# Class 11 NCERT Solutions – Chapter 10 Straight Lines – Exercise 10.3 | Set 2

### Question 11. Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x â€“ x1) + B (y â€“ y1) = 0.

Solution:

Let’s assume that the slope of line Ax + By + C = 0 be m,

Ax + By + C = 0

Therefore, y = -A/B x â€“ C/B

m = -A / B

By using the formula,

As we know that the equation of the line passing through point (x1, y1) and having slope m = -A/B is,

y â€“ y1 = m (x â€“ x1)

y â€“ y1= -A/B (x â€“ x1)

B (y â€“ y1) = -A (x â€“ x1)

Hence, A(x â€“ x1) + B(y â€“ y1) = 0

Therefore, the line through point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x â€“ x1) + B (y â€“ y1) = 0

Hence, proved.

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

Solution:

Given that, m1 = 2

Let’s assume that the slope of the first line be m1 and

The slope of the other line be m2.

Angle between the two lines is 60Â° (Given)

Therefore,

tanÎ¸ = |m1 – m2| / |1 + m1m2|

tan 60o = |2 – m2| / |1 + 2m2|

âˆš3 = Â±((2 – m2) / (1 + 2m2))

After rationalization, we got,

m2 = (2 – âˆš3) / (2âˆš3 + 1) and m2 = -(2 – âˆš3) / (2âˆš3 + 1)

Case 1: When m2 = (2 – âˆš3) / (2âˆš3 + 1)

Therefore, the equation of line passing through point (2, 3) and having slope m2 = (2 – âˆš3) / (2âˆš3 + 1) is :

y – 2 = ((2 – âˆš3) / (2âˆš3 + 1)) x (x – 2)

After solving above equation we got,

(âˆš3 – 2)x + (2âˆš3 + 1) y = 8âˆš3 – 1

Equation of line is (âˆš3 – 2)x + (2âˆš3 + 1) y = 8âˆš3 – 1.

Case 2: When m2 = -(2 – âˆš3) / (2âˆš3 + 1)

Therefore, the equation of line passing through point (2, 3) and having slope m2 = -(2 – âˆš3) / (2âˆš3 + 1) is :

y – 3 = -(2 – âˆš3) / (2âˆš3 + 1) x (x – 2)

After solving above equation we got,

(âˆš3 + 2)x + (2âˆš3 – 1) y = 8âˆš3 + 1

Equation of line is (âˆš3 + 2)x + (2âˆš3 – 1) y = 8âˆš3 + 1.

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

Solution:

Given that,

The right bisector of a line segment bisects the line segment at 90Â° and

End-points of the line segment AB are given as A (3, 4) and B (â€“1, 2).

Let’s assume that the mid-point of AB be (x, y)

x = (3-1)/2 = 2/2 = 1

y = (4+2)/2 = 6/2 = 3

(x, y) = (1, 3)

Let’s the slope of line AB be m1

m1 = (2 â€“ 4)/(-1 â€“ 3)

= -2/(-4) = 1/2

Let’s the slope of the line perpendicular to AB be m2

m2 = -1/(1/2) = -2

The equation of the line passing through (1, 3) and having a slope of â€“2 is,

(y â€“ 3) = -2 (x â€“ 1)

y â€“ 3 = â€“ 2x + 2

2x + y = 5

Hence, the required equation of the line is 2x + y = 5

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

Solution:

Let us consider the co-ordinates of the foot of the perpendicular from (-1, 3) to the line 3x â€“ 4y â€“ 16 = 0 be (a, b)

Therefore, let the slope of the line joining (-1, 3) and (a, b) be m1

m1 = (b-3)/(a+1) ,

and let the slope of the line 3x â€“ 4y â€“ 16 = 0 be m2

y = 3/4x â€“ 4

m2 = 3/4

Since these two lines are perpendicular, m1 Ã— m2 = -1 (Given)

(b-3) / (a+1) Ã— (3/4) = -1

(3b-9) / (4a+4) = -1

3b â€“ 9 = -4a â€“ 4

4a + 3b = 5 ————(i)

Point (a, b) lies on the line 3x â€“ 4y = 16

3a â€“ 4b = 16 ———-(ii)

after solving equations (i) and (ii), we get

a = 68/25 and b = -49/25

Hence, the co-ordinates of the foot of perpendicular is (68/25, -49/25)

### Question 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.

Solution:

Given that,

The perpendicular from the origin meets the given line at (â€“1, 2).

As we know that the equation of line is y = mx + c

The line joining the points (0, 0) and (â€“1, 2) is perpendicular to the given line.

therefore, the slope of the line joining (0, 0) and (â€“1, 2) = 2/(-1) = -2

Slope of the given line is m. (Assumption)

m Ã— (-2) = -1

m = 1/2

Since, point (-1, 2) lies on the given line,

y = mx + c

2 = 1/2 Ã— (-1) + c

c = 2 + 1/2 = 5/2

Hence, the values of m and c are 1/2 and 5/2 respectively.

