Straight Line Exe-12.1 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-12 | Equation of Straight Line |

Writer / Book | Understanding |

Topics | Solutions of Exe-12.1 |

Academic Session | 2024-2025 |

### Equation of Straight Line Exe-12.1

ML Aggarwal Class 10 ICSE Maths Solutions

**Question -1. ****Find the slope of a line whose inclination is**

(i) 45°

(ii) 30°

**Answer:**

(i) tan 45° = 1

(ii) tan 30° = 1/√3

**Question- 2. ****Find the inclination of a line whose gradient is**

**(i) 1**

**(ii) √3**

**(iii) 1/√3**

**Answer:**

(i) tan θ = 1 ⇒ θ = 45°

(ii) tan θ = √3 ⇒ θ = 60°

(iii) tan θ = 1/√3 ⇒ θ = 30°

**Question -3. ****Find the equation of a straight line parallel 1 to x-axis which is at a distance**

**(i) 2 units above it**

**(ii) 3 units below it.**

**Answer -3**

**(i) A line which is parallel to x-axis is y = a**

⇒ y = 2

⇒ y – 2 = 0

**(ii) A line which is parallel to x-axis is y = a**

⇒ y = -3

⇒ y + 3 = 0

**Question -4. ****Find the equation of a straight line parallel to y-axis which is at a distance of:**

**(i) 3 units to the right**

**(ii) 2 units to the left.**

**Answer:**

(i) The equation of line parallel to y-axis is at a distance of 3 units to the right is x = 3 ⇒ x – 3 = 0

(ii) The equation of line parallel to y-axis at a distance of 2 units to the left is x = -2 ⇒ x + 2 = 0

**Question -5. ****Find the equation of a straight line parallel to y-axis and passing through the point ( – 3, 5).**

**Answer:**

The equation of the line parallel to y-axis passing through ( – 3, 5) to x = -3

⇒ x + 3 = 0

**Question -6. ****Find the equation of the a line whose**

(i) slope = 3, y-intercept = – 5

(ii) slope = -2/7, y-intercept = 3

(iii) gradient = √3, y-intercept = -4/3

(iv) inclination = 30°, y-intercept = 2

**Answer:**

Equation of a line whose slope and y-intercept is given is

y = mx + c

where m is the slope and c is the y-intercept

**(i) y = mx + c**

⇒ y = 3x + (-5)

⇒ y = 3x – 5

**(ii) Given: slope = -2/7, y-intercept = 3**

⇒ y = (-2/7)x + 3

y = (-2x + 21)/7

7y = -2x + 21

Hence, the equation of line is 2x + 7y – 21= 0.

**(iii) Given: gradient = √3, y-intercept = -4/3**

⇒ y = √3x + (-4/3)

y = (3√3x – 4)/3

3y = 3√3x – 4

Hence, the equation of line is 3√3x – 3y – 4 = 0.

**(iv) Given: inclination = 30°, y-intercept = 2**

Slope = tan 30^{o} = 1/√3

⇒ y = (1/√3)x + 2

y = (x + 2√3)/ √3

√3y = x + 2√3

Hence,

the equation of line is x – √3y + 2√3 = 0.

**Question- 7. ****Find the slope and y-intercept of the following lines:**

(i) x – 2y – 1 = 0

(ii) 4x – 5y – 9 = – 0

(iii) 3x +5y + 7 = 0

(iv) (x/3)+(y/4)=1

(v) y – 3 = 0

(vi) x – 3 = 0

**Answer:**

We know that in the equation

y = mx + c, m is the slope and c is the y-intercept.

Now using this, we find,

**(i) x – 2y – 1 = 0**

2y = x – 1

⇒ y = (½) x + (-½)

Hence, slope = ½ and y-intercept = – ½

**(ii) 4x – 5y – 9 = 0**

5y = 4x – 9

⇒ y = (4/5) x + (-9/5)

Hence, slope = 4/5 and y-intercept = -9/5

**(iii) 3x + 5y + 7 = 0**

5y = -3x – 7

⇒ y = (-3/5) x + (-7/5)

Hence, slope = -3/5 and y-intercept = -7/5

**(iv) x/3 + y/4 = 1**

(4x + 3y)/ 12 = 1

4x + 3y = 12

3y = -4x + 12

⇒ y = (-4/3) x + 4

Hence, slope = -4/3 and y-intercept = 4

**(v) y – 3 = 0**

y = 3

⇒ y = (0) x + 3

Hence, slope = 0 and y-intercept = 3

(vi) x – 3 = 0

Here, the slope cannot be defined as the line does not meet y-axis.

