If two lines are parallel, then the slopes are equal.

m_{1} = m_{2}

If two lines are perpendicular, then the product of the slopes is equal to -1.

m_{1}⋅ m_{2 }= -1

**Example 1 :**

Find the equation of a straight line passing through the point P(-5, 2) and parallel to the line joining the points Q(3, -2) and R(-5, 4).

**Solution :**

The required line is passing through the point P(-5, 2) and parallel to the line joining the points Q and R.

Since the line joining the points QR is parallel to the required line, the slopes will be equal.

Slope of QR = (y_{2} - y_{1})/(x_{2} - x_{1})

= (4 - (-2))/(-5 - 3)

= (4 + 2)/(-8)

= 6/(-8)

Slope of QR = -3/4

Slope of the required line = -3/4

Equation of the line passing through the point P.

(y - y_{1}) = m(x - x_{1})

(y - 2) = (-3/4) (x - (-5))

4(y - 2) = -3 (x + 5)

4y - 8 = -3x - 15

3x + 4y - 8 + 15 = 0

3x + 4y + 7 = 0

So, the equation of the required line is 3x + 4y + 7 = 0

**Example 2 :**

Find the equation of a line passing through (6, –2) and perpendicular to the line joining the points (6, 7) and (2, -3).

**Solution :**

The required line is perpendicular to the line joining the points (6, 7) and (2, -3).

Slope of the line joining the above points

m = (y_{2} - y_{1})/(x_{2} - x_{1})

m = (-3-7)/(2-6)

m = -10/(-4)

m = 5/2

Slope of the required line = -1/(5/2)

= -2/5

The required line is passing through the point (6, -2)

(y - y_{1}) = m(x - x_{1})

(y - (-2)) = (-2/5) (x - 6)

5(y + 2) = -2 (x - 6)

5y + 10 = -2x + 12

2x + 5y + 10 - 12 = 0

2x + 5y - 2 = 0

**Example 3 :**

Find the equation of the line in the xy-plane that contains the point (3, 2) and that is parallel to the line y = 4x − 1.

**Solution :**

Since the required line is parallel to the line y = 4x - 1,

m = 4

(y - y_{1}) = m(x - x_{1})

(y - 2) = 4(x - 3)

y - 2 = 4x - 12

4x - y - 12 + 2 = 0

4x - y - 10 = 0

**Example 4 :**

Find the equation of the line that contains the point (2, 3) and that is parallel to the line containing the points (7, 1) and (5, 6).

**Solution :**

Slope of the line containing the points (7, 1) and )(5, 6)

m = (y_{2} - y_{1})/(x_{2} - x_{1})

m = (6 - 1)/(5 - 7)

m = 4/(-2)

m = -2

Equation of the line :

(y - y_{1}) = m(x - x_{1})

(y - 3) = -2(x - 2)

y - 3 = -2x + 4

2x + y - 3 - 4 = 0

2x + y - 7 = 0

**Example 5 :**

Find the equation of the line in the xy-plane that contains the point (4, 1) and that is perpendicular to the line whose equation is y = 3x + 5.

**Solution :**

The required line is perpendicular to the line y = 3x + 5

Slope (m) = 3

Slope of required line = -1/3

Equation of the line :

(y - y_{1}) = m(x - x_{1})

Point => (4, 1) and slope m = -1/3

(y - 1) = (-1/3)(x - 4)

3(y - 1) = -1(x - 4)

3y - 3 = -x + 4

x + 3y - 3 - 4 = 0

x + 3y - 7 = 0

**Example 6 :**

Find a number t such that the line in the xy plane containing the points (t, 4) and (2,−1) is perpendicular to the line y = 6x − 7.

**Solution :**

Slope of the line containing the points (t, 4) and (2, -1)

m = (y_{2} - y_{1})/(x_{2} - x_{1})

m = (-1 - 4) / (2 - t)

m = -5/(2 - t)

Slope of the given line :

y = 6x − 7

m = 6

Slope of line perpendicular to the line y = 6x - 7 is -1/6

-1/6 = -5/(2 - t)

-1(2 - t) = -5(6)

-2 + t = -30

t = -30 + 2

t = -30

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