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

**Equation of Straight Line** ML Aggarwal Solutions ICSE Class-10 Maths Chapter-12. We Provide Step by Step Answer of Exercise-12.1 , Exercise-12.2 , **Equation of Straight Line**** **, with MCQs and Chapter-Test Questions / Problems related for ICSE Class-10 APC Understanding Mathematics . Visit official Website **CISCE ** for detail information about ICSE Board Class-10.

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 10th |

Chapter-12 | Equation of Straight Line |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exe-12.1, Exe-12.2, MCQ and Chapter Test Questions |

Academic Session | 2021-2022 |

**Equation of Straight Line** ML Aggarwal Solutions ICSE Class-10 Maths Chapter-12

-: Select Topic :-

Exe-12.1 ,

Exe-12.2 ,

MCQS ,

**How to Solve Equation of Straight Line Problems/Questions / Exercise of ICSE Class-10 Mathematics**

Before viewing Answer of Chapter-12 **Equation of Straight Line** of ML Aggarwal Solution. Read the Chapter Carefully and then solve all example given in your text book.

For more practice on **Equation of Straight Line** related problems /Questions / Exercise try to solve **Equation of Straight Line** exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE) / Concise Selina Publications ICSE Mathematics. Get the formula of **Equation of Straight Line** for ICSE Class 10 Maths to understand the topic more clearly in effective way.

**Exercise-12.1, Equation of Straight Line Solutions of ML Aggarwal for ICSE Class-10**

page-237

**Question -1**

**Find the slope of a line whose inclination is**

**(i) 45°**

**(ii) 30°**

**Answer- 1**

(i) tan 45° = 1

(ii) tan 30° =

**Question- 2**

**Find the inclination of a line whose gradient is**

**(i) 1**

**(ii) √3**

**(iii) **

**Answer- 2**

(i) tan θ = 1 ⇒ θ = 45°

(ii) tan θ = √3 ⇒ θ = 60°

(iii) tan θ = ⇒ 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- 4**

(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- 5**

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 = , y-intercept = 3**

**(iii) gradient = √3, y-intercept = **

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

**Answer -6**

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

**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) **

**(v) y – 3 = 0**

**(vi) x – 3 = 0**

**Answer- 7**

We know that in the equation

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

Now using this, we find,

**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- 8**

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

Writing in form of y = mx + c

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 -9**

Slope of the line y = x + 1 after comparing

it with y = mx + c, m = 1

**Question- 10**

**Find the value of p, given that the line passes through the point ( – 4, 4) **

**Answer -10**

Equation of line is

It passes through the points (-4, 4)

and It will satisfy the equation

**Question -11**

**Given that (a, 2a) lies on the line find the value of a.**

**Answer- 11**

∵ Point (a, 2a) lies on the line

∴this point will satisfy the equation

**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- 12**

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- 13**

∴ The line intersects y-axis making an intercept of -3

∴ the co-ordinates of point of intersection will be (0, -3)

Now the slope of line (m) =

**Question- 14**

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

**Answer -14**

Equation of the line will be

⇒ 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- 15**

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 -16**

m =

Given

(i) (0, -2), (3, 4)

(ii) (3, -7), (-1, 8)

**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- 17**

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

(i) The gradient of EF

**Question- 18**

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

**Answer- 18**

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- 19**

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

∴ Slope of the line (m)

**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 -20**

Given That

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

Slope of AB = Slope of BC

**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 -21**

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

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

**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 -22**

Given that

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

**Question- 23**

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

**Answer- 23**

AD is median

⇒ D is mid point of BC

∴ D is

i.e (4, -1)

slope of AD

**Question -24**

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

**Answer -24**

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)

**Question -25**

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

**Answer -25**

x-intercept = 6

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

y -intercept = -4 ⇒ c = -4

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- 26**

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- 27**

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

x = 0,

Now substituting the value of x in the equation x – 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 -28**

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

**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-29**

** **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

**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 -30**

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

**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-31**

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-32**

**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-33**

(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-34**

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 -35**

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,

**Exercise 12.2 , ****Equation of Straight Line ML Aggarwal ICSE Class-10 **

page-245

**Question- 1**

**State which one of the following is true : The straight lines y = 3x – 5 and 2y = 4x + 7 are**

**(i) parallel**

**(ii) perpendicular**

**(iii) neither parallel nor perpendicular.**

**Answer- 1**

Slope of line y = 3x – 5 = 3

and slope of line 2y = 4x + 7

⇒ y = 2x + = 2.