**Question -8. ****The equation of the line PQ is 3y – 3x + 7 = 0**

**(i) Write down the slope of the line PQ.**

**(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.**

**Answer:**

Equation of line PQ is 3y – 3x + 7 = 0

Writing in form of y = mx + c

3y = 3x – 7

⇒ y = x + (-7/3)

Here,

(i) Slope = 1

(ii) As tan θ = 1

θ = 45^{o}

Equation of Straight Line Exe-12.1

**ML Aggarwal Class 10 ICSE Maths Solutions**

Page-238

**Question -9. ****The given figure represents the line y = x + 1 and y = √3x – 1. Write down the angles which the lines make with the positive direction of the x-axis. Hence determine θ.**

**Answer:**

Slope of the line y = x + 1 after comparing

it with y = mx + c, m = 1

So, tan θ = 1 ⇒ θ = 45^{o}

And,

The slope of the line: y = √3x – 1 is √3 as m = √3

So, tan θ = √3 ⇒ θ = 60^{o}

Now, in triangle formed by the given two lines and x-axis

Ext. angle = Sum of interior opposite angle

60^{o} = θ + 45^{o}

θ = 60^{o} – 45^{o}

Thus, θ = 15^{o}

**Question- 10. ****Find the value of p, given that the line y/2 = x – p passes through the point ( – 4, 4) **

**Answer :**

Equation of line is **y/2 = x – p**

It passes through the points (-4, 4)

and It will satisfy the equation

4/2 = (-4) – p

2 = -4 – p

p = -4 – 2

Thus, p = -6

**Question -11. ****Given that (a, 2a) lies on the line y/2 = 3x-6 find the value of a.**

**Answer:**

∵ Point (a, 2a) lies on the line

**y/2 = 3x-6**

∴this point will satisfy the equation

2a/2 = 3(a) – 6

a = 3a – 6

2a = 6

Thus, a = 3

**Question- 12. ****The graph of the equation y = mx + c passes through the points (1, 4) and ( – 2, – 5). Determine the values of m and c.**

**Answer:**

Equation of the line is y = mx + c

∴ it passes through the points (1, 4)

∴ 4 = m x 1 + c

⇒ 4 = m + c

hence ⇒ m + c = 4 … (i)

Again it passes through the point (-2, -5)

∴ 5 = m (-2) + c

⇒ 5 = -2 m + c

so ⇒ 2m – c = 5 …(ii)

Adding (i) and (ii)

3m = 9

⇒ m = 3

Substituting the value of m in (i)

3 + c = 4

⇒ c = 4 – 3 = 1

Hence m = 3, c = 1

**Question -13. ****Find the equation of the line passing through the point (2, – 5) and making an intercept of – 3 on the y-axis.**

**Answer:**

A line equation passes through point (2, -5) and makes a y-intercept of -3

We know that,

The equation of line is y = mx + c, where m is the slope and c is the y-intercept

So, we have

y = mx – 3

Now, this line equation will satisfy the point (2, -5)

-5 = m (2) – 3

-5 = 2m – 3

2m = 3 – 5 = -2

⇒ m = -1

Hence, the equation of the line is y = -x + (-3)

⇒ x + y + 3 = 0

**Question- 14. ****Find the equation of a straight line passing through ( – 1, 2) and whose slope is 2/5.**

**Answer :**

Equation of the line will be

(y-2) = 2/5(x+1)

⇒ 5y – 10 = 2x + 2

so ⇒ 2x – 5y + 2 + 10 = 0

hence ⇒ 2x – 5y + 12 = 0

**Question -15. ****Find the equation of a straight line whose inclination is 60° and which passes through the point (0, – 3).**

**Answer:**

The equation of line whose slope is wand passes through a given point is

Here m = tan 60° = √3 and point is (0, -3)

∴ y + 3 = √3 (x – 0)

⇒ y + 3 = √3x

⇒ √3x – y – 3 = 0

**Question- 16. ****Find the gradient of a line passing through the following pairs of points.**

**(i) (0, – 2), (3, 4)**

**(ii) (3, – 7), ( – 1, 8)**

**Answer :**

Gradient of a line (m) = y_{2 }– y_{1} / x_{2 }– x_{1}

**(i) (0, – 2), (3, 4)**

m = (4 + 2)/(3 – 0) = 6/3 = 2

Hence, gradient = 2

**(ii) (3, – 7), (– 1, 8)**

m = (8 + 7)/(-1 – 3) = 15/-4

Hence, gradient = -15/4

**Question- 17. ****The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find :**

**(i) The gradient of EF**

**(ii) The equation of EF**

**(iii) The coordinates of the point where the line EF intersects the x-axis.**

**Answer:**

Co-ordinates of points E (0, 4) and F (3, 7) are given, then

**(i) The gradient of EF**

m = y_{2 }– y_{1} / x_{2 }– x_{1} = (7 – 4)/(3 – 0) = 3/3

⇒ m = 1

**(ii) Equation of line EF is given by,**

y – y_{1} = m (x – x_{1})

