**Properties of perpendicular lines :**

(i) Let m₁ and m₂ be the slopes of two lines.

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

That is,

**m₁ x m₂ = -1 **

(ii) Let us consider the general form of equation of a straight line ax + by + c = 0.

If the two lines are perpendicular, then their general form of equations will differ as given in the figure below.

(iii) Let us consider the slope intercept form of equation of a straight line y = mx + b.

If the two lines are perpendicular, then their slope-intercept form equations will differ as given in the figure below

(iv) Let l₁ and l₂ be two lines.

**If the two lines are perpendicular, the angle between them will be 90**°

**The figure given below illustrates the above situation.**

**Problem 1 :**

The slopes of the two lines are 7 and (3k +2). If the two lines are perpendicular, find the value of "k"

**Solution :**

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

7(3k + 2) = - 1

21k + 14 = -1

21k = -15

k = -15/21

**k = -5/7**

**Problem 2 :**

The equations of the two perpendicular lines are

3x + 2y - 8 = 0

(5k+3) - 3y + 1 = 0

Find the value of "k"

**Solution :**

If the two lines are perpendicular, then the coefficient "y" term in the first line is equal to the coefficient of "x" term in the second line.

So, we have

5k + 3 = 2

5k = -1

**k = -1/5**

**Problem 3 :**

Find the equation of a straight line is passing through (2, 3) and perpendicular to the line 2x - y + 7 = 0.

**Solution :**

Since the required line is perpendicular to 2x - y + 7 = 0,

then, equation of the required line is x + 2y + k = 0 ------> (1)

The required line is passing through (2, 3).

So, we can plug x = 2 and y = 3 in the equation of the required line.

2 + 2(3) + k = 0

2 + 6 + k = 0

8 + k = 0

k = - 8

**Hence, the equation of the required line is x + 2y - 8 = 0.**

**Problem 4 :**

Verify, whether the two lines 3x - 2y - 7 = 0 and y = - (2x/3) + 4 are perpendicular.

**Solution :**

In the equations of the given two lines, the equation of the second line is not in general form.

Let us write the equation of the second line in general form.

y = - (2x/3) + 4

Multiply by 3 on both sides,

3y = - 2x + 12

2x + 3y - 12 = 0

Now, let us compare the equations of two lines,

3x - 2y - 7 = 0

2x + 3y - 12 = 0

When we look at the general form of equations of the above two lines, we get the following points.

(i) The sign of y- terms are different.

(ii) The coefficient of "x" term in the first equation is the coefficient of "y" term in the second equation.

(iii) The coefficient of "y" term in the first equation is the coefficient of "x" term in the second equation.

(iv) The above equations differ in constant terms.

Considering the above points, it is clear that the given two lines are perpendicular.

**Hence, the equations of the given two lines are perpendicular.**

**Problem 5 :**

Verify, whether the two lines 5x + 7y - 1 = 0 and 14x - 10y + 5 = 0 are perpendicular.

**Solution :**

In the equation of the second line 14x-10y + 5 = 0, the coefficients of "x" and "y" have the common divisor 2.

So, let us divide the second equation by 2

(14x/2) - (10y/2) + (5/2) = (0/2)

7x - 5y + 2.5 = 0

Now, let us compare the equations of two lines,

5x + 7y - 1 = 0

7x - 5y + 2.5 = 0

When we look at the general form of equations of the above two lines, we get the following points.

(i) The sign of y- terms are different.

(ii) The coefficient of "x" term in the first equation is the coefficient of "y" term in the second equation.

(iii) The coefficient of "y" term in the first equation is the coefficient of "x" term in the second equation.

(iv) The above equations differ in constant terms.

Considering the above points, it is clear that the given two lines are perpendicular.

**Hence, the equations of the given two lines are perpendicular.**

After having gone through the stuff given above, we hope that the students would have understood "Properties of perpendicular lines".

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