# If the lines y = 3x + 1 and 2y = x + 3 are equally indicated to the line y = mx + 4 , find the value of m

**Solution:**

The equations of the given lines are:

y = 3x + 1 ....(1)

2y = x + 3 ....(2)

y = mx + 4 ....(3)

Slope of line (1), m₁ = 3

Slope of line (2), m₂ = 1/2

Slope of line (3), m₃ = m

It is given that lines (1) and (2) are equally inclined to line (3). This means that the given angle between lines (1) and (3) equals the angle between lines (2) and (3).

Therefore,

|(m₁ - m₃)/(1 + m₁m₃)| = |(m₂ - m_{3})/(1 + m₂m₃)|

|(3 - m)/(1 + 3m)| = |(1/2 - m)/(1 + (1/2)m))|

|(3 - m)/(1 + 3m)| = |(1 - 2m)/(m + 2)|

(3 - m)/(1 + 3m) = (1 - 2m)/(m + 2) or (3 - m)/(1 + 3m) = - [(1 - 2m)/(m + 2)]

Case I:

(3 - m)/(1 + 3m) = (1 - 2m)/(m + 2)

⇒ (3 - m)(m + 2) = (1 - 2m)(1 + 3m)

⇒ - m^{2} + m + 6 = 1 + m - 6m^{2}

⇒ 5m^{2} + 5 = 0

⇒ m^{2} + 1 = 0

⇒ m^{2} = - 1

m = √- 1

Here, m = √- 1, which is not real.

Hence, this case is not possible.

Case II:

(3 - m)/(1 + 3m) = - [(1 - 2m)/(m + 2)]

⇒ (3 - m)(m + 2) = - (1 - 2m)(1 + 3m)

⇒ - m^{2} + m + 6 = - (1 + m - 6m^{2})

⇒ 7m^{2} - 2m - 7 = 0

⇒ m = [2 ± √4 - 4(7)(- 7)] / 2(7)

⇒ m = [2 ± 2 √1 + 49] / 14

⇒ m = [2 ± 2 √50] / 14

⇒ m = [2 ± 10 √2] / 14

⇒ m = (1 ± 5√2)/7

Thus, the required value of m = (1 ± 5√2)/7

NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 19

## If the lines y = 3x + 1 and 2y = x + 3 are equally indicated to the line y = mx + 4 , find the value of m.

**Summary:**

If the lines y = 3x + 1 and 2y = x + 3 are equally indicated to the line y = mx + 4 , then the value of m is (1 ± 5√2)/7

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