# 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 60**^{o}. If slope of one line is 2, find the **equation of the other line.**

^{o}. If slope of one line is 2, find

**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 60

^{o}= |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 p^{2} + 4q^{2} = k^{2}

**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| / âˆšA

^{2}+ B^{2}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| / âˆšA

^{2}+ B^{2}= |-k cos 2Î¸| / âˆšcos

^{2}Î¸ + sin^{2}Î¸ = k cos 2Î¸p = k cos 2Î¸

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

p

^{2}= k^{2}cos^{2}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| / âˆšA

^{2}+ B^{2}= |C| / âˆšA^{2}+ B^{2}= |-k| / âˆšsec

^{2}Î¸ + cosec^{2}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

4q

^{2}= k^{2 }sin^{2}2Î¸ ————–(iv)Now add (iii) and (iv) we get

p

^{2}+ 4q^{2}= k^{2}cos^{2 }2Î¸ + k^{2}sin^{2}2Î¸p

^{2}+ 4q^{2}= k^{2}(cos^{2}2Î¸ + sin^{2}2Î¸) (As we know that cos^{2}2Î¸ + sin^{2}2Î¸ = 1)Hence, p

^{2}+ 4q^{2}= k^{2}

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

AD is perpendicular to BC

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| / âˆšA

^{2}+ B^{2}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| / âˆš1

^{2}+ 1^{2}= 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/p**^{2} = 1/a^{2} + 1/b^{2}

^{2}= 1/a

^{2}+ 1/b

^{2}

**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| / âˆšA

^{2}+ B^{2}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| / âˆša

^{2}+ b^{2}= |-ab| / âˆša^{2}+ b^{2}Now square on both the sides we get

p

^{2}= (-ab)^{2}/ a^{2}+ b^{2}1 / p

^{2}= (a^{2}+ b^{2}) / a^{2}b^{2}Hence, 1/p

^{2}= 1/a^{2}+ 1/b^{2}

Hence, proved.

## Please

Loginto comment...