Question

Find the image of (1,-2) with respect to the straight line 2x-3y+5=0​

Answer 1

Answer:

(h,k) is the image of (1 ,-2) w.r.t. the line 2x-3y+5=0.

h-1/2=k+2/-3=-2(2+6+5)/4+9=-2

h=-3?k=4...

(-3,4) IS THE image of(-1,2)in the line 2x-3y+5=0

Answer 2

Answer:

(-3,4)

Step-by-step explanation:

• A point that is at the same perpendicular distance from a straight line is considered to be the image of that point.
• A reflected duplicate of a point that appears to be identical but in reverse is called a mirror image.
• example: In the x-axis, the mirror image of (x, y) is equal to ( x , -y)
• In this case, the given point is (h,k), and the image of that point is found by using the following formula: (x,y)
• $\frac{x-h}{a}=\frac{y-k}{b} =-2\frac{ah+bk+c}{a^2+b^2}$
• At this point: (1,-2)
• Line equation: 2x - 3y + 5 = 0.
• $\frac{x-1}{2}=\frac{y-(-2)}{-3}=-2\frac{2(1)-3(-2)+5}{2^2+3^2}$
• $=-2\frac{13}{13} =-2$
• x coordinate:
• $\frac{x-1}{2} =-2\\x=-3$
• y coordinate:
• $\frac{y+2}{-3}=-2\\ y=4$
• Image of point (1,-2) is (-3,4) wrt line 2x-3y+5=0

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