∴ Slope of both the lines are neither equal nor their product is – 1.

∴ These line are neither parallel nor perpendicular.

**Question- 2**

**If 6x + 5y – 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.**

**Answer -2**

In equation

6x + 5 y – 7 = 0

⇒ 5y = -6x + 7

**Question -3**

**Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b. **

**Answer- 3**

In equation 2x – by + 5 = 0

⇒ – by = – 2x – 5

⇒ y = +

**Question -4**

If the straight lines 3x – 5y = 7 and 4x + ay + 9 = O are perpendicular to one another, find the value of a (2018)

**Answer-4**

Given lines are

3x – 5y = 1 ……….(i) and 4x + ay + 9 = 0 …………(ii)

Slope of line (i) (m_{1}) = −(3/−5)=3/5

Slope of line (ii) (m_{2}) = −(4/𝑎)

Also, given that two lines are perpendicular to one and another

∴ (m_{1}) (m_{2}) = – 1

Hence, the value of a = 12/5 .

**Question- 5**

**If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.**

**Answer -5**

Given

In the equation 3x + by + 5 = 0

**Question-6**

**Is the line through ( – 2, 3) and (4, 1) perpendicular to the line 3x = y + 1 ?**

**Does the line 3x = y + 1 bisect the join of ( – 2, 3) and (4, 1). **

**Answer -6**

Slope of the line passing through the points

(-2, 3) and (4, 1) =

Hence the line 3x = y + 1 bisects the line joining the points (-2, 3), (4, 1).

**Question -7**

**The line through A ( – 2, 3) and B (4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b. (2012)**

**Answer -7**

Gradient (m_{1}) of the line passing through the

points A (-2, 3) and B (4, b)

**Question-8**

**If the lines 3x + y = 4, x – ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the value of a and b.**

**Answer -8**

In the line 3x + y = 4 …(i)

⇒ y = – 3x + 4

Slope (m_{1}) = – 3

In the line x – ay + 7 = 0…..(ii)

⇒ -b = -6 ⇒ b = 6

Hence a = 3, b = 6

**Question -9**

**Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x – 3y – 7 = 0. Find the coordinates of the point where it cuts the x-axis. **

**Answer -9**

In the given line 2x – 3y – 7 = 0

⇒ 3y = 2x – 7

⇒

**Question -10**

**Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept – 3 **

**Answer -10**

In the line 2x + 5y + 7 = 0

⇒ 5y = – 2x – 7

⇒

**Question -11**

**Find the equation of a straight. line perpendicular to the line 3x – 4y + 12 = 0 and having same y-intercept as 2x – y + 5 = 0.**

**Answer -11**

In the given line 3x – 4y + 12 = 0

⇒ 4y = 3x + 12

⇒ y =

**Question-12**

**Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0. **

**Answer -12**

In the given equation 3x + 5y + 15 = 0

⇒ 5y = – 3x – 15

⇒ y =

**Question -13**

**Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point ( – 1, – 2).**

**Answer -13**

In the given line 3x + 8y = 12

⇒ 8y = -3x + 12

⇒

**Question-14**

**(i) The line 4x – 3y + 12 = 0 meets the x-axis at A. Write down the co-ordinates of A.**

**(ii) Determine the equation of the line passing through A and perpendicular to 4x – 3y + 12 = 0.**

**Answer-14**

(i) In the line 4x – 3y + 12 = 0 …(i)

3y = 4x + 12

⇒ y =

**Question -15**

**Find the equation of the line that is parallel to 2x + 5y – 7 = 0 and passes through the mid-point of the line segment joining the points (2, 7) and ( – 4, 1).**

**Answer -15**

The given line 2x + 5y – 7 = 0

5y = -2x + 7

⇒

**Question -16**

**Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes through the mid-point of the line segment joining the points (5, – 2), (2, 2).**

**Answer -16**

In the given line 3x + 2y – 8=0

⇒ 2y = – 3x + 8

so ⇒ y =

**Question -17**

**Find the equation of a straight line passing through the intersection of 2x + 5y – 4 = 0 with x-axis and parallel to the line 3x – 7y + 8 = 0.**

**Answer -17**

Let the point of intersection of the line 2x + 5y – 4 = 0 and x-axis be (x, 0)

