# Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.15

### Question 1. Find the distance of the point (4, 5) from the straight line 3x – 5y + 7 = 0

**Solution:**

From question, we have,

The line: 3x – 5y + 7 = 0

By comparing ax + by + c = 0 and 3x − 5y + 7 = 0,

a = 3, b = − 5 and c = 7

So, the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0 is

d = |(3 × 4 – 5 × 5 + 7)/√3

^{2}+ (-5^{2}) |Hence, the distance (d) is 6/√34

### Question 2. Find the perpendicular distance of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) from the origin.

**Solution:**

According to the question

Points are: (cos θ, sin θ) and (cosϕ, sin ϕ).

The equation of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ),

y – sin θ = (sinϕ – sin θ) / (cosϕ – cos θ) (x – cos θ)

(cosϕ – cos θ)y – sin θ(cosϕ – cos θ) = (sinϕ – sin θ)x – (sinϕ – sin θ)cos θ

Let us considered D be the perpendicular distance from the origin to the line

(sinϕ – sin θ)x – (cosϕ – cos θ)y + sinθcosϕ – sinϕcos θ = 0

D = |( sin θ – ϕ)/ √ (sinϕ – sin θ)

^{2}+ (cosϕ – cos θ)^{2})|= |(sin θ – ϕ) / sin

^{2}ϕ + sin^{2}θ – 2sinϕsin θ + cos^{2}ϕ + cos^{2}θ – 2cos ϕcos θ|= |1/√2 (sin (θ – ϕ)) / √1 – (cos (θ – ϕ))|

On solving these we get,

D = cos (θ – ϕ/2)

Hence, the distance is cos (θ – ϕ/2)

### Question 3. Find the length of the perpendicular from the origin to the straight line joining the two points whose coordinates are (a cos α, a sin α) and (a cos β, a sin β).

**Solution:**

According to the question

The coordinates are (a cos α, a sin α) and (a cos β, a sin β).

So, the equation line passing through (a cos α, a sin α) and (a cos β, a sin β) is,

y – a sin α = ((a sin β – a sin α) / (a cos β – a cos α)) (x – a cos α)

y – a sin α = ((sin β – sin α)/(cos β – cos α))(x – a cos α)

y – a sin α = – cot((α – β)/2)(x – a cos α)

xcot((α – β)/2) + y – a sin α – a cos α cot((α – β)/2) = 0

So, the distance of the line from the origin is

D = |(- a sin α – a cos α cot((α – β)/2)) / √cot

^{2}((α – β)/2) +1|= a|sin((α – β)/2)sin α + cos αcos((α – β)/2)|

= a |cos(((α – β)/2) – α)|

D = acos((α- β)/2)

Hence, the distance is acos((α- β)/2)

### Question 4. Show that the perpendicular let fall from any point on the straight line 2x + 11y – 5 = 0 upon the two straight lines 24x + 7y = 20 and 4x – 3y – 2 = 0 are equal to each other.

**Solution:**

According to the question

The equations of lines are

24x + 7y = 20

4x – 3y – 2 = 0

Now let us considered that Q(a, b) be any point lies on 2x + 11y − 5 = 0

So,

2a + 11b − 5 = 0

b = (5 – 2a)/11 ….. (1)

Let D1 and D2 be the perpendicular distance from point P on line 24x + 7y = 20 and 4x – 3y – 2 = 0,

So, D1 =| (24a + 7b – 20)/ √24

^{2 }+ 7^{2}|D1 = |((24a + 7) × ((5 – 2a)/11) – 20)/25)|

From (1),

D1 = | (50a – 37)/55 |

Similarly,

D2 = | (4a – 3b – 2)/ √3

^{2 }+ (-4)^{2}|From (1) we get,

D2 = | (50a – 37)/55 |

D1 = D2

Hence proved

### Question 5. Find the distance of the point of intersection of the lines 2x + 3y = 21 and 3x – 4y + 11 = 0 from the line 8x + 6y + 5 = 0.

**Solution:**

According to the question

The equations of lines are

2x + 3y = 21 ……….. (1)

3x – 4y + 11 = 0 ………… (2)

By solving the eq(1) and (2) we get :-

x/(33 – 84) = y/(-63 – 22) = 1/(-8 – 9)

x = 3,

y = 5

Hence, the point is (3, 5).