### Question 16. If p and q are the lengths of perpendiculars from the origin to the lines x cos Î¸ âˆ’ y sin Î¸ = k cos 2Î¸ and x sec Î¸ + y cosec Î¸ = k, respectively, prove that p2 + 4q2 = k2

Solution:

Given that,

The equations of given lines are

x cos Î¸ â€“ y sin Î¸ = k cos 2Î¸ ———–(i)

x sec Î¸ + y cosec Î¸ = k —————(ii)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

d = |Ax1 + By1 + C| / âˆšA2 + B2

After comparing equation (i) and (ii) we get,

A = cos Î¸, B = -sin Î¸ and C = -k cos 2Î¸

Given that p is length of perpendicular from (0, 0) to line (i)

p = |A Ã— 0 + B Ã— 0 + C| / âˆšA2 + B2

= |-k cos 2Î¸| / âˆšcos2 Î¸ + sin2 Î¸ = k cos 2Î¸

p = k cos 2Î¸

Let’s square on both side, and we get,

p2 = k2 cos2 2Î¸ ————(iii)

Now compare eqn. (ii) with general equation of line i.e. Ax + By + C = 0, and we get,

A = sec Î¸, B = cosec Î¸ and C = -k

As we know that q is length of perpendicular from (0, 0) to line (ii)

q = |A x 0 + B x 0 + C| / âˆšA2 + B2 = |C| / âˆšA2 + B2

= |-k| / âˆšsec2 Î¸ + cosec2 Î¸ = k cos Î¸ sin Î¸

q = k cos Î¸ sin Î¸

Multiply both sides by 2, we get

2q = 2k cos Î¸ sin Î¸ = k Ã— 2sin Î¸ cos Î¸

2q = k sin 2Î¸

Squaring both sides, we get

4q2 = k2 sin22Î¸ ————–(iv)

Now add (iii) and (iv) we get

p2 + 4q2 = k2 cos2 2Î¸ + k2 sin2 2Î¸

p2 + 4q2 = k2 (cos2 2Î¸ + sin2 2Î¸) (As we know that cos2 2Î¸ + sin2 2Î¸ = 1)

Hence, p2 + 4q2 = k2

Hence, proved.

### Question 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 vertex A.

Solution:

Let’s assume that AD be the altitude of triangle ABC from vertex A.

Therefore, AD is perpendicular to BC

Given that,

Vertices A (2, 3), B (4, â€“1) and C (1, 2)

Let’s slope of line BC = m1

m1 = (-1 â€“ 2) / (4 â€“ 1)

m1 = -1

Let’s slope of line AD be m2

m1 Ã— m2 = -1

-1 Ã— m2 = -1

m2 = 1

The equation of the line passing through point (2, 3) and having a slope of 1 is,

y â€“ 3 = 1 Ã— (x â€“ 2)

y â€“ 3 = x â€“ 2

y â€“ x = 1

Equation of the altitude from vertex A = y â€“ x = 1

Length of AD = Length of the perpendicular from A (2, 3) to BC

Equation of BC is

y + 1 = -1 Ã— (x â€“ 4)

y + 1 = -x + 4

x + y â€“ 3 = 0 ————-(i)

Perpendicular distance d of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |Ax1 + By1 + C| / âˆšA2 + B2

Now compare equation (i) to the general equation of line i.e., Ax + By + C = 0, and we get,

Length of AD = |1 Ã— 2 + 1 Ã— 3 – 3| / âˆš12 + 12 = âˆš2 units (A = 1, B = 1 and C = -3)

Hence, the equation and the length of the altitude from vertex A are y â€“ x = 1 and âˆš2 units respectively.

### Question 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/p2 = 1/a2 + 1/b2

Solution:

Equation of a line whose intercepts on the axes are a and b is x/a + y/b = 1

bx + ay = ab

bx + ay â€“ ab = 0 ————-(i)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

d = |Ax1 + By1 + C| / âˆšA2 + B2

After comparing eqn. (i) with general equation of line i.e. Ax + By + C = 0 we get,

A = b, B = a and C = -ab

Let’s assume that if p is length of perpendicular from point (x1, y1) = (0, 0) to line (i), we get

p = |A x 0 + B x 0 – ab| / âˆša2 + b2 = |-ab| / âˆša2 + b2

Now square on both the sides we get

p2 = (-ab)2 / a2+ b2

1 / p2 = (a2 + b2) / a2b2

Hence, 1/p2 = 1/a2 + 1/b2

Hence, proved.

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