# This note is designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students.

# Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

# These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the IPE examinations.

**Straight Lines**

**QUESTION1.**

Prove that the points (1, 11), (2, 15), and (– 3, – 5) are collinear, and find the equation of the line containing them.

Sol:

Let A (1, 11), B (2, 15), and C (– 3, – 5)

** The slope of the line segment joining the points (x**_{1}, y_{1}) and (x_{2}, y_{2}) is

Slope of AB = = 4

Slope of BC = = = 4

**QUESTION2.**

Find the condition for the points (a, 0), (h, k), and (0, b) to be collinear.

Sol:

Let A (a, 0), B (h, k) and C (0, b)

**The slope of the line segment joining the points (x**_{1}, y_{1}) and (x_{2}, y_{2}) is Given that A, B, and C are collinear points

The slope of AB = The slope of BC

⟹

⟹

⟹ – hk = (h – a) ( b – k)

– hk = hb – hk – ab + ak

⟹ 0 = hb + ak – ab

⟹ hb + ak = ab or

**QUESTION3.**

Find the equations of the straight lines which makes intercepts whose sum is sum is 5 and product is 6.

Sol:

The equation of the line in the intercept form is

Given that, a + b = 5 and ab = 6

⟹ b = 5 – a

a(5 – a) = 6

5a – a^{2} = 6

a^{2} – 5a + 6 = 0

a – 3a – 2a + 6 = 0

a (a – 3) – 2(a – 3) = 0

(a – 3) (a – 2) = 0

a = 3 or a = 2

case (i) if a = 3 ⟹ b = 2

⟹ 2x + 3y – 6 = 0

case (ii) if a = 2 ⟹ b = 3

= 1 ⟹ 3x + 2y – 6 = 0

**QUESTION4.**

Find the equation of the straight line which makes an angle 135^{0} with the positive X – axis measured countered clockwise and passing through the point (– 2, 3).

Sol:

Slope of the line m = tan 135^{0} = – 1

The point is (– 2, 3)

**The equation of the straight line in slope point form is (y – y**_{1}) = m (x – x_{1})

The equation of the line passing through the point (– 2, 3) with slope – 1 is

y – 3 = – 1 (x + 2)

y – 3 =– x – 2

x + y – 1 = 0

**QUESTION5.**

Find the equation of the straight line passing through the points (1, – 2) and (– 2, 3).

Sol:

Given points are (1, – 2), (– 2, 3)

**The equation of the straight line in two points form is (y – y**_{1}) = ** (x – x**_{1})

The equation of required straight line is

(y + 2) = (x – 1)

(y + 2) = (x – 1)

– 3 (y + 2) = 5 (x – 1)

– 3y – 6 = 5x – 5

5x + 3y + 1 = 0

**QUESTION6.**

Find the slopes of the line x + y = 0 and x – y = 0

Sol:

**The slope of the line ax + by + c = 0 is ** ** **

The slope of the line x + y = 0 is = – 1

The slope of the line x – y = 0 is = 1

**QUESTION7.**

Find the angle which the straight-line y = x – 4 makes with the Y-axis.

Sol:

Given equation is y = x – 4

Compare with y = mx + c

m = ⟹ tan θ =

θ =

Angle made by the line with X-axis is

Angle made by the line with Y-axis is

**QUESTION8.**

Find the equation of the reflection of the line x = 1 in the Y-axis.

Sol:

Given equation is x = 1

Reflection about the Y-axis is x =– 1

Required equation of the line is x + 1 = 0

**QUESTION9.**

Write the equations of the straight lines parallel to X-axis and (i) at a distance of 3 units above the X-axis and (ii) at a distance of 4 units below the X-axis.

Sol:

(i) The equation of the straight line parallel to X-axis which is at a distance of 3 units above the X-axis is y = 3

⟹ y – 3 = 0

(ii) The equation of the straight line parallel to X-axis which is at a distance of 4 units below the X-axis is y = – 4

⟹ y + 4 = 0

**QUESTION10.**

Write the equations of the straight lines parallel to the Y-axis and (i) at a distance of 2 units from the Y-axis to the right of it (ii) at a distance of 5 units from the Y-axis to the left of it.

Sol:

(i) The equation of the straight line parallel to the Y-axis which is at a distance of 2 units from the Y-axis to the right of it is x = 2

⟹ x – 2 = 0

(ii) The equation of the straight line parallel to the Y-axis which is at a distance of 5 units from the Y-axis to the left of it is x =– 5

⟹ x + 5 = 0

**QUESTION 11.**

Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2.

Sol:

Given the slope of the line passing through (2, 5) and (x, 3) is 2.

**The slope of the line segment joining the points (x**_{1}, y_{1}) and (x_{2}, y_{2}) is

⟹ = 2

** **⟹ 2(x – 2) = – 2

x – 2 = – 1

x = 1

**QUESTION 12.**

Find the equation of the straight line passing through (– 4, 5) and making non – zero intercepts on the coordinate axes whose sum is zero.