Now, we find the perpendicular distance D

of the line8x + 6y + 5 = 0from the point (3, 5) is:d = |(24 + 30 + 5)/√8

^{2 }+ 6^{2}| = 59/10 = 5.9Hence, the distance is 5.9

### Question 6. Find the length of the perpendicular from the point (4, -7) to the line joining the origin and the point of intersection of the line 2x – 3y + 14 and 5x + 4y – 7 = 0.

**Solution:**

According to the question

The equations of lines are

2x – 3y + 14 = 0

5x + 4y – 7 = 0

On solving both the equation we get

x = -35/23

y = 252/69

So, the point is (-35/23, 252/69)

and the equation of line joining origin and the point is y = mx,

where m = (252/69) / (-35/23) = -12/5

So, the equation of required line is y = -12x/5

12x + 5y = 0

Now, the perpendicular distance from (4, -7) to 12x + 5y = 0 is

D = |(12(4) + 5(-7)) / √12

^{2 }+ (-5)^{2}|D = 13/13

D = 1

Hence, the distance is 1

### Question 7. What are the points on x-axis whose perpendicular distance from the straight line x/a + y/b = 1 is a?

**Solution:**

Let us considered a point on x-axis is (a, 0)

So, the perpendicular distance from a line bx + ay = ab is,

a = |(ax

_{1 }+ by_{1 }+ c)√a^{2 }+ b^{2}|Where,

a = b, b = a, x = -ab, x

_{1}= ±a, y_{1 }= 0a = |(b(x) + a(0) – ab)/√a

^{2 }+ b^{2}|a = 0

or

a = (b(x) + a(0) – ab)/√a

^{2 }+ b^{2}a = (b/a)x = ±√a

^{2 }+ b^{2}x = 0

### Question 8. Show that the product of perpendiculars on the line (x/a)cos θ + (y/b)sin θ – 1 from the points (±√a^{2 }– b^{2}, 0) is b^{2}.

**Solution:**

According to the question

We have to prove that the perpendicular distance from (±√a

^{2 }– b^{2}, 0) to (x/a)cos θ + (y/b)sin θ – 1 is b^{2}So, D1 is the perpendicular distance from (√a

^{2 }– b^{2}, 0) to (x/a)cos θ + (y/b)sin θ – 1 isD1 = | ((√a

^{2 }– b^{2}/a)cos θ + (0x/b)sin θ – 1) / (cos^{2}θ/a^{2}) + (sin^{2}θ/b^{2}) |= ((√a

^{2 }– b^{2}/a)cos θ – 1) / (cos^{2 }θ/a^{2}) + (sin^{2}θ/b^{2}) ….(1)Also, D2 is the perpendicular distance from (-√a

^{2 }– b^{2}, 0) to (x/a)cos θ + (y/b)sin θ – 1 isD2 = |((-√a

^{2 }– b^{2}/a)cos θ + (0x/b)sin θ – 1) / (cos^{2}θ/a^{2}) + (sin^{2}θ/b^{2}) | ……(2)So, from (1) and (2) we get

((√a

^{2}-b^{2}/a^{2})cos^{2}θ-1) / (cos^{2}θ/a^{2})+(sin^{2}θ/b^{2}) = b^{2}Hence proved

### Question 9. Find the perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line x – √3y + 4 = 0.

**Solution:**

According to the question

It is given that the perpendicular of (1, 2) on the straight line x – √3y = -4

So, the equation is

y – y

_{1}= m'(x – x_{1})1 x

_{1}= 1, y₁ = 2, m = 1/√3, m’ = -√3y – 2 = -√3 (x – 1)

y + √3x – (2 + √3) = 0…… (1)

It is given that the perpendicular distance from (0, 0) to eq(1) is

D = |(ax

_{1}+ by_{1}+ c)| / √a^{2}+ b^{2}a = √3, b = 1, c = -(2 + √3)

x

_{1}= 0, y_{1 }= 0D = |√3 (0) + 1 (0) + (−2 − √3) | / (√3)

^{2}+ 1^{2}Hence, the distance is (2 + √3)/2

### Question 10. Find the distance of the point(1, 2) from the straight line with slope 5 and passing through the point of intersection of x + 2y = 5 and x – 3y = 7.

