Question

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

Answer 1

18 Find the distance of the point (3, 5) from the line 3x 4y 26 = 0. We know that distance (d) of a point (x1, y1) from a line Ax + By + C = 0 is d = | _1 + _2 + |/ ( ^2 + ^2 ) Now, our equation is 3x 4y 26 = 0 The above equation is of the form Ax + By + C = 0 where A = 3 , B = 4 , C = 26 & we have to find the distance of the point (3, 5) from the line So, x1 = 3 , y1 = -5 Now finding distance d = | _1 + _2 + |/ ( ^2 + ^2 ) Putting values = |3(3) + ( 4)( 5) 26|/ (32 + ( 4)2) = |9 + 20 26|/ (9 + 16) = |29 26|/ 25 = |3|/ (5 5) = |3|/5 = 3/5 Required distance = d = 3/5 units

Answer 2

$hello \: frnd$

line is :

→ 3x-4y-26=0.....( 1 )

comparing with Ax+ßy+c=0

A=3
ß=(-4)
c=(-26)

point is (3,-5)

x1=3,y1=(-5)

↪the distance of point P (x1,y1) to Ax+say+c=0 is⤵⤵

$d = |\frac{ax1 + by1 + c}{ \sqrt[]{a { }^{2} + b {}^{2} } } |$

$d = \frac{9 + 20 - 26}{25}$

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

distance is
$\frac{3}{5}$

$hope \: it \: helps$