y – 7 = 1 (x – 3)

y – 7 = x – 3

x – y + 7 – 3 = 0

Hence, the equation of line EF is x – y + 4 = 0

**(iii) It’s seen that the co-ordinates of point of intersection of EF and the x-axis will be y = 0**

So, substituting the value y = 0 in the above equation

x – y + 4 = 0

x – 0 + 4 = 0

x = -4

Hence,

the co-ordinates are (-4, 0).

**Question- 18. ****Find the intercepts made by the line 2x – 3y + 12 = 0 on the co-ordinate axis.**

**Answer:**

Putting y = 0, we will get the intercept made on x-axis,

2x – 3y + 12 = 0

⇒ 2x – 3 × 0 + 12 = 0, ⇒ 2x – 0 + 2 = 0,⇒ 2x = -12

and ⇒ x = -6

and putting x = 0, we get the intercepts made on y-axis,

2x – 3y + 12 = 0

⇒ 2 × 0 – 3y + 12 = 0

so ⇒ -3y = -12

therefore ⇒ = 4

**Question -19. ****Find the equation of the line passing through the points P (5, 1) and Q (1, – 1). Hence, show that the points P, Q and R (11, 4) are collinear.**

**Answer:**

The two given points are P (5, 1), Q(1, -1).

∴ Slope of the line (m)

= y_{2} – y_{1}/ x_{2} – x_{1}

= -1 – 1/ 1 – 5

= -2/-4 = ½

So, the equation of the line is

y – y_{1} = m (x – x_{1})

y – 1 = ½ (x – 5)

2y – 2 = x – 5

x – 2y – 3 = 0

Now, if point R (11, 4) is collinear to points P and Q then, R (11, 4) should satisfy the line equation

On substituting, we have

11 – 2(4) – 3 = 11 – 8 – 3 = 0

As point R satisfies the line equation,

Hence, P, Q and R are collinear.

**Question- 20. ****Find the value of ‘a’ for which the following points A (a, 3), B (2,1) and C (5, a) are collinear. Hence find the equation of the line. (2014)**

**Answer :**

Given That

A(a, 3), B (2, 1) and C (5, a) are collinear.

Slope of AB = Slope of BC

⇒ -6 = (a – 1) (2 – a) [On cross multiplying]

-6 = 2a – 2 – a^{2} + a

-6 = 3a – a^{2} – 2

a^{2} – 3a – 4 = 0

a^{2} – 4a + a – 4 = 0

a (a – 4) + (a – 4) = 0

(a + 1) (a – 4) = 0

a = -1 or 4

As a = -1 doesn’t satisfy the equation

⇒ a = 4

Now,

Slope of BC = (a – 1)/(5 – 2) = (4 – 1)/3 = 3/3 = 1 = m

So, the equation of BC is

(y – 1) = 1 (x – 2)

y – 1 = x – 2

x – y = -1 + 2

Thus, the equation of BC is x – y = 1.

**Question -21. ****Use a graph paper for this question. The graph of a linear equation in x and y, passes through A ( – 1, – 1) and B (2, 5). From your graph, find the values of h and k, if the line passes through (h, 4) and (½, k). (2005)**

**Answer :**

Points (h, 4) and (½, k) lie on the line passing

through A(-1, -1) and B(2, 5)

From the graph, its clearly seen that

h = 3/2 and

k = 2

**Question -22. ****ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, – 4). Find:**

**(i) the coordinates of A**

**(ii) the equation of the diagonal BD.(2011)**

**Answer :**

Given that

ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4)

O in the point of intersection of the diagonals of the parallelogram.

So, the co-ordinates of O = ([5 + 2]/2, [8 – 4]/2) = (3.5, 2)

Now, for the line AC we have

3.5 = (x + 4)/2 and 2 = (y + 7)/2

7 = x + 4 and 4 = y + 7

x = 7 – 4 and y = 4 – 7

x = 3 and y = -3

Thus, the co-ordinates of A are (3, -3).

**(ii) Equation of diagonal BD is given by**

y – 8 = (-4 – 8)/(2 – 5) × (x – 5)

y – 8 = (-12/-3) × (x – 5)

y – 8 = 4 (x – 5)

y – 8 = 4x – 20

4x – y – 20 + 8 = 0

Hence, the equation of the diagonal is 4x – y – 12 = 0.