Substituting the value of y in the equation

2x + 5 × 0 – 4 = 0

⇒ 2x – 4 = 0

and ⇒ 2x = 4

so ⇒ x = = 2

Coordinates of the points of intersection will be (2, 0)

**Question -18**

**The equation of a line is 3x + 4y – 7 = 0. Find**

**(i) the slope of the line. .**

**(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0. (2010)**

**Answer -18**

#### (i) Equation of the line is 3x + 4y – 1 = 0

⇒ 4y = 7 – 3x

**Question -19**

**Find the equation of the line perpendicular from the point (1, – 2) on the line 4x – 3y – 5 = 0. Also find the co-ordinates of the foot of perpendicular.**

**Answer -19**

In the equation 4x – 3y – 5 = 0,

⇒ 3y = 4x – 5

⇒

**Question -20**

**Prove that the line through (0, 0) and (2, 3) is parallel to the line through (2, – 2) and (6, 4).**

**Answer -20**

Given that

Slope of the line through (0, 0) and (2, 3)

page-246

**Question -21**

**Prove that the line through,( – 2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (4, 24).**

**Answer -21**

Given that

Slope of the line through (-2, 6) and (4, 8)

**Question-22**

**Show that the triangle formed by the points A (1, 3), B (3, – 1) and C ( – 5, – 5) is a right angled triangle by using slopes.**

**Answer -22**

Slope (m_{1}) of line by joining the points

A(1, 3), B (3, -1) =

**Question -23**

**Find the equation of the line through the point ( – 1, 3) and parallel to the line joining the points (0, – 2) and (4, 5).**

**Answer -23**

Slope of the line joining the points (0, -2) and (4, 5) =

=

Slope of the line parallel to it passing through (-1, 3) =

**Question-24**

**A ( – 1, 3), B (4, 2), C (3, – 2) are the vertices of a triangle.**

**(i) Find the coordinates of the centroid G of the triangle.**

**(ii) Find the equation of the line through G and parallel to AC**

**Answer-24**

Given, A (-1, 3), B (4, 2), C (3, -2)

(i) Coordinates of centroid G

**Question-25**

**Find the equation of the line through (0, – 3) and perpendicular to the line joining the points (– 3, 2) and (9, 1).**

**Answer -25**

The slope (m_{1}) of the line joining the points (-3, 2) and (9, 1)

**Question-26**

The vertices of a ∆ABC are A(3, 8), B(-1, 2) and C(6, -6). Find :

(i) Slope of BC.

(ii) Equation of a line perpendicular to BC and passing through A.(2019)

**Answer-26**

Vertices of a ∆ABC are A(3, 8), B(-1, 2) and C(6, -6)

Slope of the line perpendicular to BC = 7/8

Now, equation of the line perpendicular to BC and passing through A is

8y – 64 = 7x – 21

7x – 8y + 43 = 0

**Question-27**

**The vertices of a triangle are A (10, 4), B (4, – 9) and C ( – 2, – 1). Find the equation of the altitude through A. **

(The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.)

**Answer-27**

**Question-28**

**A (2, – 4), B (3, 3) and C ( – 1, 5) 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**

**Answer -28**

(i) D is the mid-point of BC

Co-ordinates of D will

**Question-29**

**Find the equation of the right bisector of the line segment joining the points (1, 2) and (5, – 6).**

**Answer -29**

Slope of the line joining the points (1, 2) and (5, -6)

**Question -30**

**Points A and B have coordinates (7, – 3) and (1, 9) respectively. Find**

**(i) the slope of AB.**

**(ii) the equation of the perpendicular bisector of the line segment AB.**

**(iii) the value of ‘p’ if ( – 2, p) lies on it.(2008)**

**Answer -30**

Coordinates of A are (7, -3), of B = (1, 9)

(i) ∴ slope (m)

**Question-31**

**The points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.**

**Answer -31**

Slope of BD (m_{1}) =

=

Diagonal AC is perpendicular bisector of diagonal BD

**Question-32**

**ABCD is a rhombus. The co-ordinates of A and C are (3, 6) and ( – 1, 2) respectively. Write down the equation of BD. (2000)**

**Answer-32**

Co-ordinates of A (3, 6), C (-1, 2)

Slope of AC (m_{1}) =

=

**Question -33**

**Find the image of the point (1, 2) in the line x – 2y – 7 = 0**

**Answer -33**

Draw a perpendicular from the point P(1, 2) on the line, x – 2y – 7 = 0

Let P’ is the image of P and let its

co-ordinates sue (α, β) slope of line x – 2y – 7 = 0

**Question -34**

**If the line x – 4y – 6 = 0 is the perpendicular bisector of the line segment PQ and the co-ordinates of P are (1, 3), find the co-ordinates of Q.**

**Answer -34**

Let the co-ordinates of Q be (α, β) and let the line x – 4y – 6 = 0 is the

perpendicular bisector of PQ and it intersects the line at M.

