Question

Find the equation of a line through the point (1,2) whose distance from the point (3,1) has the greatest value .

Answer 1

Equation of the line passing through (1, 2) is

⇒ (y - 2) = m ( x - 1)

⇒ y - 2 = mx - m

⇒mx - y - ( m - 2) = 0.

Distance of the line L(x, y) = ax + by + c = 0 from (m, n) is

$\frac{L ( m, n) }{ \sqrt{ {a}^{2} + {b}^{2} } } = \frac{am + bn + c}{ \sqrt{ {a}^{2} + {b}^{2} } }$

So, Distance of mx - y - (m - 2) = 0 from (3,1) is

$= \frac{m(3) - 1 - m + 2}{ \sqrt{ {m} ^{2} + {1}^{2} } } \\ \\ = \frac{3m - m + 1}{ \sqrt{ {m}^{2} + 1 } } \\ \\ = \frac{2m + 1}{ \sqrt{ { m}^{2} + 1 } }$

For this value to be maximum, It's derivative must be 0.

Let's say this is f(m) then

f'(m) = 0

$f'(m) = \frac{ \sqrt{ { m}^{2} + 1 } (2) - (2m + 1)( \frac{2m}{2 \sqrt{m ^{2} + 1} } )}{ {m}^{2} + 1}$

Since we need f'(m) = 0, The denominator can be ignored. Let's simplify the numerator.

$= \sqrt{ {m}^{2} + 1 } (2 - m \frac{2m + 1}{ {m}^{2} + 1} ) \\ \\ = \sqrt{ {m}^{2} + 1} (2 {m}^{2} + 2 - 2m^2 - m) \div ( {m}^{2} + 1) \\ \\ = \sqrt{m ^{2} + 1 } (2 - m) \times \frac{1}{ {m}^{2} + 1 } \\ \\ = \frac{2 - m}{ \sqrt{ {m}^{2} + 1 } }$

Now f'(m) = 0

2 - m = 0

m = 2.

The required line is mx - y - ( m - 2) = 0.

⇒ 2x - y = 0 is the required line.

2x - y = 0 is the required line.

Answer 2

Answer:

Step-by-step explanation:

Equation of the line passing through (1, 2) is

⇒ (y - 2) = m ( x - 1)

⇒ y - 2 = mx - m

⇒mx - y - ( m - 2) = 0.

Distance of the line L(x, y) = ax + by + c = 0 from (m, n) is

So, Distance of mx - y - (m - 2) = 0 from (3,1) is

For this value to be maximum, It's derivative must be 0.

Let's say this is f(m) then

f'(m) = 0

Since we need f'(m) = 0, The denominator can be ignored. Let's simplify the numerator.

Now f'(m) = 0

2 - m = 0

m = 2.

The required line is mx - y - ( m - 2) = 0.

⇒ 2x - y = 0 is the required line.

2x - y = 0 is the required line.