Question

what is the image of the point (1,2) on the line 3x + 4y -1=0​

Answer 1

Answer:

The answer is

$( \frac{ - 7}{5} \: \frac{ - 6}{5} \: )$

Step-by-step explanation:

hope it helped

Answer 2

Image of (1,2) on the line 3x+4y-1=0 is $(\frac{-7}{5} ,\frac{-6}{5})$

Step-by-step explanation:

Given,

Point => (1,2)

Line => 3x + 4y - 1 =0

=> Image of a point (x₁ , y₁) on ax + by + c =0 can be calculated from the below formula :

$\frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} = -2\frac{ax_{1}+by_{1}+c}{a^{2}+ b^{2} }$

In our question ,

x₁ = 1

y₁ = 2

a = 3

b = 4

c = -1

Substituing all these values in the image equation,

$\frac{x-1}{3} = \frac{y-2}{4} = -2(\frac{3(1)+4(2)-1}{(3)^{2}+ (4)^{2}})\\\\\frac{x-1}{3} = \frac{y-2}{4} = -2(\frac{3+8-1}{(9+16)})\\\\\frac{x-1}{3} = \frac{y-2}{4} = -2(\frac{10}{25})\\\\\frac{x-1}{3} = \frac{y-2}{4} = (\frac{-20}{25})\\\\\frac{x-1}{3} = \frac{y-2}{4} = \frac{-4}{5}\\\\\frac{x-1}{3} = \frac{-4}{5} \ and \ \frac{y-2}{4} = \frac{-4}{5}\\\\Solving \ for \ x \ and \ y, we \ get:\\\\x-1 = \frac{(-4) * 3}{5} \ and \ y-2= \frac{(-4)*4}{5}$

$x-1 = \frac{-12}{5} \ and \ y-2= \frac{-16}{5}\\\\x = \frac{-12}{5}+1 \ and \ y= \frac{-16}{5}+2\\\\x = \frac{-12+5}{5} \ and \ y= \frac{-16+10}{5}\\\\x = \frac{-7}{5} \ and \ y= \frac{-6}{5}$

=> Hence image of (1,2) on the line 3x+4y-1=0 is $(\frac{-7}{5} ,\frac{-6}{5})$