M is the mid point of PQ Now slope of line x – 4y – 6 = 0

**Question -35**

**OABC is a square, O is the origin and the points A and B are (3, 0) and (p, q). If OABC lies in the first quadrant, find the values of p and q. Also write down the equations of AB and BC.**

**Answer -35**

OA =

**MCQ of ****Equation of Straight Line of ML Aggarwal Solutions for ICSE Class-10**

**Choose the correct answer from the given four options (1 to 13) :**

page-246

**Question- 1**

**The slope of a line parallel to y-axis is**

**(a) 0**

**(b) 1**

**(c) – 1**

**(d) not defined**

**Answer -1**

Slope of a line parallel to y-axis is not defined. (d)

page-247

**Question -2**

**The slope of a line which makes an angle of 30° with the positive direction of x-axis is**

**(a) 1**

**(b) **

**(c) √3**

**(d) **

**Answer -2**

Slope of a line which makes an angle of 30°

with positive direction of x-axis = tan 30°

= (b)

**Question -3**

**The slope of the line passing through the points (0, – 4) and ( – 6, 2) is**

**(a) 0**

**(b) 1**

**(c) – 1**

**(d) 6**

**Answer -3**

Slope of the line passing through the points (0, -4) and (-6, 2)

(c)

**Question- 4**

**The slope of the line passing through the points (3, – 2) and ( – 7, – 2) is**

**(a) 0**

**(b) 1**

**(c) **

**(d) not defined**

**Answer- 4**

Slope of the line passing through the points (3, -2) and (-7, -2)

(a)

**Question- 5**

**The slope of the fine passing through the points (3, – 2) and (3, – 4) is**

**(a) – 2**

**(b) 0**

**(c) 1**

**(d) not defined**

**Answer- 5**

The slope of the line passing through (3, -2) and (3, -4)

(d)

**Question -6**

**The inclination of the line y = √3x – 5 is**

**(a) 30°**

**(b) 60°**

**(c) 45°**

**(d) 0°**

**Answer -6**

The inclination of the line y = √3x – 5 is

√3 = tan 60° = 60° (b)

**Question- 7**

**If the slope of the line passing through the points (2, 5) and (k, 3) is 2, then the value of k is**

**(a) -2**

**(b) -1**

**(c) 1**

**(d) 2**

**Answer- 7**

Slope of the line passing through the points (2, 5) and (k, 3) is 2, then

**Question- 8**

**The slope of a line parallel to the line passing through the points (0, 6) and (7, 3) is**

**(a) **

**(b) **

**(c) -5**

**(d) 5**

**Answer -8**

Slope of the line parallel to the line passing through (0, 6) and (7, 3)

Slope of the line = (b)

**Question -9**

**The slope of a line perpendicular to the line passing through the points (2, 5) and ( – 3, 6) is**

**(a) **

**(b) **

**(c) -5**

**(d) 5**

**Answer -9**

Slope of the line joining the points (2, 5), (-3, 6)

**Question- 10**

**The slope of a line parallel to the line 2x + 3y – 7 = 0 is**

**(a) **

**(b) **

**(c) **

**(d) **

**Answer -10**

The slope of a line parallel to the line 2x + 3y – 7 = 0

slope of the line

**Question -11**

**The slope of a line perpendicular to the line 3x = 4y + 11 is**

**(a) **

**(b) **

**(c) **

**(d) **

**Answer -11**

slope of a line perpendicular to the line 3x = 4y + 11 is

**Question -12**

**If the lines 2x + 3y = 5 and kx – 6y = 7 are parallel, then the value of k is**

**(a) 4**

**(b) – 4**

**(c) **

**(d) **

**Answer -12**

lines 2x + 3y = 5 and kx – 6y = 7 are parallel

Slope of 2x + 3y = 5 = Slope of kx – 6y = 7

⇒ 3y – 2x + 5

so option (b) is correct

**Question -13**

**If the line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0 are perpendicular to each other, then the value of k is**

