## NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.3

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1. Reduce the following equations into slope intercept form and find their slopes and intercepts on the axes.

(i) x + 7y = 0

(ii) 6x + 3y â€“ 5 = 0

(iii) y = 0

(i) x + 7y = 0 or y = - (1/7) x + 0

âˆ´ Slope = -(1/7)Â and y-intercept = 0

(ii) 6x + 3y â€“ 5 = 0 or 3y = â€“ 6x + 5

y = â€“ 2x + 5/3

âˆ´ Slope = â€“ 2 and y-intercept = 5/3

(iii) y = 0 or y = 0.x + 0

âˆ´ Slope = 0, y-intercept = 0.

2. Reduce the following equations into intercepts form and their intercepts on the axes.

(i) 3x + 2y â€“ 12 = 0

(ii) 4x â€“ 3y = 6

(iii) 3y + 2 = 0

(i) 3x + 2y â€“ 12 = 0

or 3x + 2y = 12

or 3x/12 + 2y/12 = 1

or x/4 + y/6 = 1 (intercepts form)

where a = 4, b = 6

(ii) 4x + 3y = 6

4x/6 â€“ 3y/6 = 1

x/3/2 + y/-2 = 1 (intercepts form)

where a = 3/2, b = -2

(iii) 3y + 2 = 0

3y = 2

3y/-2 = 1 => y/-2/3 = 1

Now b = -(2/3) and no intercept along x- axis.

3. Reduce the following equations into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive x-axis.

(i) x â€“ âˆš3y + 8 = 0

(ii) y â€“ 2 = 0

(iii) x â€“ y = 4

-x +Â âˆš3y = 8

Now a = 1, b =Â âˆš3

HereÂ âˆš

Thus, -(x/2) +Â âˆš3/2 y = 8/2

-(x/2) +Â âˆš3/2 y = 4

Or x cos 120Â°Â + y sin 120Â°Â = 4 (normal form)

Where p = 4 andÂ Ï‰Â = 120Â°

y = 2

now a = 0, b = 1

hereÂ âˆš

thus, 0x + 1y = 2

or x cos 90Â°Â + ysin 90Â°Â = 2 (normal form)

whereÂ Ï‰Â = 90Â°Â and p = 2

Now a = 1, b = -1

Here,Â âˆš

Thus, 1/âˆš2 x â€“ 1/âˆš2y = 4

Or xcos 315Â°Â + ysin(315Â°) = 4(normal form)

WhereÂ Ï‰Â = 315Â°Â and p = 4

4. Find the distance of the point (â€“1, 1) from the line 12(x + 6) = 5 (y â€“ 2)

5. Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

The given line is x/3 + y/4 = 1

Multiplying by 12

4x + 3y = 12 or 4x + 3y â€“ 12 = 0

Any point on the x-axis is (x, 0)

Perpendicular distance from (x, 0) to the given line

(4x â€“ 12)/âˆš

Or | 4x - 12 |/âˆš

Taking (+ve)sign

4x â€“ 12 = 20 or 4x = 20 + 12 = 32

=> x = 32/4 = 8

Taking (-ve) sign 4x â€“ 12 = -20

4x = -20 + 12 = -8

x = -2

âˆ´Â required points on the x â€“ axis are (8, 0) and (-2, 0)

6. Find the distance between parallel lines

(i) 15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0

(ii) l(x + y) + p = 0 and l(x + y) â€“ r = 0.

(i) One of the lines is 15x + 8y â€“ 34 = 0

Putting y = 0, 15x = 34 or x = 34/15

âˆ´Â The point (34/15, 0) lies on

15x + 8y â€“ 34 = 0

âˆ´Â perpendicular distance from (34/15 , 0)

to 15x + 8y + 31 = 0 is

= |15 . 34/15 + 0 + 31|/âˆš

= 65/âˆš

(ii) l(x + y) + p = 0

lx + ly â€“ r = 0

Now A = l, B = l, C

Distance between parallel lines

d = |C1Â â€“ C2|/âˆš

= |p + r|/âˆš

7. Find equation of the line parallel to the line 3x â€“ 4y + 2 = 0 and passing through the point (â€“2, 3).

The given line is 3x â€“ 4y + 2 = 0

Slope of the line = 3/4

Let equation of the parallel line to the given line

3x â€“ 4y + 2 is 3x â€“ 4y + Î» = 0

Required line passing through point (â€“2, 3)

