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Ncert Solutions Maths class 11th Maths Class 11th Straight Lines

50. Find the coordinates of the foot of perpendicular from the point (–1, 3) to theline 3x – 4y – 16 = 0.

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alok kumar singh | Contributor-Level 10

4 months ago

50. Let P(-1, 3) be the given point and Q(x, y,) be the Co-ordinate of the foot of perpendicular

So, slope of line 3x - 4y - 16 = 0 is

$m_{1} = \frac{- 3}{- 4} = \frac{3}{4}$

And slope of line segment joining P(-1, 3) and Q(x, y,) is

$m_{2} = \frac{y_{1} - 3}{x_{1} - (- 1)} = \frac{y_{1} - 3}{x_{1} + 1}$

As they are perpendicular we can write as,

$m_{1} = \frac{- 1}{m_{2}}$

$\frac{3}{4} = - \frac{1}{(y_{1} - {3 / x}_{} + 1)}$

$\Rightarrow \frac{3}{4} = - \frac{(x_{1} + 1)}{(y_{1} - 3)}$

(y₁-3)3 = - 4(x₁ +1)

3y₁- 9 = - 4x₁- 4.

4x₁ + 3y₁-9 + 4 = 0

4x₁ + 3y₁-5 = 0 ___ (1)

As point Q(x₁, y₁) lies on the line 3x- 4y - 16 = 0 it must satisfy the equation hence,

3x₁- 4y₁- 16 = 0 ____ (2)

Now, multiplying equation (1) by 4 and equation (2) by 3 and adding then,

4× (4x₁ + 3y₁- 5) + 3(3x₁- 4y₁- 16) = 0.

16x₁ + 12y₁- 20 + 9x₁- 12y₁- 48 = 0

25x₁ = 48 + 20

$\Rightarrow x_{1} = \frac{68}{25}$ .

Putting value of x₁ in equation (1) w

...more

50. Let P(-1, 3) be the given point and Q(x, y,) be the Co-ordinate of the foot of perpendicular

So, slope of line 3x - 4y - 16 = 0 is

$m_{1} = \frac{- 3}{- 4} = \frac{3}{4}$

And slope of line segment joining P(-1, 3) and Q(x, y,) is

$m_{2} = \frac{y_{1} - 3}{x_{1} - (- 1)} = \frac{y_{1} - 3}{x_{1} + 1}$

As they are perpendicular we can write as,

$m_{1} = \frac{- 1}{m_{2}}$

$\frac{3}{4} = - \frac{1}{(y_{1} - {3 / x}_{} + 1)}$

$\Rightarrow \frac{3}{4} = - \frac{(x_{1} + 1)}{(y_{1} - 3)}$

(y₁-3)3 = - 4(x₁ +1)

3y₁- 9 = - 4x₁- 4.

4x₁ + 3y₁-9 + 4 = 0

4x₁ + 3y₁-5 = 0 ___ (1)

As point Q(x₁, y₁) lies on the line 3x- 4y - 16 = 0 it must satisfy the equation hence,

3x₁- 4y₁- 16 = 0 ____ (2)

Now, multiplying equation (1) by 4 and equation (2) by 3 and adding then,

4× (4x₁ + 3y₁- 5) + 3(3x₁- 4y₁- 16) = 0.

16x₁ + 12y₁- 20 + 9x₁- 12y₁- 48 = 0

25x₁ = 48 + 20

$\Rightarrow x_{1} = \frac{68}{25}$ .

Putting value of x₁ in equation (1) we get,

$4 \times \frac{68}{25} + 3 y_{1} - 5 = 0$

4× 68 + 25× 3y₁ = 25 ×5.

25× 3y₁ = 125 - 272

$y_{1} = \frac{- 147}{25 \times 3}$

$\Rightarrow y_{1} = - \frac{49}{25}$

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50. Find the coordinates of the foot of perpendicular from the point (–1, 3) to theline 3x – 4y – 16 = 0.

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