Sol:

The equation of a line in the intercept form is

Given a = b

⟹ = 1

### ⟹

x + y = a

but it is passing through (– 4, 5)

– 4+ 5 = a

a = 1

∴ The equation of the required straight line is x + y = 1

x + y – 1 = 0

**QUESTION13.**

Find the equation of the straight line passing through (– 2, 5) and cutting off equal and non – zero intercepts on the coordinate axes.

Sol:

The equation of a line in the intercept form is

Given a + b = 0

b = – a

⟹

⟹

x – y = a

but it is passing through (– 2, 4)

– 2– 4 = a

a = –6

∴ The equation of the required straight line is x – y = –6

x + y + 6 = 0

**QUESTION14.**

### Find the equation of the straight line whose distance from the origin is 4, if the normal ray from the origin to the straight line makes an angle of 135^{0} with the positive direction of X – axis

Sol:

The equation of the straight line in the normal form is x cos α + y sin α = p

p = 4 and α = 135^{0}

x cos 135^{0} + y sin 135^{0} = 4

x + y = 4

= 4

– x + y = 4

x – y + 4 = 0

**QUESTION15.**

### Write the parametric equation of the straight line passing through the point (3, 2) and making an angle 135^{0} with the positive direction of the X – axis.

Sol:

Given θ = 135^{0} and (h, k) = (3, 2)

The parametric equations are: x = h + r Cos θ, y = k +r Sin θ

x = 3 + r Cos 135^{0}, y = 2 + r Sin 135^{0}

x = 3 – , y = 2 +

**QUESTION16.**

### Transform the following equation into normal form.

(i) x + y +1 = 0

Sol:

Given equation is x + y +1 = 0

x + y =– 1

– x – y = 1

Dividing on both sides by

(ii) x + y = 2

Sol:

Given equation is x + y = 2

Dividing on both sides by

**QUESTION 17.**

### If the area of the triangle formed by the straight lines x = 0, y = 0 and 3x + 4y =a (a > 0) is 6 square units. Find the value of ‘a’.

Sol:

Given equation is 3x + 4y = a

The area of the triangle formed by the straight-line ax + by + c = 0 with the coordinate axes is

⟹ = 6

= 6

a^{2 }= 24 ⟹ a = 12(∵ a > 0)

**QUESTION18.**

Find the sum of the squares of the intercepts of the line 4x – 3y = 12

Sol:

Given equation is 4x – 3y = 12

It is in the form of

a = 3 and b =– 4

the sum of the squares of intercepts = a^{2} + b ^{2}

= 3^{2} + (– 4)^{2}

= 9 + 16 = 25

**QUESTION 19.**

Find the angle made by the straight – line y = – x + 3 with the positive direction of the X-axis measured in the counter-clockwise direction.

Sol:

Give equation of straight line is y = x + 3

It is in the form of y = mx + c

m =

tan θ =

⟹ θ =

**QUESTION 20.**

Find the equation of the straight line in the symmetric form, given the slope and point on the line (2, 3).

Sol:

Equation of the line in the symmetric form is

Given Point (x_{1}, y_{1}) = (2, 3) and slope m =

tan θ =

⟹ θ = 60^{0}

∴ The equation of the line is

**QUESTION 21.**

If the product of the intercepts made by the straight-line x tan α + y sec α = 1 (0≤ α < π/2) on the coordinate axes is equal to sin α, find α.

Sol:

Given equation is x tan α + y sec α = 1

It is in the form of

a = cot α, b = cos α

product of intercepts = sin α

⟹ ab = sin α

cot α.cos α = sin α

(cos α)/ (sin α). cos α = sin α

⟹ cos^{2} α = sin^{2} α

⟹ tan^{2} α = 1

### tan α = 1(∵ 0≤ α < π/2)

∴ α = 45^{0}

**QUESTION22.**

If the sum of the reciprocals of the intercepts made by a variable straight–line on the axes of coordinates is a constant, then prove that the line always passes through a fixed point.

Sol:

The equation of the line in the intercept form is

Let the sum of the reciprocals of intercepts is k

⟹

∴ the line always passes through a fixed point (1/k, 1/k).

**QUESTION23.**

Find the ratio in which the straight-line 2x + 3y – 20 = 0 divides the join of the points (2, 3) and (2, 10).

Sol:

Given equation is L ≡ 2x + 3y – 20 = 0

L_{11} = 2(2) + 3(3) = 4 + 9– 20 = – 7

L_{22} = 2(2) + 3(10) = 4 + 30 – 20 = 14

We know that the line L = 0 divides the line joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) in the ratio – L_{11} : L_{22}

= – (– 7) : 14

= 1 : 2

**QUESTION24.**

State whether (3, 2) (– 4, – 3) are on the same or opposite sides of the straight-line 2x – 3y + 4 = 0.

