Question

The coordinates of the point Q symmetric to the point P(-5,13) with respect to the line 2x-3y-3=0

Answer 1

Answer:

$\textsf{The point symmetric to (-5,13) is (11,-11)}$

Step-by-step explanation:

The coordinates of the point symmetric to (-5,13) with respect to the line 2x-3y-3=0 is given by

$\boxed{\bf\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2(ax_1+by_1+c)}{a^2+b^2}}$

$\text{Here, }$

$(x_1,y_1)=(-5,13)$

$a=2,\,b=-3,\,c=-3$

$\frac{x+5}{2}=\frac{y-13}{-3}=\frac{-2(2(-5)-3(13)-3)}{2^2+(-3)^2}$

$\frac{x+5}{2}=\frac{y-13}{-3}=\frac{-2(-10-39-3)}{4+9}$

$\frac{x+5}{2}=\frac{y-13}{-3}=\frac{-2(-52)}{13}$

$\frac{x+5}{2}=\frac{y-13}{-3}=8$

$\implies\frac{x+5}{2}=8\;\;\&\;\;\frac{y-13}{-3}=8$

$\implies\;x+5=16\;\;\&\;\;y-13=-24$

$\implies\;x=11\;\;\&\;\;y=-11$

$\therefore\textsf{The point symmetric to (-5,13) is (11,-11)}$

Answer 2

Answer:

The coordinates of the point symmetric to (-5,13) with respect to the line 2x-3y-3=0 is given by

\boxed{\bf\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2(ax_1+by_1+c)}{a^2+b^2}}

a

x−x

1

=

b

y−y

1

=

a

2

+b

2

−2(ax

1

+by

1

+c)

\text{Here, }Here,

(x_1,y_1)=(-5,13)(x

1

,y

1

)=(−5,13)

a=2,\,b=-3,\,c=-3a=2,b=−3,c=−3

\frac{x+5}{2}=\frac{y-13}{-3}=\frac{-2(2(-5)-3(13)-3)}{2^2+(-3)^2}

2

x+5

=

−3

y−13

=

2

2

+(−3)

2

−2(2(−5)−3(13)−3)

\frac{x+5}{2}=\frac{y-13}{-3}=\frac{-2(-10-39-3)}{4+9}

2

x+5

=

−3

y−13

=

4+9

−2(−10−39−3)

\frac{x+5}{2}=\frac{y-13}{-3}=\frac{-2(-52)}{13}

2

x+5

=

−3

y−13

=

13

−2(−52)

\frac{x+5}{2}=\frac{y-13}{-3}=8

2

x+5

=

−3

y−13

=8

\implies\frac{x+5}{2}=8\;\;\&\;\;\frac{y-13}{-3}=8⟹

2

x+5

=8&

−3

y−13

=8

\implies\;x+5=16\;\;\&\;\;y-13=-24⟹x+5=16&y−13=−24

\implies\;x=11\;\;\&\;\;y=-11⟹x=11&y=−11

\therefore\textsf{The point symmetric to (-5,13) is (11,-11)}∴The point symmetric to (-5,13) is (11,-11)