Question

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

Answer 1

Answer:

DISTANCE  = 3/5  UNITS

Step-by-step explanation:

********** HEY DEAR **********

HERE IS YOUR ANSWER:

AS WE KNOW THE DISTANCE FORMULA IS APPLIED TO FIND THE DISTANCE BETWEEN A LINE AND A POINT..

SO, HERE ALSO WE WILL APPLY DISTANCE FORMULA THAT IS:

  {MODE OF   (AX + BY + C)}/ ROOT[ A^2 +B^2]  = DISTANCE

IN THIS CASE PUT THE POINT IN PLACE OF X AND Y INT HE GIVEN EQUATION...

DISTANCE IN THIS QUESTION

=>{ MODE OF { 3(3) - 4 (-5) -26}} / ROOT ( 3^2 + 4^2)

=> {MODE OF { 9 + 20 -26}} / ROOT ( 9 + 16)

=> 3 /5 UNITS ................. ANS.

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HOPE IT HELPS YOU..

THANKS.

^_^

Answer 2

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Here is your answer.

Point is (3 , -5)
X1 = 3 , Y1 = -5

Equation is 3x - 4y - 26 = 0
0n comparing Ax + Bx +C = 0
A = 3, B = -4 and C = -26

Now Distance of a point from the line is
$= \frac{ax1 + by1 + c}{ \sqrt{a {}^{2} + b{}^{2} } } \\ \\ = \frac{(3 \times 3 + ( - 4)( - 5) + ( - 26)}{ \sqrt{3 {}^{2} + ( - 4) {}^{2} } } \\ \\ = \frac{(9 + 20 - 26)}{ \sqrt{9 + 16} } \\ \\ = \frac{3}{ \sqrt{25} } \\ \\ = \frac{3}{5}$
Therefore Distance is 3/5 units

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