Sol:

Given equation is L ≡ 2x – 3y + 4 = 0

L_{11} = 2(3) – 3(2) + 4 = 6 – 6 + 4 = 4 > 0

L_{22} = 2(– 4) – 3(–3) + 4 = – 8 + 9 + 4 = 5 > 0

L_{11}, L_{22} has the Same sign

∴ Given points lies the same side of the line 2x – 3y + 4 = 0.

**QUESTION25.**

Find the ratio in which (i) X-axis and (ii) Y-axis divide the line segment joining A (2, –3) and B (3, – 6)

Sol:

(i) X – axis divide in the ratio – y_{1} : y_{2 } = – ( –3) : – 6 = – 1 : 2

(ii) Y – axis in the ratio – x_{1} : x_{2 = } – 2 : 3 = – 2 : 3

**QUESTION26.**

Find the equation of the straight line passing through the point of intersection of the lines x + y + 1 = 0 and 2x – y + 5 = 0 and containing the point (5, – 2).

Sol:

Given equations are

x + y + 1 = 0 ……… (1)

2x – y + 5 = 0……. (2)

The equation of the line passing through the point of intersection of the lines L_{1} = 0 and L_{2} = 0 is L_{1} + λ L_{2} = 0

The equation of the line passing through the point of intersection of the lines (1) and (2) is

x + y + 1 + λ (2x – y + 5) = 0

but the line passes through (5, – 2)

5 – 2 + 1 + λ (2(5) – (– 2) + 5) = 0

4 + λ (10 + 2 + 5) = 0

4 + λ (17) = 0

λ = –4 /17

∴ the equation of required line is

x + y + 1 + (–4 /17) (2x – y + 5) = 0

17(x + y + 1) –4 (2x – y + 5) = 0

17x + 17y + 17 – 8x + 4y – 20 = 0

9x + 21y – 3 = 0

3x + 7y – 1 = 0

**QUESTION27.**

If a, b, and c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.

Sol:

Given, a, b, and c are in A.P. and the equation is ax + by + c = 0

a, b, and c are in A.P.

⟹ b = (a + c)/2

Substitute b value in the equation ax + by + c = 0

⟹ ax + [ (a + c)/2]y + c = 0

ax + ay/2 + cy/2 + c = 0

a(x + y/2) + c(y/2 + 1) = 0

dividing on both sides by a

⟹ (x + y/2) + (c/a) (y/2 + 1) = 0

It is in the form of L_{1} + λ L_{2} = 0

∴ given equation represent a family of concurrent lines

x + y/2 = 0 and y/2 + 1 = 0 are the lines

y/2 + 1 = 0 ⟹ y/2 = – 1

y =– 2 ⟹ x = 1

The point of concurrency is (– 2, 1)

**QUESTION28.**

Find the point of intersection of the line 7x + y + 3 = 0 and x + y = 0

Sol:

Given equations are 7x + y + 3 = 0 and x + y = 0

∴ The point of intersection is

**QUESTION29.**

Transform the following equations into the form L_{1} + λ L_{2} = 0 and find the point of concurrency of the family of straight lines represented by the equation.

(i) (2 + 5k) x – 3 (1 + 2k) y + (2 – k) = 0

Given equation is

(2 + 5k) x – 3 (1 + 2k) y + (2 – k) = 0

2x + 5kx – 3y – 6ky + 2 – k = 0

2x – 3y + 2 + k (5x – 6y – 1) = 0

It is in the form of L_{1} + λ L_{2} = 0 L_{1 }≡ 2x – 3y + 2 = 0 and L_{2}≡ 5x – 6y – 1 = 0

By solving above equations, we get point of concurrency

∴ The point of concurrency is (5, 4)

(ii) ( k + 1) x + (k + 2) y + 5 = 0

Given equation is (k + 1) x + (k + 2) y + 5 = 0

kx + x + ky + 2y + 5 = 0

x + 2y + 5 + k (x + y) = 0

It is in the form of L_{1} + λ L_{2} = 0

L_{1 }≡ x + 2y + 5 = 0 and L_{2}≡ x + y = 0

By solving the above equations, we get point of concurrency

∴ The point of concurrency is (5, – 5)

**QUESTION30.**

Find the value of ‘p’, if the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent.

Sol:

Given that x + p = 0 —— (1), y + 2 = 0 —— (2) and 3x + 2y + 5 = 0 —— (3) are concurrent.