**Question- 23. ****In ∆ABC, A (3, 5), B (7, 8) and C (1, – 10). Find the equation of the median through A.(2013)**

**Answer:**

AD is median

⇒ D is mid point of BC

∴ D is ([7 + 1]/2, [8 – 10]/2)

i.e (4, -1)

slope of AD

∆ABC and their vertices A (3, 5), B (7, 8) and C (1, – 10).

And, AD is median

So, D is mid-point of BC

Hence, the co-ordinates of D is ([7 + 1]/2, [8 – 10]/2) = (4, -1)

Now,

Slope of AD, m = y_{2} – y_{1}/ x_{2} – x_{1}

m = (5 + 1)/ (3 – 4) = 6/-1 = -6

Thus, the equation of AD is given by

y – y_{1} = m (x – x_{1})

y + 1 = -6 (x – 4)

y + 1 = -6x + 24

⇒ 6x + y – 23 = 0

**Question -24. ****Find the equation of a line passing through the point ( – 2, 3) and having x-intercept 4 units. (2002)**

**Answer :**

x-intercept = 4

∴ Co-ordinates of the point will be (4, 0)

Now slope of the line passing through the points (-2, 3) and (4, 0)

m = y_{2} – y_{1}/ x_{2} – x_{1}

m = (0 – 3)/(4 + 2) = -3/6 = -1/2

Hence, the equation of the line will be

y – y_{1} = m (x – x_{1})

y – 0 = -½ (x – 4)

2y = -x + 4

⇒ x + 2y = 4

**Question -25. ****Find the equation of the line whose x-intercept is 6 and y-intercept is – 4.**

**Answer:**

x-intercept = 6

∴ The line will pass through the point (6, 0)

y -intercept = -4

⇒ c = -4

So, the line will pass through the point (0, -4)

Now,

Slope, m = m = y_{2} – y_{1}/ x_{2} – x_{1}

m = (-4 – 0)/(0 – 6) = -4/-6 = 2/3

Thus, the equation of the line is given by

y = mx + c

y = (2/3)x + (-4)

3y = 2x – 12

⇒ 2x – 3y – 12 = 0

Equation of Straight Line Exe-12.1

**ML Aggarwal Class 10 ICSE Maths Solutions**

Page-239

**Question -26. ****A(2, 5), B(-1, 2) and C(5, 8) are the vertices of a triangle ABC, ‘M’ is a point on AB such that AM : MB = 1:2. Find the co-ordinates of M’. Hence find the equation of the line passing through the points C and M. (2018)**

**Answer:**

Coordinates of the vertices of a ∆ ABC are A(2, 5), B(- 1, 2) and C (5, 8). Since M is a point on AB such that AM : MB = 1 : 2

**Question -27. ****Find the equation of the line passing through the point (1, 4) and intersecting the line x – 2y – 11 = 0 on the y-axis.**

**Answer:**

line x – 2y – 11 = 0 passes through y-axis

x = 0,

Now substituting the value of x in the equation x – 2y – 11 = 0

y = -11/2

The co-ordinates are (0, -11/2)

Now, the slope of the line joining the points (1, 4) and (0, -11/2) is given by

m = y_{2} – y_{1}/ x_{2} – x_{1}

= (-11/2 – 4)/ (0 – 1)

= 19/2

Thus, the line equation will be

y – y_{1} = m (x – x_{1})

y + 11/2 = 19/2 (x – 0)

2y + 11 = 19x

⇒ 19x – 2y – 11 = 0

**Question -28. ****Find the equation of the straight line containing the point (3, 2) and making positive equal intercepts on axes.**

**Answer :**

Let the line containing the point P (3, 2)

passes through x-axis at A (x, 0) and y-axis at B (0, y)

OA = OB given

∴ x = y

Now, the slope of the line (m) = y_{2} – y_{1}/ x_{2} – x_{1}

= 0 – y/ x – 0

= -x/x = -1

Hence, the equation of the line will be

y – y_{1} = m (x – x_{1})

y – 2 = -1 (x – 3)

y – 2 = -x + 3

⇒ x + y – 5 = 0

**Question- 29. ****Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2) find:**

**(i) the coordinates of the fourth vertex D.**

**(ii) length of diagonal BD.**

**(iii) equation of side AB of the parallelogram ABCD. (2015)**

**Answer:**

** **Three vertices of a parallelogram ABCD taken in order are

A (3, 6), B (5, 10) and C (3, 2)