âˆ´ 3 Ã— (â€“2) â€“ 4 Ã— 3 + Î» = 0

=>Â â€“ 6 â€“ 12 + Î» = 0

âˆ´Â Î» = 18

âˆ´ Required equation is 3x â€“ 4y + 18 = 0

8. Find the equation of the line perpendicular to the line x â€“ 7y + 5 = 0 and having x-intercept 3.

Equation of required line is 7x + y + Î» = 0

It passes through the point (3, 0)

âˆ´ 7 Ã— 3 + 0 + Î» = 0

Î»=-21

âˆ´ Required equation is 7x + y â€“ 21 = 0.

9. Find the angles between the lines âˆš3x + y = 1 and x + âˆš3y =1.

The first line is âˆš3x +y = 1 or y = -âˆš3 . x + 1

The slope m1 of the line âˆš3x + y = 1 is - âˆš3

or, m1 = - âˆš3

The other line is

x + âˆš3y = 1 => y = -(1/âˆš3) x + 1/âˆš3

The slope m2 of the line x + âˆš3y = 1 is â€“(1/âˆš3)

âˆ´ m2 = -(1/âˆš3)

if Î¸ is the angle between two lines, then

tan Î¸ =Â

tanÎ¸ = 2/âˆš3 . 1/2 = 1/âˆš3 => Î¸ = Ï€/6

10. The line through the points (h, 3) and (4, 1) intersects the line 7x â€“ 9y â€“ 19 = 0 at right angle. Find the value of h.

Slope of the line PQ passing through P(h, 3) and Q (4, 1) is 2/(h â€“ 4)

Slope of the line AB 7x â€“ 9y â€“ 19 = 0 is 7/9.

The lines AB and PQ are perpendicular to each other

âˆ´ Product of their slopes = â€“1

Or 2/(h â€“ 4) Ã— 7/9 = -1Â Â âˆ´ 14 = -9(h â€“ 4)

Or 9h = 36 â€“ 14 or 9h = 22

Or h = 22/9.

11. Prove that the line through the point (x

The given line is Ax + By + C = 0

or y = -(A/B)x â€“ C/B

Slope of the parallel line to the line

Ax + By + C = 0 is â€“A/B

The parallel line passes through (x 1, y 1)

âˆ´ Equation of the parallel line passing through (x 1, y 1) is

âˆ´Â y â€“ y1 = -(A/B)(x â€“ x1)

=> B(y â€“ y1) = -A(x â€“ x1)

A(x â€“ x1) + B(y â€“ y1) = 0 Hence proved.

12. Two lines passing through the point (2, 3) intersects each other at an angle 60Â°. If slope of one line is 2, find equation of the other line.

Let m be the slope of the other line and given slope of one line = 2

and angle between them = 60Â°

âˆ´ tan 60 = Â± (m â€“ 2)/(1 + 2m)

=> (m â€“ 2)/(1 + 2m) = Â± âˆš3

For +ve sign,

m â€“ 2 = âˆš3(1 + 2m) = âˆš3 + 2âˆš3m

(2âˆš3 â€“ 1)m = -2-âˆš3

âˆ´ m = -((2 + âˆš3)/(2âˆš3 â€“ 1))

Equation of the line passing through P(2, 3) with slope m is

y â€“ 3 = -((2 + âˆš3)/(2âˆš3 â€“ 1))( x â€“ 2)

=> (2âˆš3 â€“ 1)y â€“ 3(2âˆš3 â€“ 1)

= -(2 + âˆš3)x + 2(2 + âˆš3)

=> (2âˆš3 â€“ 1)y â€“ 6âˆš3 + 3

= -(2 + âˆš3)x + 4 + 2âˆš3

=> (2 + âˆš3)x + (2âˆš3 â€“ 1)y â€“ 6âˆš3 + 3 â€“ 4 â€“ 2âˆš3 = 0

=> (2 + âˆš3)x + (2âˆš3 â€“ 1)y â€“ 8âˆš3 â€“ 1 = 0

For â€“ve sign (m â€“ 2)/(1 + 2m) = - âˆš3

=> m â€“ 2 = -âˆš3(1 + 2m) = -âˆš3 â€“ 2âˆš3m

(2âˆš3 + 1)m = 2 â€“ âˆš3

âˆ´ m = - ((2 â€“ âˆš3)/(1 + 2âˆš3))

