Question

Find the distance between the point (3, -5) from the line 3x - 4y - 26 = 0.

Answer 1

Answer:- 0.6 units

Given the linear equation,

• 3x - 4y - 26 = 0

We have to find the perpendicular distance of the point (3, -5) from the line 3x - 4y - 26 = 0

Comparing the given linear with the standard form of quadratic equation i.e., Ax + By + C = 0

We get,

• A = 3 , B = -4 , C = -26

And regarding the point (3, -5), we have

• x₁ = 3 , y₁ = -5

Using the formula to calculate distance of a point from a line as.

⇒ d = | Ax₁ + By₁ + C | / {√(A² + B²) }

⇒ d = | 3×3 + -4 × -5 + (-26) | / { √(3² + (-4)²) }

⇒ d = | 9 + 20 - 26 | / √(9 + 16)

⇒ d = | 29 - 26 | / √25

⇒ d = 3 / 5

⇒ d = 0.6 units

Answer 2

Now finding distance

$d = \frac{ |ax1 + by2 + c| }{ \sqrt{ {a}^{2} + {b}^{2} } }$

$putting \: \: values$

$= \frac{ |3(3) + ( - 4)( - 5) - 26| }{ \sqrt{ {3}^{2} + ( { - 4}^{2} } }$

$= \frac{ |9 + 20 - 26| }{ \sqrt{9 + 16} }$

$= \frac{ |29 - 26| }{ \sqrt{25} }$

$= \frac{ |3| }{ \sqrt{5 \times 5} }$

$= \frac{ |3| }{5}$

$= \frac{3}{5}$

$= \frac{3}{5} answer$