**Solution:**

According to the question

The equations of lines are

x + 2y = 5

x – 3y = 7

On solving both the equations we get a point A(29/5, -2/5)

So, the equation of line passing through point A(29/5, -2/5) which has slop 5 is

y + (2/5) = 5(x – 29/5)

5y + 2 = 25x – 145

25x – 5y – 147 = 0

So, D is the distance of (1, 2) from 25x – 5y – 147 = 0 is

D = | (25(1) – 5(2) – 147) / √25

^{2}+ 5^{2}||-132 √650|

Hence, the distance is |132 √650|

### Question 11. What are the points on the y-axis whose distance from the line is x/3 + y/4 = 1 is 4 units?

**Solution:**

Let us assume that the point on the y-axis is (0, a)

It is given that the distance of (0, a) from line 4x + 3y – 12 = 0 is 4 units.

So, by using distance formula, we get

d = |(ax₁ + by + c)/√a

^{2}+ b^{2}|4 = |(4(0) + 3(a) – 12)/√4

^{2}+ 32 |4 =|(3а – 12)/5|

4 = (-3a + 12)/5

-3a = 20 – 12

a = -8/3

And 4 =(3a – 12)/5

3a = 20 + 12

a = 32/3

Hence, the points are

(0, 32/3), (0, -8/3)

### Question 12. In the triangle ABC with vertices A(2, 3), B(4, -1), and C(1, 2). Find the equation and the length of the altitude from vertex A.

**Solution:**

According to the question

ABC is a triangle whose vertices are A(2, 3), B(4, -1), and C(1, 2)

So, the equation of BC is

y + 1 = ((2 + 1/()1 – 4))(x – 4)

y + 1 = -x + 4

x + y – 3 = 0

and AL = |(2 + 3 – 3)/√1 + 1 |

AL = √2

Here we conclude that the slope of line BC -1. So, the slope of AL is 1.

And it passes through A(2, 3) so, its equation is

y – 3 = 1(x – 2)

x – y + 1 = 0

### Question 13. Show that the path of a moving point such that its distances from two lines 3x – 2y = 5 and 3x + 2y = 5 are equal is a straight line.

**Solution:**

According to the question

The equations of lines are

3x – 2y = 5

3x + 2y = 5

Let us assume point P(h, k) is the moving point and equidistant from both the lines

So,

|(3h – 2k – 5)/√9 + 4| = |(3h + 2k – 5)/√9 + 4|

|3h – 2k – 5| = |3h + 2k – 5|

4k = 0 or 6h – 10 = 0

k = 0

3h = 5

So, the locus of (h, k) is :

y = 0

or

3x = 5, which are straight lines.

**Question 14. If sum of perpendicular distances of a variable point P(x, y) from the lines x + y – 5 = 0 and 3x – 2y = 0 is always 10. Show that P must move on a line.**

**Solution:**

According to the question

The sum of the perpendicular distances of a variable point P(x, y)

from the lines (x + y – 5) = 0 and 3x – 2y + 7 = 0 is always 10

So,

((x + y – 5)/√2) + ((3x – 2y+7) +√13) = 10

(3√2 + √13)x + (√13 – 2√2) + (7√2 – 5√13 – 10√26) = 0

So, from here we conclude that it is a straight line.

### Question 15. If the length of the perpendicular form the point (1, 1) to the line ax – by + c to unity. Show that 1/c + 1/a – 1/b = c/2ab.

**Solution:**

Given that, the length of perpendicular from point (1, 1) to ax – by + c is 1

So, we have to prove that 1/c + 1/a – 1/b = c/2ab

By using the distance formula, we get

|(a (1) – b (1) + c)/√a

^{2}+ b^{2}| = 1a – b + c = √a

^{2}+ b^{2}(a – b + c)

^{2}= a^{2}+ b^{2}a

^{2}+ b^{2}+ c^{2}+ 2ac – 2bc – 2ab = a^{2}+ b^{2}c

^{2}+ 2ac – 2bc = 2abc + 2a – 2b = 2ab/c

(c/2ab) + (2a/2ab) – (2b/2ab) = 1/c

(c/2ab) = 1/c + 1/a – 1/b

Hence proved

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