Problems on Condition of Perpendicularity

Here we will solve various types of problems on condition of perpendicularity of two lines. 

1. Prove that the lines 5x + 4y = 9 and 4x – 5y – 1 = 0 are perpendicular to each other.

Solution:

Equation of the 1st line 5x + 4y = 9.

Now we need to express the above equation in the form y = mx + c.

5x + 4y = 9

4y = -5x + 9

y = -\(\frac{5}{4}\)x + \(\frac{9}{4}\)

Therefore, the slope (m \(_{1}\)) of the 1st line = -5/4

Equation of the second line 4x - 5y - 1 = 0

Now we need to express the above equation in the form y = mx + c.

4x – 5y – 1 = 0

⟹ -5y = -4x + 1

⟹ y = \(\frac{4}{5}\)– \(\frac{1}{5}\) 

Therefore, the slope (m \(_{2}\)) of the 2nd line = \(\frac{4}{5}\)

Now,

m \(_{1}\) × m \(_{2}\) = \(\frac{-5}{4}\)  ×  \(\frac{4}{5}\)= -1

Therefore, the given lines are perpendicular to each other.


2. Find the value of k if the lines 7y = kx + 4 and x + 2y = 3 are perpendicular.

Solution:

The slope of the lines can be found by comparing the equations with y = mx + c.

Equation of the first straight line 7y = kx + 4

Now we need to express the given equation in the form y = mx + c.

7y = kx + 4

⟹ y = \(\frac{k}{7}\)x + \(\frac{4}{7}\)

Therefore, the slope (m\(_{1}\)) of the given line = \(\frac{k}{7}\)

Equation of the second line x + 2y = 3

Now we need to express the given equation in the form y = mx + c.

x + 2y = 3

⟹ 2y = -x + 3

⟹ y = -\(\frac{1}{2}\)x + \(\frac{3}{2}\)

Therefore, the slope (m\(_{2}\)) of the given line = -\(\frac{1}{2}\)

Now according o the problem the two given lines are perpendicular.

i.e., m\(_{1}\) × m\(_{2}\) = -1

⟹ \(\frac{k}{7}\) × -\(\frac{1}{2}\) = -1

⟹ -\(\frac{k}{14}\) = -1

⟹ k = 14

Therefore, the value of k = 14

 Equation of a Straight Line










10th Grade Math

From Problems on Condition of Perpendicularity to HOME




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Perimeter of a Square | How to Find the Perimeter of Square? |Examples

    Apr 25, 24 05:34 PM

    Perimeter of a Square
    We will discuss here how to find the perimeter of a square. Perimeter of a square is the total length (distance) of the boundary of a square. We know that all the sides of a square are equal. Perimete…

    Read More

  2. Perimeter of a Triangle | Perimeter of a Triangle Formula | Examples

    Apr 25, 24 05:13 PM

    Perimeter of a Triangle
    We will discuss here how to find the perimeter of a triangle. We know perimeter of a triangle is the total length (distance) of the boundary of a triangle. Perimeter of a triangle is the sum of length…

    Read More

  3. Perimeter of a Rectangle | How to Find the Perimeter of a Rectangle?

    Apr 25, 24 03:45 PM

    Perimeter of a Rectangle
    We will discuss here how to find the perimeter of a rectangle. We know perimeter of a rectangle is the total length (distance) of the boundary of a rectangle. ABCD is a rectangle. We know that the opp…

    Read More

  4. Dividing 3-Digit by 1-Digit Number | Long Division |Worksheet Answer

    Apr 24, 24 03:46 PM

    Dividing 3-Digit by 1-Digit Number
    Dividing 3-Digit by 1-Digit Numbers are discussed here step-by-step. How to divide 3-digit numbers by single-digit numbers? Let us follow the examples to learn to divide 3-digit number by one-digit nu…

    Read More

  5. Symmetrical Shapes | One, Two, Three, Four & Many-line Symmetry

    Apr 24, 24 03:45 PM

    Symmetrical Figures
    Symmetrical shapes are discussed here in this topic. Any object or shape which can be cut in two equal halves in such a way that both the parts are exactly the same is called symmetrical. The line whi…

    Read More