Equation of the line passing through P(2, 3) with slope m is

y â€“ 3 = -((2 â€“ âˆš3)/(1 + 2âˆš3))(x â€“ 2)

=> (1 + 2âˆš3)y â€“ 3(1 + 2âˆš3)

= (2 â€“ âˆš3)x â€“ 2(2 â€“ âˆš3)

=> (1 + 2âˆš3)y â€“ 3 â€“ 6âˆš3 = (2 â€“ âˆš3)x â€“ 4+ 2âˆš3

=> (2 â€“ âˆš3)x â€“ (1 + 2âˆš3)y â€“ 4 + 2âˆš3 + 3 + 6âˆš3 = 0

=> (2 â€“ âˆš3)x â€“ (1 + 2âˆš3)y â€“ 1 + 8âˆš3 = 0

âˆ´ Required lines are

(âˆš3 + 2)x + (2âˆš3 â€“ 1)y â€“ 8âˆš3 â€“ 1 = 0

And (2 â€“ âˆš3)x â€“ (1 + 2âˆš3)y + 8âˆš3 â€“ 1 = 0.

13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (â€“1, 2).

Slope of the line joining the points A(â€“1, 2) and B(3, 4)

= (4 â€“ 2)/(3 + 1) = 2/4 = 1/2

Right bisector PQ is perpendicular to AB

âˆ´ Slope of PQ = â€“ 2

middle point of AB is ((-1 + 3)/2 , (2 + 4)/2 ) i.e. (1, 3)

Right bisector passes through M(1, 3)

Equation of right bisector PQ is y â€“ 3

= â€“ 2 (x â€“ 1) = â€“2x + 2

=>Â 2x + y â€“ 3 â€“ 2 = 0 Ãž 2x + y â€“ 5 = 0.

14. Find the coordinates of the foot of perpendicular from the point (â€“1, 3) to the line 3x â€“4y â€“ 16 = 0.

The equation of the given line is

3x â€“ 4y â€“ 16 = 0

Let the equation of a line perpendicular to the given line is 4x + 3y + k = 0, where k is a constant .If this line passes through the point (â€“1, 3), then

â€“ 4 + 9 + k = 0 => k = â€“5

âˆ´ The equation of a line passing through the point (â€“1, 3) and perpendicular to the given line is

4x + 3y â€“ 5 = 0

âˆ´ The required point of the foot of the perpendicular is the point of the intersection of the lines

3x â€“ 4y â€“ 16 = 0 â€¦ (i)

and 4x + 3y â€“ 5 = 0 â€¦ (ii)

Solving (i) and (ii) by cross-multiplication, we have

x/(20 + 48) = y/(-64 + 15) = 1/(9 + 16)

=> x/68 = y/-49 = 1/25 => x = 68/25 , y = -(49/25)

âˆ´ The required point is (68/25 , -49/25).

15. The perpendicular from the origin to the line y = mx + c meets it at the point (â€“ 1, 2). Find the values of m and c.

Let the perpendicular OM is drawn from the origin to AB.

M is the foot of the perpendicular

Slope of OM = (2 â€“ 0)/(-1 â€“ 0) = 2/-1 ;

Slope of AB = m

OM âŠ¥ AB âˆ´ m Ã— (-2) = -1Â âˆ´ m = 1/2

M(â€“1, 2) lies on AB whose equation is

y = mx + c or y = 1/2 x + c

2 = 1/2 Ã— (-1) + c => c = 2 + 1/2 = 5/2

âˆ´ m = 1/2 or c = 5/2.

