Question

Find the image of (3,8) with respect to the line x + 3y =7 assuming line to be a plane mirror.

Answer 1

Answer:

Step-by-step explanation:The equation of the given line is

x + 3y = 7 … (1)

Let point B (a, b) be the image of point A (3, 8).

Accordingly, line (1) is the perpendicular bisector of AB.

Since line (1) is perpendicular to AB,

The mid-point of line segment AB will also satisfy line (1).

Hence, from equation (1), we have

On solving equations (2) and (3), we obtain a = –1 and b = –4.

Thus, the image of the given point with respect to the given line is (–1, –4).

Answer 2

The equation of the given line is
x+3y=7
x
+
3
y
=
7
-----(1)
Let a point B(a, b) be the image of the point A(3, 8).
It is given that equation (1) in the perpendicular bisector of AB.

Hence the slope of AB = y2−y1x2−x1
y
2
−
y
1
x
2
−
x
1
i.e., m1=b−8a−3
m
1
=
b
−
8
a
−
3
Slope of the line (1) is −(coefficientofx)(coefficientofy)
−
(
c
o
e
f
f
i
c
i
e
n
t
o
f
x
)
(
c
o
e
f
f
i
c
i
e
n
t
o
f
y
)
(i.e.,) m2=−13
m
2
=
−
1
3
Since line (1) is perpendicular to AB, the product of their slopes is -1.
(i.e) m1m2=−1
m
1
m
2
=
−
1
⇒b−8a−3
⇒
b
−
8
a
−
3
×−13
×
−
1
3
=−1
=
−
1
On simplifying we get,
b−83a−9
b
−
8
3
a
−
9
=1
=
1
⇒b−8=3a−9
⇒
b
−
8
=
3
a
−
9
⇒3a−b=1
⇒
3
a
−
b
=
1
---------(2)
Mid point of AB = (a+32
(
a
+
3
2
,b+82)
,
b
+
8
2
)
The midpoint of the line segment AB will also satisfy line (1).
Hence (a+32)
(
a
+
3
2
)
+3(b+82)
+
3
(
b
+
8
2
)
=7
=
7
On simplifying we get,
a+3+3b+24=14
a
+
3
+
3
b
+
24
=
14
⇒a+3b=−13
⇒
a
+
3
b
=
−
13
-------(3)
Let us solve equation (2) and (3) for a
a
and b
b
.
3a−b=1
3
a
−
b
=
1
(×3)a+3b=−13
(
×
3
)
a
+
3
b
=
−
13
3a−b=1
3
a
−
b
=
1
3a+9b=−39
3
a
+
9
b
=
−
39
−10b=40⇒b=−4
−
10
b
=
40
⇒
b
=
−
4
∴a=−1
∴
a
=
−
1
Hence a=−1andb=−4
a
=
−
1
a
n
d
b
=
−
4
Thus the image of the given point with respect to the given line is (-1, -4)