Question

Find the length of the perpendicular from the point
(2,-3) to the line 4x + 3y + 16 = 0
lol 3​

Answer 1

Answer:

distance of a point (c,d ) fron a line

ax +by + e = o

is given as

distance = (|ac +bd +e|)/(a.a +b.b)

therfore according to question

dist = (8-9+16)/5

=. 3 units

Answer 2

The length of the perpendicular is  (x - 2)² + (y + 3)² = $(\frac{15}{\sqrt{13}})^2$.

To find : Length of the perpendicular.

Given :

Point (2, -3) to the line 4x + 3y + 16 = 0.

Formula used :

                      $d=\frac{\left|a x_{0}+b y_{0}+c\right|}{\sqrt{a^{2}+b^{2}}}$    

Here,

d is the distance.

${\left|a x_{0}+b y_{0}+c\right|}$ is the equation of the line.      

${\sqrt{a^{2}+b^{2}}}$ is the point.

Applying the given data in the above formula we get,

Point : (2, -3)

Line : 4x + 3y + 16 = 0

$d=\frac{\left|a x_{0}+b y_{0}+c\right|}{\sqrt{a^{2}+b^{2}}}$

Here, a = 2  ;  b = -3                        

$d=\frac{|(2)\times(4) +(-3)\times(3) +16|}{\sqrt{(2)^{2}+(-3)^{2}}}$

$d=\frac{|8+(-9)+16|}{\sqrt{4+9}}$          

$d=\frac{|-1+16|}{\sqrt{13}}$    

$d=\frac{|15|}{\sqrt{13}}$

$d=\frac{15}{\sqrt{13}}$

Length = (x - 2)² + (y - ( -3))² = $(\frac{15}{\sqrt{13}})^2$.

Length = (x - 2)² + (y + 3)² = $(\frac{15}{\sqrt{13}})^2$.            

Hence, the length (x - 2)² + (y + 3)² = $(\frac{15}{\sqrt{13}})^2$.

To learn more...

1. Find the value of k for which the length of perpendicular from the point (4, 1) on the line 3x - 4y + k = 0 is 2 units.

brainly.in/question/7294771        

2. A) Find length of perpendicular from the point p(3,4) on the line 3x+4y-5=0

B) Find the equation of line passing through (1, 7) and having slope 2 units.

brainly.in/question/1850282