16. If p and q are the lengths of perpendiculars from the origin to the lines x cosÂ Î¸Â â€“ y sinÂ Î¸Â = k cos 2Î¸Â and x secÂ Î¸Â + y cosecÂ Î¸Â = k respectively, prove that p

The perpendicular distance from the origin to the line x cosÎ¸Â â€“ y sinÎ¸Â = k cos2Î¸Â â€¦(i)

p = (kcos 2Î¸)/cos2Î¸Â + sin2Î¸Â = kcos 2Î¸,

the other line is x secÎ¸Â + y cosecÎ¸Â = k

or x/cosÎ¸Â + y/sinÎ¸Â = k

xsinÎ¸Â + ycosÎ¸Â = ksinÎ¸Â cosÎ¸

xsinÎ¸Â + ycosÎ¸Â = k/2 sin 2Î¸ â€¦â€¦.(ii)

q = (k/2 sin 2Î¸)/(sin

now p

=> p

Hence, p

1. Reduce the following equations into slope intercept form and find their slopes and intercepts on the axes.

(i) x + 7y = 0

(ii) 6x + 3y â€“ 5 = 0

(iii) y = 0

**Answer**(i) x + 7y = 0 or y = - (1/7) x + 0

âˆ´ Slope = -(1/7)Â and y-intercept = 0

(ii) 6x + 3y â€“ 5 = 0 or 3y = â€“ 6x + 5

y = â€“ 2x + 5/3

âˆ´ Slope = â€“ 2 and y-intercept = 5/3

(iii) y = 0 or y = 0.x + 0

âˆ´ Slope = 0, y-intercept = 0.

2. Reduce the following equations into intercepts form and their intercepts on the axes.

(i) 3x + 2y â€“ 12 = 0

(ii) 4x â€“ 3y = 6

(iii) 3y + 2 = 0

**Answer**(i) 3x + 2y â€“ 12 = 0

or 3x + 2y = 12

or 3x/12 + 2y/12 = 1

or x/4 + y/6 = 1 (intercepts form)

where a = 4, b = 6

(ii) 4x + 3y = 6

4x/6 â€“ 3y/6 = 1

x/3/2 + y/-2 = 1 (intercepts form)

where a = 3/2, b = -2

(iii) 3y + 2 = 0

3y = 2

3y/-2 = 1 => y/-2/3 = 1

Now b = -(2/3) and no intercept along x- axis.

3. Reduce the following equations into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive x-axis.

(i) x â€“ âˆš3y + 8 = 0

(ii) y â€“ 2 = 0

(iii) x â€“ y = 4

**Answer****(i)**Â x â€“Â âˆš3y + 8 = 0-x +Â âˆš3y = 8

Now a = 1, b =Â âˆš3

HereÂ âˆš

*a*2*Â + b*2Â =Â âˆš*1 + 3*Â =Â âˆš4 = 2Thus, -(x/2) +Â âˆš3/2 y = 8/2

-(x/2) +Â âˆš3/2 y = 4

Or x cos 120Â°Â + y sin 120Â°Â = 4 (normal form)

Where p = 4 andÂ Ï‰Â = 120Â°

**(ii)**Â y â€“ 2 = 0y = 2

now a = 0, b = 1

hereÂ âˆš

*a*2*Â + b*2 =Â âˆš*0 + 1*= 1thus, 0x + 1y = 2

or x cos 90Â°Â + ysin 90Â°Â = 2 (normal form)

whereÂ Ï‰Â = 90Â°Â and p = 2

**(iii)**Â x â€“ y = 4Now a = 1, b = -1

Here,Â âˆš

*a*2*Â + b*2Â =Â âˆš*1 + 1*Â =Â âˆš2Thus, 1/âˆš2 x â€“ 1/âˆš2y = 4

Or xcos 315Â°Â + ysin(315Â°) = 4(normal form)

WhereÂ Ï‰Â = 315Â°Â and p = 4

4. Find the distance of the point (â€“1, 1) from the line 12(x + 6) = 5 (y â€“ 2)

**Answer**5. Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

**Answer**The given line is x/3 + y/4 = 1

Multiplying by 12

4x + 3y = 12 or 4x + 3y â€“ 12 = 0

Any point on the x-axis is (x, 0)

Perpendicular distance from (x, 0) to the given line

(4x â€“ 12)/âˆš

*4*2*Â + 3*2Â = 4Or | 4x - 12 |/âˆš

*25*Â = 4 or | 4x - 12 | = 5Â Ã—Â 4 = 20Taking (+ve)sign

4x â€“ 12 = 20 or 4x = 20 + 12 = 32

=> x = 32/4 = 8

Taking (-ve) sign 4x â€“ 12 = -20

4x = -20 + 12 = -8

x = -2

âˆ´Â required points on the x â€“ axis are (8, 0) and (-2, 0)

6. Find the distance between parallel lines

(i) 15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0

(ii) l(x + y) + p = 0 and l(x + y) â€“ r = 0.

