## Equation of line solution12

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In this page equation of line solution12 we are going to see solution of
each problem with detailed explanation of the worksheet slope of the
line.

(8) Find the
equation of the straight line parallel to the line 3x – y + 7 =0 and passing
through the point (1,-2)

**Solution:**

The required
line is parallel to the line 3x – y + 7 = 0 and passing through the point
(1,-2)

Slope of
required line = -3/(-1)

= 3

Equation of
required line:

(y - y1)= m (x - x1)

(y – (-2)) = 3 (x – 1)

y + 2 = 3 x – 3

3x – y – 3 – 2 = 0

3x – y – 5 = 0

(9)Find the
equation of the straight line perpendicular to the straight line x – 2 y + 3 =
0 and passing through the point (1,-2).

**Solution:**

The required
line is perpendicular to the straight line x – 2 y + 3 = 0 and passing through
the point (1,-2)

Slope of the
given line (m1) = -1/(-2)

=
1/2

Slope of the
required line = -1/(1/2)

= -2

Equation of
required line:

(y - y1) = m (x - x1)

(y – (-2)) = -2 (x – 1)

y + 2 = -2 x + 2

2x + y + 2 - 2 = 0

2x + y = 0

(10) Find the
equation of the perpendicular bisector of the straight line segment joining the
points (3, 4) and (-1, 2).

**Solution:**

The required
line being a perpendicular bisector of the straight line segment joining the
points (3, 4) and (-1, 2)

Midpoint of
the line segment = (x₁+x₂)/2 , (y₁+y₂)/2

= (3 +
(-1))/2,(4 + 2)/2

= 2/2 , 6/2

=
(1,3)

Slope of the
line segment joining the points (3, 4) and (-1, 2)

= (y₂ - y₁)/(x₂ – x₁)

=
(2 – 4)/(-1-3)

=
-2/(-4)

=
½

Equation of
the required line:

(y – y1) = m
(x – x1)

(y – 3) =
(½)(x – 1)

2(y – 3) =
1(x – 1)

2 y – 6 = x –
1

x – 2 y – 1 +
6 = 0

x – 2 y + 5 =
0

equation of line solution12 equation of line solution12