(i) We need to find the co-ordinates of D

We know that the diagonals of a parallelogram bisect each other

Let (x, y) be the co-ordinates of D

Mid-point of diagonal AC = ((3 + 3)/2, (6 + 2)/2) = (3, 4)

Mid-point of diagonal BD = ((5 + x)/2, (10 + y)/2)

And, these two should be the same

On equating we get,

(5 + x)/2 = 3 and (10 + y)/2 = 4

5 + x = 6 and 10 + y = 8

x = 1 and y = -2

Thus, the co-ordinates of D = (1, -2)

##### (ii) Length of diagonal BD

##### (iii) Equation of the side joining A (3, 6) and D (1, -2) is given by

4 (x – 3) = y – 6

4x – 12 = y – 6

4x – y = 6

Thus, the equation of the side joining A (3, 6) and D (1, -2) is 4x – y = 6.

**Question -30. ****A and B are two points on the x-axis and y-axis respectively. P (2, – 3) is the mid point of AB. Find the**

**(i) the co-ordinates of A and B.**

**(ii) the slope of the line AB.**

**(iii) the equation of the line AB. (2010)**

**Answer :**

Points A and B are on x-axis and y-axis respectively

Let co-ordinates of A be (X, O) and of B be (O, Y)

P (2, -3) is the midpoint of AB

So, we have

2 = (x + 0)/2 and -3 = (0 + y)/2

x = 4 and y = -6

##### (i) Hence, the co-ordinates of A are (4, 0) and of B are (0, -6).

##### (ii) Slope of AB = y_{2} – y_{1}/ x_{2} – x_{1}

= (-6 – 0)/ (0 – 4)

= -6/-4 = 3/2 = m

**(iii) Equation of AB will be**

y – y_{1} = m (x – x_{1})

y – (-3) = 3/2 (x – 2) [As P lies on it]

y + 3 = 3/2 (x – 2)

2y + 6 = 3x – 6

3x – 2y – 12 = 0

**Question- 31. ****M and N are two points on the X axis and Y axis respectively. P(3, 2) divides the line segment MN in the ratio 2 : 3. Find :**

(i) the coordinates of M and N

(ii) slope of the line MN

**Answer:**

Let the coordinates of M and N be (x, 0) and (0, y)

Thus, the coordinates of M and N are M(5,0) and N(0, 5).

Hence, the slope of the line MN is – 1.

**Question-32. 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

**Answer:**

**Question-33**

(i) write down the coordinate of point P that divide the line joining A (-4, 1) an B (17, 10) in the ratio 1:2

(ii) Calculate the distance OP, where O is the origin.

(iii) In what ratio does the Y-axis divide the line AB?

**Answer:**

**(i) Co-ordinates of point P are**

**(iii) Let AB be divided by the point P(0, y) lying on y-axis in the ratio**

Thus, the ratio in which the y-axis divide the line AB is 4: 17.

**Question-34. ****Find the equations of the diagonals of a rectangle whose sides are x = – 1, x = 2 , y = – 2 and y = 6.**

**Answer:**

The equations of sides of a rectangle whose equations are

x_{1} = -1, x_{2} = 2, y_{1} = -2, y_{2} = 6.

These lines form a rectangle when they intersect at A, B, C, D respectively

Co-ordinates of A, B, C and D will be

(-1, -2), (2, -2), (2, 6) and (-1, 6) respectively.

AC and BD are its diagonals

**(i) Slope of the diagonal AC**

**Question-35. ****Find the equation of a straight line passing through the origin and through the point of intersection of the lines 5x +7y = 3 and 2x – 3y = 7**

**Answer :**

5x + 7y = 3 …(i)

2x – 3 y = 7 …(ii)

Multiply (i) by 3 and (ii) by 7,

15x + 21y = 9

14x – 21y = 49

Adding we get,

29x = 58

x = 58/29 = 2

Substituting the value of x in (i), we get

5(2) + 7y = 3

10 + 7y = 3

7y = 3 – 10

y = -7/7 = -1

Hence, the point of intersection of lines is (2, -1)

Now, the slope of the line joining the points (2, -1) and (0, 0) will be

m = y_{2} – y_{1}/ x_{2} – x_{1}

= (0 + 1)/ (0 – 2)

= -1/2

Equation of the line is given by:

y – y_{1} = m (x – x_{1})

y – 0 = -1/2 (x – 0)

2y = -x

Thus, the required line equation is x + 2y = 0.

— : End of ML Aggarwal Equation of Straight Line Exe-12.1 Class 10 ICSE Maths : –

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Sum number 29 (iii) it’s said to find the equation of line AB you have found the equation of AD pls kindly see to it..

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