**Answer**(i) One of the lines is 15x + 8y â€“ 34 = 0

Putting y = 0, 15x = 34 or x = 34/15

âˆ´Â The point (34/15, 0) lies on

15x + 8y â€“ 34 = 0

âˆ´Â perpendicular distance from (34/15 , 0)

to 15x + 8y + 31 = 0 is

= |15 . 34/15 + 0 + 31|/âˆš

*15*2*Â + 8*2Â =| 34 + 31/âˆš*225 + 64*= 65/âˆš

*289*Â = 65/17(ii) l(x + y) + p = 0

lx + ly â€“ r = 0

Now A = l, B = l, C

_{1}Â = p, C_{2}Â = â€“rDistance between parallel lines

d = |C1Â â€“ C2|/âˆš

*A*2*Â + B*2= |p + r|/âˆš

*l*2*Â + l*2Â = |p + r|/âˆš*2*Â l = 1/âˆš*2*Â |(p + r)/l|7. Find equation of the line parallel to the line 3x â€“ 4y + 2 = 0 and passing through the point (â€“2, 3).

**Answer**The given line is 3x â€“ 4y + 2 = 0

Slope of the line = 3/4

Let equation of the parallel line to the given line

3x â€“ 4y + 2 is 3x â€“ 4y + Î» = 0

Required line passing through point (â€“2, 3)

âˆ´ 3 Ã— (â€“2) â€“ 4 Ã— 3 + Î» = 0

=>Â â€“ 6 â€“ 12 + Î» = 0

âˆ´Â Î» = 18

âˆ´ Required equation is 3x â€“ 4y + 18 = 0

8. Find the equation of the line perpendicular to the line x â€“ 7y + 5 = 0 and having x-intercept 3.

**Answer**Equation of required line is 7x + y + Î» = 0

It passes through the point (3, 0)

âˆ´ 7 Ã— 3 + 0 + Î» = 0

Î»=-21

âˆ´ Required equation is 7x + y â€“ 21 = 0.

9. Find the angles between the lines âˆš3x + y = 1 and x + âˆš3y =1.

**Answer**The first line is âˆš3x +y = 1 or y = -âˆš3 . x + 1

The slope m1 of the line âˆš3x + y = 1 is - âˆš3

or, m1 = - âˆš3

The other line is

x + âˆš3y = 1 => y = -(1/âˆš3) x + 1/âˆš3

The slope m2 of the line x + âˆš3y = 1 is â€“(1/âˆš3)

âˆ´ m2 = -(1/âˆš3)

if Î¸ is the angle between two lines, then

tan Î¸ =Â

tanÎ¸ = 2/âˆš3 . 1/2 = 1/âˆš3 => Î¸ = Ï€/6

10. The line through the points (h, 3) and (4, 1) intersects the line 7x â€“ 9y â€“ 19 = 0 at right angle. Find the value of h.

**Answer**Slope of the line PQ passing through P(h, 3) and Q (4, 1) is 2/(h â€“ 4)

Slope of the line AB 7x â€“ 9y â€“ 19 = 0 is 7/9.

The lines AB and PQ are perpendicular to each other

âˆ´ Product of their slopes = â€“1

Or 2/(h â€“ 4) Ã— 7/9 = -1Â Â âˆ´ 14 = -9(h â€“ 4)

Or 9h = 36 â€“ 14 or 9h = 22

Or h = 22/9.

11. Prove that the line through the point (x

_{1}, y_{Â 1}) and parallel to the line Ax + By + C = 0 is A(x â€“ x_{1}) + B(y â€“ y_{Â 1}Â )= 0.**Answer**The given line is Ax + By + C = 0

or y = -(A/B)x â€“ C/B

Slope of the parallel line to the line

Ax + By + C = 0 is â€“A/B

The parallel line passes through (x 1, y 1)

âˆ´ Equation of the parallel line passing through (x 1, y 1) is

âˆ´Â y â€“ y1 = -(A/B)(x â€“ x1)

=> B(y â€“ y1) = -A(x â€“ x1)

A(x â€“ x1) + B(y â€“ y1) = 0 Hence proved.

12. Two lines passing through the point (2, 3) intersects each other at an angle 60Â°. If slope of one line is 2, find equation of the other line.

