# Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.7

### Question 1: Find the equation of a line for which

(i) p = 5, Î± = 60Â°

(ii) p = 4, Î± = 150Â°

(iii) p = 8, Î± = 225Â°

(iv) p = 8, Î± = 300Â°

Solution:

(i) p = 5, Î± = 60Â°

Given: p = 5, Î± = 60Â°

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

x cos 60Â° + y sin 60Â° = 5

x/2 + âˆš3y/2 = 5

x + âˆš3y = 10

Therefore, the equation of line is x + âˆš3y = 10.

(ii) p = 4, Î± = 150Â°

Given: p = 4, Î± = 150Â°

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

x cos 150Â° + y sin 150Â° = 4

x cos(180Â° â€“ 30Â°) + y sin(180Â° â€“ 30Â°) = 4 [As, cos (180Â° â€“ Î¸) = â€“ cos Î¸ , sin (180Â° â€“ Î¸) = sin Î¸]

â€“ x cos 30Â° + y sin 30Â° = 4

â€“âˆš3x/2 + y/2 = 4

-âˆš3x + y = 8

Therefore, the equation of line is -âˆš3x + y = 8.

(iii) p = 8, Î± = 225Â°

Given: p = 8, Î± = 225Â°

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

x cos 225Â° + y sin 225Â° = 8

– x/âˆš2 – y/âˆš2  = 8

x + y + 8âˆš2  = 0

Therefore, the equation of line is x + y + 8âˆš2  = 0

(iv) p = 8, Î± = 300Â°

Given: p = 8, Î± = 300Â°

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

x cos 300Â° + y sin 300Â° = 8

x/2 – yâˆš3/2 = 8

x – âˆš3y = 16

Therefore, the equation of line is x – âˆš3y = 16

### Question 2: Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x-axis is 30Â°.

Solution:

Given: p = 4, Î± = 30Â°

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

x cos 30Â° + y sin 30Â° = 4

xâˆš3/2 + y1/2 = 4

âˆš3x + y = 8

Therefore, the equation of line is âˆš3x + y = 8.

### Question 3: Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x-axis is 15Â°.

Solution:

Given: p = 4, Î± = 15Â°

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

x cos 15Â° + y sin 15Â° = 4

Now as, cos 15Â° = cos (45Â° â€“ 30Â°) = cos45Â°cos30Â° + sin45Â°sin30Â°  [Since, cos (A â€“ B) = cos A cos B + sin A sin B ]

= 1/âˆš2 Ã— âˆš3/2 + 1/âˆš2 Ã— 1/2

= 1/2âˆš2( âˆš3 + 1 )

And sin 15 = sin (45Â° â€“ 30Â°) = sin 45Â° cos 30Â° â€“ cos 45Â° sin 30Â°  [Since, Sin (A â€“ B) = sin A cos B â€“ cos A sin B ]

= 1/âˆš2 Ã— âˆš3/2 – 1/âˆš2 Ã— 1/2

= 1/2âˆš2(âˆš3 – 1)

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

x Ã— [1/2âˆš2(âˆš3 + 1)] + y Ã— [1/2âˆš2(âˆš3 – 1)]

(âˆš3+1)x +(âˆš3-1) y = 8âˆš2

Therefore, the equation of line is (âˆš3+1)x +(âˆš3-1) y = 8âˆš2.

### Question 4: Find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle Î± given by tan Î± = 5/12 with the positive direction of xâ€“axis.

Solution:

Given: p = 3, Î± = tan-1 (5/12)

tan Î± = 5/12

So,

sin Î± = 5/13

cos Î± = 12/13

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

12x/13 + 5y/13 = 3

12x + 5y = 39

Therefore, the equation of line is 12x + 5y = 39.

### Question 5: Find the equation of the straight line on which the length of the perpendicular from the origin is 2 and the perpendicular makes an angle Î± with xâ€“axis such that sin Î± = 1/3.

Solution:

Given: p = 2, sin Î± = 1/3

As, cos Î± = âˆš(1 â€“ sin2 Î±)

= âˆš(1 â€“ 1/9)

= 2âˆš2/3

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

x2âˆš2/3 + y/3 = 2

2âˆš2x + y = 6

Therefore, the equation of line is 2âˆš2x + y = 6.

### Question 6: Find the equation of the straight line upon which the length of the perpendicular from the origin is 2 and the slope of this perpendicular is 5/12.

Solution:

Given: p = Â±2

tan Î± = 5/12

Therefore, sin Î± = 5/13

cos Î± = 12/13

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

12x/13 + 5y/13 = Â±2

12x + 5y = Â±26

Therefore, the equation of line is 12x + 5y = Â±26.

### Question 7: The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150Â° with the positive direction of y-axis. Find the equation of the line.

Solution:

Given: p = 7 (perpendicular distance from origin)

Also given that the angle made with y-axis is 150Â°

therefore, the angle made with x-axis is 180Â° – 150Â° = 30Â°

sin 30Â° = 5/13

cos 30Â° = 12/13

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

x(âˆš3/2) + y(1/2) = 7

âˆš3x + y = 14

Therefore, the equation of line is âˆš3x + y = 14

### Question 8: Find the value of Î¸ and p if the equation xcosÎ¸ + ysinÎ¸ = p id the normal of the line âˆš3x + y + 2 = 0

Solution:

Given equation of line = âˆš3x + y + 2 = 0

Which can also be written as -âˆš3x – y = 2

(-âˆš3/2)x + (-1/2)y = 1

This is same as the equation of line i.e. x cos Î± + y sin Î± = p

Therefore, cosÎ¸ = -âˆš3/2

sinÎ¸ = -1/2

p = 1

Hence, Î¸ = 210Â° = 7Ï€/6 and p =1

### Question 9: Find the equation of the straight line which makes a triangle of area 96âˆš3 with the axes and perpendicular from the origin to it makes an angle of 30Â° with y-axis.

Solution:

Given: Perpendicular from origin makes an angle of 30Â° with y-axis.

Therefore, it makes 60Â° with the x-axis.

Also given area of triangle = 96âˆš3

1/2 Ã— 2p Ã— 2p/âˆš3 = 96âˆš3

p2 = (96âˆš3 Ã— âˆš3) / 2 = 48 Ã— 3 = 144

p = 12

By using the formula of the equation of line, x cos Î± + y sin Î± = p, and substituting the values, we get

x cos60Â° + y sin60Â° = 12

(1/2)x + (âˆš3/2) = 12

x + âˆš3y = 24

Therefore, the equation of line is x + âˆš3y = 24

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