**(a) **

**(b) **

**(c) **

**(d) **

**Answer-13**

line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0

are perpendicular to each other

**Chapter – Test ****Equation of Straight Line Solutions of ML Aggarwal**

**Chapter – Test**

page-249

**Question- 1**

**Find the equation of a line whose inclination is 60° and y-intercept is – 4.**

**Answer- 1**

Angle of inclination = 60°

Slope = tan θ = tan 60° = √3

Equation of the line will be,

y = mx + c = √3x + ( – 4)

⇒ y – √3x – 4

**Question -2**

**Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.**

**Answer- 2**

Slope of the line 3y + 2x = 12

⇒ 3y = 12 – 2x

⇒ 3y = -2x + 12

**Question -3**

**If the equation of a line is y – √3x + 1, find its inclination.**

**Answer -3**

In the line

y = √3 x + 1

Slope = √3

⇒ tan θ = √3

⇒ θ = 60° (∵ tan 60° = √3)

**Question -4**

**If the line y = mx + c passes through the points (2, – 4) and ( – 3, 1), determine the values of m and c.**

**Answer- 4**

The equation of line y = mx + c

∵ it passes through (2, – 4) and ( – 3, 1)

Now substituting the value of these points -4 = 2m + c …(i)

and 1 = -3m + c …(ii)

Subtracting we get,

**Question -5**

**If the point (1, 4), (3, – 2) and (p, – 5) lie on a st. line, find the value of p.**

**Answer- 5**

Let the points to be A (1, 4), B (3, -2) and C (p, -5) are collinear and let B (3, -2)

divides AC in the ratio of m_{1} : m_{2}

**Question- 6**

**Find the inclination of the line joining the points P (4, 0) and Q (7, 3).**

**Answer -6**

Slope of the line joining the points P (4, 0) and Q (7, 3)

**Question -7**

**Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to **

**Answer -7**

Equation of lines are

2x + y = 5 …(i)

x – 2y = 5 …(ii)

Multiply (i) by 2 and (ii) by 1, we get

4x + 2y = 10

x – 2y = 5

Adding we get,

**Question -8**

**If the lines and 3x + ky = 11 are perpendicular to each other, find the value of k.**

**Answer -8**

Given Equation of lines are

**Question -9**

**Write down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2).**

**Answer -9**

The equation of the line is x – 2y + 8 = 0

⇒ 2y = x + 8

**Question-10**

**Write down the equation of the line passing through ( – 3, 2) and perpendicular to the line 3y = 5 – x.**

**Answer-10**

Equations of the line is

3y = 5 – x ⇒ 3y = -x + 5

**Question-11**

**Find the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.**

**Answer-11**

Let slope of the line joining the points A (1, 2) and B (6, 7) be m_{1}

**Question-12**

**The points A (7, 3) and C (0, – 4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.**

**Answer-12**

Slope of line AC (m_{1})

**Question-13**

**A straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find:**

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

**(ii) the equation of the line AB**

**Answer-13**

A lies on x-axis and B lies on y-axis

Let co-ordinates of A be (x, 0) and B be (0, y)

and P (2, 1) divides BA in the ratio 3 : 1.

**Question-14**

**A straight line makes on the co-ordinates axes positive intercepts whose sum is 7. If the line passes through the point ( – 3, 8), find its equation.**

**Answer -14**

Let the line make intercept a and b with the

x-axis and y-axis respectively then the line passes through

Hence, the equation of the line is x3+y4=1 or 4x + 3y = 12

**Question-15**

**If the coordinates of the vertex A of a square ABCD are (3, – 2) and the equation of diagonal BD is 3 x – 7 y + 6 = 0, find the equation of the diagonal AC. Also find the co-ordinates of the centre of the square.**

**Answer -15**

Co-ordinates of A are (3, -2).

Diagonals AC and BD of the square ABCD

bisect each other at right angle at O.

∴ O is the mid-point of AC and BD Equation

The co-ordinate of intersection of diagonal -O

3x-7y= -6 ……….(i)

7x+3y=15 ……..(ii)

multiply (i) by 3 and .(ii) by 7

9x – 21y = -18 …….(iii)

49x + 21y = 105 ………(iv)

Adding 3rd and 4th equation

— : End of Equation of Straight Line Solutions :–

Return to – ML Aggarwal Solutions for ICSE Class-10

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