Question

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

Answer 1

$hello \: frnd$

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
$\frac{3}{5}$

$hope \: it \: helps$

Answer 2

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