Question

Find the distance between line 3x + 4y - 12 and point P(2,3).​

Answer 1

Given line

• 3x + 4y - 12 = 0

⠀

Given point

• P(2,3)

⠀

To find

• Distance between the given line and the point.

⠀

Solution

• We have equation of the line and a point P(2,3).

⠀

$\: \: \: \: \: \: \: \: \: \:\underline{\sf{\red{Using\: Formula}}}$

⠀

$\large{\boxed{\boxed{\sf{d = \dfrac{|\: Ax + By + C\: |}{\sqrt{A^2 + B^2}}}}}}$

⠀

Here,

• A = 3
• B = 4
• C = -12
• x = 2
• y = 3

⠀

$\: \: \: \: \: \: \: \: \: \:\underline{\sf{\red{Putting\: the\: values}}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {d = \dfrac{|\: (3)(2) + (4)(3) - 12\: |}{\sqrt{(3)^2 + (4)^2}}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {d = \dfrac{|\: 6 + 12 - 12\: |}{\sqrt{9 + 16}}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {d = \dfrac{|\: 6 \: |}{\sqrt{25}}}$

⠀

$\bf:\implies\: \: \: \: \: \: \: \: {d = \dfrac{6}{5}}$

⠀

Hence,

• The required distance is $\sf{\dfrac{6}{5}}$ units.