**Answer**Let m be the slope of the other line and given slope of one line = 2

and angle between them = 60Â°

âˆ´ tan 60 = Â± (m â€“ 2)/(1 + 2m)

=> (m â€“ 2)/(1 + 2m) = Â± âˆš3

For +ve sign,

m â€“ 2 = âˆš3(1 + 2m) = âˆš3 + 2âˆš3m

(2âˆš3 â€“ 1)m = -2-âˆš3

âˆ´ m = -((2 + âˆš3)/(2âˆš3 â€“ 1))

Equation of the line passing through P(2, 3) with slope m is

y â€“ 3 = -((2 + âˆš3)/(2âˆš3 â€“ 1))( x â€“ 2)

=> (2âˆš3 â€“ 1)y â€“ 3(2âˆš3 â€“ 1)

= -(2 + âˆš3)x + 2(2 + âˆš3)

=> (2âˆš3 â€“ 1)y â€“ 6âˆš3 + 3

= -(2 + âˆš3)x + 4 + 2âˆš3

=> (2 + âˆš3)x + (2âˆš3 â€“ 1)y â€“ 6âˆš3 + 3 â€“ 4 â€“ 2âˆš3 = 0

=> (2 + âˆš3)x + (2âˆš3 â€“ 1)y â€“ 8âˆš3 â€“ 1 = 0

For â€“ve sign (m â€“ 2)/(1 + 2m) = - âˆš3

=> m â€“ 2 = -âˆš3(1 + 2m) = -âˆš3 â€“ 2âˆš3m

(2âˆš3 + 1)m = 2 â€“ âˆš3

âˆ´ m = - ((2 â€“ âˆš3)/(1 + 2âˆš3))

Equation of the line passing through P(2, 3) with slope m is

y â€“ 3 = -((2 â€“ âˆš3)/(1 + 2âˆš3))(x â€“ 2)

=> (1 + 2âˆš3)y â€“ 3(1 + 2âˆš3)

= (2 â€“ âˆš3)x â€“ 2(2 â€“ âˆš3)

=> (1 + 2âˆš3)y â€“ 3 â€“ 6âˆš3 = (2 â€“ âˆš3)x â€“ 4+ 2âˆš3

=> (2 â€“ âˆš3)x â€“ (1 + 2âˆš3)y â€“ 4 + 2âˆš3 + 3 + 6âˆš3 = 0

=> (2 â€“ âˆš3)x â€“ (1 + 2âˆš3)y â€“ 1 + 8âˆš3 = 0

âˆ´ Required lines are

(âˆš3 + 2)x + (2âˆš3 â€“ 1)y â€“ 8âˆš3 â€“ 1 = 0

And (2 â€“ âˆš3)x â€“ (1 + 2âˆš3)y + 8âˆš3 â€“ 1 = 0.

13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (â€“1, 2).

**Answer**Slope of the line joining the points A(â€“1, 2) and B(3, 4)

= (4 â€“ 2)/(3 + 1) = 2/4 = 1/2

Right bisector PQ is perpendicular to AB

âˆ´ Slope of PQ = â€“ 2

middle point of AB is ((-1 + 3)/2 , (2 + 4)/2 ) i.e. (1, 3)

Right bisector passes through M(1, 3)

Equation of right bisector PQ is y â€“ 3

= â€“ 2 (x â€“ 1) = â€“2x + 2

=>Â 2x + y â€“ 3 â€“ 2 = 0 Ãž 2x + y â€“ 5 = 0.

14. Find the coordinates of the foot of perpendicular from the point (â€“1, 3) to the line 3x â€“4y â€“ 16 = 0.

**Answer**The equation of the given line is

3x â€“ 4y â€“ 16 = 0

Let the equation of a line perpendicular to the given line is 4x + 3y + k = 0, where k is a constant .If this line passes through the point (â€“1, 3), then

â€“ 4 + 9 + k = 0 => k = â€“5

âˆ´ The equation of a line passing through the point (â€“1, 3) and perpendicular to the given line is

4x + 3y â€“ 5 = 0

âˆ´ The required point of the foot of the perpendicular is the point of the intersection of the lines

3x â€“ 4y â€“ 16 = 0 â€¦ (i)

and 4x + 3y â€“ 5 = 0 â€¦ (ii)

Solving (i) and (ii) by cross-multiplication, we have

x/(20 + 48) = y/(-64 + 15) = 1/(9 + 16)

=> x/68 = y/-49 = 1/25 => x = 68/25 , y = -(49/25)

âˆ´ The required point is (68/25 , -49/25).

