Question

find the length of the perpendicular drawn from the points (-2,3) to the straight line 5x-2y+4=0​

Answer 1

Question :-

Find the length of the perpendicular drawn from the points (-2,3) on the straight line 5x - 2y +4 = 0.

Answer :-

$\mapsto \: \boxed{ \sf \: D = \frac{ 12 }{ \sqrt{ 29 } } } \:$

To Find :-

→ Perpendicular distance from the given points on the given line .

Solution :-

Given point from where perpendicular is drawn on line 5x - 2y + 4 = 0 is P(-2,3)

We know that,

If any point $\sf \: P(x_1, \: y_1)$ drawn on a straight line Ax + By + C = 0

then perpendicular distance from point is -

$\mapsto \boxed{ \sf \: D = \bigg|\: \frac{ Ax_1 + By_1 + C }{ \sqrt{ {A}^{2} + {B}^{2} } } \bigg| \ \: }$

Hence ,

We have straight line → 5x - 2y + 4 = 0

compare with the given standard equation of straight line .

→ A = 5 , B = -2 and C = 4

$\sf P(-2,3) \: \to \: x_1 = -2 ,\: y_1 = 3$

Hence ,put these values in the above formula

$\mapsto \: \sf \: D = \bigg|\: \frac{ 5 \times ( - 2) + ( - 2) \times 3 + 4}{ \sqrt{ {( 5)}^{2} + {( - 2)}^{2} } } \bigg| \ \: \\ \\ \mapsto \sf \: D = \bigg|\: \frac{ - 10 - 6 + 4 }{ \sqrt{ 25 + 4 } } \bigg| \ \: \\ \\ \mapsto \: \sf \: D = \bigg|\: \frac{ - 16 + 4 }{ \sqrt{ 29 } } \bigg| \ \: \\ \\ \mapsto \: \sf \: D = \bigg|\: \frac{ - 12 }{ \sqrt{ 29} } \bigg| \ \: \\ \\ \sf \to \: distance \: can \: not \: be \: negative \: \\ \\ \mapsto \: \boxed{ \sf \: D = \frac{ 12 }{ \sqrt{ 29 } } }$

Answer 2

ÂÑŠWÊŘ

12/√29

hope this is correct please give and comments ☁