Question

find the image of (2, 3) in the line 3x+4y=4​

Answer 1

Answer:

Step-by-step explanation:

apply the formula

Answer 2

Given :

point (2,3) & line 3x+5y=4

To Find :

image of point (2,3) on the line 3x+5y=4

Important Points Related to this :

• There is a perpendicular line exists to the plane.
• Equation of normal between the plane and of the line passing through the points can be written as $\frac{x-x1}{a} =\frac{y-y1}{b} = \frac{z-z1}{c}$

Formula for finding coordinate:

$= > ( \frac{a + 1}{2} ),( \frac{b + 2}{2} )$

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Step by step explanation:

Let p(a,b) be the image of point (2,3) on the line 3x+5y=4

3x+5y=4.......(1)

$Now \: Slope = \frac{ - A}{B} = \frac{ - 3}{5}$

AP ⟂BC

$Slope \: of \: line \: AP= m2 = \frac{ \frac{ - 1}{ - 3} }{5} = \frac{5}{3}$

$\frac{b - 2}{a - 1} = \frac{5}{3}$

$= > 3b - 6 = 5a - 5$

$= > 5a - 3b = - 1.....(2)$

Now O is the midpoint of AP

$Coordinate \: of \: O = (\frac{a + 1}{2} )( \frac{b + 1}{2} )$

Now put O's coordinate in equation 1

$= > 3( \frac{a + 1}{2} ) + 5( \frac{b + 2}{2} ) - 4 = 0$

$= > \frac{3a + 3 + 5b + 10 - 8}{2} = 0$

$= > 3a + 5b + 5 = 0.....(3)$

Now equating equation 2&3

$we \: get \: a = - \frac{10}{17} ,b = - \frac{11}{17}$

Therefore ,image of point (2,3) on the line 3x+5y=4 is ($\bold{\red{- \frac{10}{17} ,- \frac{11}{17}}}$)