Question

The distance of the point(2, 3) from the line 2x – 3y + 9 = 0
measured along the line 2x – 2y + 5 = O is
​

Answer 1

EXPLANATION.

Distance of the point ( 2,3) from the line

=> 2x - 3y + 9 = 0 measured along the line

=> 2x - 2y + 5 = 0

=> 2x - 3y = -9 ........(1)

=> 2x - 2y = -5 .......(2)

From equation (1) and (2) we get,

=> multiply equation (1) by 2

=> multiply equation (2) by 3

we get,

=> 4x - 6y = -18

=> 6x - 6y = - 15

we get,

=> - 2x = - 3

=> x = 3/2

put the value of x = 3/2 in equation (1)

we get,

=> 2 X 3/2 - 3y = -9

=> 3 - 3y = -9

=> -3y = -9 - 3

=> -3y = -12

=> y = 4

Therefore,

point of intersection = ( 3/2 , 4 )

Distance between = ( 2,3) and ( 3/2 , 4 )

$\sf : \implies \: distance \: formula \: = \sqrt{( x_{2} - x_{1}) {}^{2} + ( y_{2} - y_{1}) {}^{2} } \\ \\ \sf : \implies \: \sqrt{( \frac{3}{2} - 2) {}^{2} + (4 - 3) {}^{2} } \\ \\ \sf : \implies \: \sqrt{( \frac{3 - 4}{2} ) {}^{2} + 1 } \\ \\ \sf : \implies \: \sqrt{( \frac{1}{4} ) + 1} \\ \\ \sf : \implies \: \sqrt{ \frac{5}{4} } = \frac{ \sqrt{5} }{2}$

Answer 2

$\bf{\gray{\underbrace{\blue{GIVEN:-}}}}$

• A point (2 , 3) .

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

• The distance of the point (2 , 3) from the line 2x – 3y + 9 = 0 measured along the line 2x – 2y + 5 = O .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

$\bf\pink{\implies\:2x\:-\:3y\:+\:9\:=\:0\:}$ ------(1)

$\bf\pink{\implies\:2x\:-\:2y\:+\:5\:=\:0\:}$ -----(2)

✨ Substract the Equation (1) from Equation (2),

➬ 2x - 2y + 5 - (2x - 3y + 9) = 0

➬ 2x - 2y + 5 - 2x + 3y - 9 = 0

➬ y - 4 = 0

➬ y = 4

✨ Put the value of y in the equation (2)

➟ 2x - 2 × 4 + 5 = 0

➟ 2x - 8 + 5 = 0

➟ 2x - 3 = 0

➟ 2x = 3

➟ x = 3/2 = 1.5

$\huge\blue\checkmark$Hence, Point of intersection = (1.5 , 4) .

✨ So, distance between (2 , 3) and (1.5 , 4) is,

$\bf{\implies\:Distance\:=\:\sqrt{(1.5\:-\:2)^2\:+\:(4\:-\:3)^2}\:}$

$\rm{\implies\:Distance\:=\:\sqrt{(-0.5)^2\:+\:(1)^2}\:}$

$\rm{\implies\:Distance\:=\:\sqrt{0.25\:+\:1}\:}$

$\rm{\implies\:Distance\:=\:\sqrt{1.25}\:}$

$\bf\green{\implies\:Distance\:=\:\sqrt{1.25}\:units}$

$\huge\pink{\therefore}$ The distance of the point (2 , 3) from the line 2x – 3y + 9 = 0 measured along the line 2x – 2y + 5 = O is "$\bf\purple{\sqrt{1.25}\:units}$"