From (1) x + p = 0 ⟹ x = – p

From (2) y + 2 = 0 ⟹ y = – 2

Point of intersection of (1) and (2) is (– p, – 2)

From (3)

3y + 2x + 5 = 0

3(– p) + 2 (– 2) + 5 = 0

– 3p – 4 + 5 = 0

– 3p + 1 = 0

3p = 1

p = 1/3

**QUESTION31.**

Find the area of the triangle formed by the following straight lines and the coordinate axes.

(i) x – 4y + 2 = 0

The area of the triangle formed by the straight-line ax + by + c = 0 with the coordinate axes is

The area of the triangle formed by the straight-line x – 4y + 2 = 0 with the coordinate axes is =

(ii) 3x – 4y + 12 = 0

The area of the triangle formed by the straight-line 3x – 4y + 12 = 0 with the coordinate axes is =

**QUESTION32.**

Find the angle between the lines 2x + y + 4 = 0 and y – 3x =7

Sol:

Given Equations are 2x + y + 4 = 0 and y – 3x =7

If θ is the angle between the lines a_{1}x + b_{1}y + c = 0 and a_{2}x + b_{2}y + c = 0, then cos θ =

⟹ cos θ =

=

=

cos θ =

∴ θ = 45^{0}

**QUESTION33.**

Find the perpendicular distance from the point (– 3, 4) to the straight line 5x – 12y = 2

Sol:

Given equation is 5x – 12y = 2 ⟹ 5x – 12y – 2 = 0

The perpendicular distance from the point (x_{1}, y_{1}) to the straight-line ax + by + c = 0 is

The perpendicular distance from the point (– 3, 4) to the straight-line 5x – 12y – 2 = 0 is

= 5

**QUESTION34.**

Find the distance between the parallel lines 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0

Sol:

Given equations are 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0

3x + 4y – 3 = 0 ⟹ 2(3x + 4y – 3) = 2(0) (multiplying on both sides by 2)

⟹ 6x + 8y – 6 = 0

The distance between two parallel lines ax + by + c_{1} =0 and ax + by + c_{2} =0 is

The distance between two parallel lines 6x + 8y – 6 = 0 and 6x + 8y – 1 = 0 is

**QUESTION35.**

Find the foot of the perpendicular from (– 1, 3) on the straight line 5x – y – 18 = 0

Sol:

Let (h, k) be the foot of the perpendicular from (– 1, 3) on the straight line 5x – y – 18 = 0

If (h, k) is the foot of the perpendicular from (x_{1}, y_{1}) on the straight-line ax + by + c = 0, then

⟹

⟹

⟹

h+ 1 = 5 and k – 3 = – 1

h = 4 and k = 2

∴ (h, k) = (4, 2)

**QUESTION36.**

Find the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0

Sol:

Let (h, k) be the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0

If (h, k) is the image of the point (x_{1}, y_{1}) with respect to the straight line ax + by + c = 0, then

If (h, k) is the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0,

⟹

⟹

⟹

h – 1 = – 4 and k + 2 = 6

h =– 3 and k = 4

∴ (h, k) = (– 3, 4)

**QUESTION37.**

If 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, –4) and (α, β), find α + β.

Sol:

Given 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, –4) and (α, β)

⟹ (α, β) is the image of (3, –4) concerning the straight line 2x – 3y – 5 = 0

⟹

α – 3 = – 4 and β + 4 = 6

α = – 1 and β = 2

α + β = – 1 + 2 = 1

**QUESTION38.**

Find the incenter of the triangle whose vertices are (1, ), (2, 0) and (0, 0)

Sol:

Let O = (0, 0), A = (1, ) and B = (2, 0)

a = AB =

b = OB =

c = OA =

in centre of a triangle with sides a, b and c, whose vertices (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is

In centre of ∆ OAB =

=

= =

**QUESTION39.**

Find the ortho centre of the triangle whose sides are given by x + y + 10 = 0, x – y – 2 = 0 and 2x + y – 7 = 0.

Sol:

Given equations are

x + y + 10 = 0 …… (1)

x – y – 2 = 0 …… (2)

2x + y – 7 = 0. …… (3)

Slope of the line (1) is – 1

Slope of the line (2) is 1

The straight lines (1) and (2) are perpendicular lines

The point of intersection of (1) and (2) is the orthocenter

x + y = – 10 ⟹ y = – 10 – x

from (2) x – y – 2 = 0

x – (– 10 – x) – 2 = 0

x + 10 + x – 2 = 0

2x + 8 = 0 ⟹ x = – 4

⟹ y = – 10 + 4 = – 6

∴ Orthocenter is (– 4, – 6)

**QUESTION40.**

Find the circum centre of the triangle whose sides are x = 1, y = 1 and x + y = 1

Sol:

Given equations are x = 1, y = 1 and x + y = 1

A = (1, 0), B = (1, 1) and C = (0, 1)

AB and BC are perpendicular lines

Circum centre is the midpoint of hypotenuse

i.e., midpoint of AC

∴ Circum centre =

=

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