15. The perpendicular from the origin to the line y = mx + c meets it at the point (â€“ 1, 2). Find the values of m and c.

**Answer**Let the perpendicular OM is drawn from the origin to AB.

M is the foot of the perpendicular

Slope of OM = (2 â€“ 0)/(-1 â€“ 0) = 2/-1 ;

Slope of AB = m

OM âŠ¥ AB âˆ´ m Ã— (-2) = -1Â âˆ´ m = 1/2

M(â€“1, 2) lies on AB whose equation is

y = mx + c or y = 1/2 x + c

2 = 1/2 Ã— (-1) + c => c = 2 + 1/2 = 5/2

âˆ´ m = 1/2 or c = 5/2.

16. If p and q are the lengths of perpendiculars from the origin to the lines x cosÂ Î¸Â â€“ y sinÂ Î¸Â = k cos 2Î¸Â and x secÂ Î¸Â + y cosecÂ Î¸Â = k respectively, prove that p

^{2}+ 4q^{2}= k^{2}Â .**Answer**The perpendicular distance from the origin to the line x cosÎ¸Â â€“ y sinÎ¸Â = k cos2Î¸Â â€¦(i)

p = (kcos 2Î¸)/cos2Î¸Â + sin2Î¸Â = kcos 2Î¸,

the other line is x secÎ¸Â + y cosecÎ¸Â = k

or x/cosÎ¸Â + y/sinÎ¸Â = k

xsinÎ¸Â + ycosÎ¸Â = ksinÎ¸Â cosÎ¸

xsinÎ¸Â + ycosÎ¸Â = k/2 sin 2Î¸ â€¦â€¦.(ii)

*âˆ´Â*the perpendicular distance q from the origin to the line (ii)q = (k/2 sin 2Î¸)/(sin

^{2}Î¸+ cos^{Â 2}Î¸) = k/2 sin 2Î¸now p

^{2}Â + 4q^{2}Â = k^{2}Â cos^{2}2 + 4(k/2 sin 2)^{2}=> p

^{2}Â + 4q^{2}Â = k^{2}Â cos^{2}Â 2Î¸Â + 4. K2/4 sin^{2}Â 2Î¸^{p2}Â + 4q^{2}Â = k^{2}(cos^{2}2Î¸Â + sinÂ^{2}2Î¸),Hence, p

^{2}Â + 4q^{2}Â = k^{2}17. In the triangle ABC with vertices A(2, 3), B(4, â€“1) and C(1, 2), find the equation and length of altitude from the vertex A.

**Answer**

The vertices of â–³ABC are A(2, 3), B(4, â€“1) and C(1, 2) and AM is the altitude.

Slope of BC = (2 + 1)/(1 â€“ 4) = 3/-3 = -1

âˆ´ Slope of altitude AM = 1

Now, altitude passes through A(2, 3) and has the slope 1.

*âˆ´Â*Equation of AM is y â€“ 3 = (x â€“ 2)

x â€“ y + 3 â€“ 2 = 0

or y â€“ x = 1

Equation of BC passing through B(4, â€“1) and C(1, 2)

y + 1 = (2 â€“ 4)/(1 + 1) (x â€“ 4) => y + 1 = -2/2 (x â€“ 4)

y + 1 = -(x â€“ 4)

or x + y = 3 or x + y â€“ 3 = 0

*âˆ´Â*length of altitude = AM

= perpendicular distance from A(2, 3) to BC.

= (2 + 3 â€“ 3)/(âˆš

*1*2

*Â + 1*2) = 2/âˆš2 =Â âˆš2

*âˆ´Â*Length of altitude =Â âˆš2

Equation of altitude is y â€“ x = 1 and length of altitude =Â âˆš2.

18. If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p

^{2}Â = 1/a

^{2}Â + 1/b

^{2}.

**Answer**

Equation of the line which makes intercepts a, b on the axes is x/a + y/b = 1

âˆ´ The perpendicular distance p from the origin

p =

=> 1/p =Â