If two lines are perpendicular to each other then product of their slopes is equal to

We will learn how to find the condition of perpendicularity of two lines.

If two lines AB and CD of slopes m\(_{1}\) and m\(_{2}\) are perpendicular, then the angle between the lines θ is of 90°.

Therefore, cot θ = 0

⇒ \(\frac{1 + m_{1}m_{2}}{m_{2} - m_{1}}\) = 0

⇒ 1 + m\(_{1}\)m\(_{2}\) = 0

⇒ m\(_{1}\)m\(_{2}\) = -1.

Thus when two lines are perpendicular, the product of their slope is -1. If m is the slope of a line, then the slope of a line perpendicular to it is -1/m.

Let us assume that the lines y = m\(_{1}\)x + c\(_{1}\) and y = m\(_{2}\) x + c\(_{2}\) make angles α and β respectively with the positive direction of the x-axis and θ be the angle between them.

Therefore, α = θ + β = 90° + β [Since, θ = 90°]

Now taking tan on both sides we get,

tan α = tan (θ + β)

tan α = - cot  β

tan α = - \(\frac{1}{tan β}\)

or,  m\(_{1}\) =  - \(\frac{1}{m_{1}}\)    

or, m\(_{1}\)m\(_{2}\) = -1

Therefore, the condition of perpendicularity of the lines y = m\(_{1}\)x + c\(_{1}\), and y = m\(_{2}\) x + c\(_{2}\) is m\(_{1}\)m\(_{2}\) = -1.

Conversely, if m\(_{1}\)m\(_{2}\) = - 1 then

tan ∙ tan β = - 1      

\(\frac{sin α sin β}{cos α cos β}\) = -1

sin α sin β = - cos α cos β

cos α cos β + sin α sin β = 0

cos (α - β) = 0        

Therefore, α - β = 90°

Therefore, θ = α - β = 90°

Thus, the straight lines AB and CD are perpendicular to each other.

Solved examples to find the condition of perpendicularity of two given straight lines:

1. Let P (6, 4) and Q (2, 12) be the two points. Find the slope of a line perpendicular to PQ.

Solution:

Let m be the slope of PQ.

Then m = \(\frac{12 - 4}{2 - 6}\) = \(\frac{8}{-4}\) = -2

Therefore the slope of the line perpendicular to PQ = - \(\frac{1}{m}\) = ½

2. Without using the Pythagoras theorem, show that P (4, 4), Q (3, 5) and R (-1, -1) are the vertices of a right angled triangle.

Solution:

In ∆ ABC, we have:

m\(_{1}\) = Slope of the side PQ = \(\frac{4 - 5}{4 - 3}\) = -1

m\(_{2}\) = Slope of the side PR = \(\frac{4 - (-1)}{4 - (-1)}\) = 1

Now clearly we see that m\(_{1}\)m\(_{2}\) = 1 × -1 = -1

Therefore, the side PQ perpendicular to PR that is ∠RPQ = 90°.

Therefore, the given points P (4, 4), Q (3, 5) and R (-1, -1) are the vertices of a right angled triangle.

3. Find the ortho-centre of the triangle formed by joining the points P (- 2, -3), Q (6, 1) and R (1, 6).

Solution:       

The slope of the side QR of the ∆PQR is  \(\frac{6 - 1}{1 - 6}\) =  \(\frac{5}{-5}\) = -1∙

Let PS be the perpendicular from P on QR; hence, if the slope of the line PS be m then,

m × (- 1) = - 1        

or, m  = 1.

Therefore, the equation of the straight line PS is

y + 3 = 1 (x + 2)         

 or, x - y = 1     …………………(1)  

Again, the slope of the side RP of the ∆ PQR is \(\frac{6 + 3}{1 + 2}\) = 3∙

Let QT be the perpendicular from Q on RP; hence, if the slope of the line QT be m1 then,

m\(_{1}\) × 3  = -1  

or, m\(_{1}\) =  -\(\frac{1}{3}\)

Therefore, tile equation of the straight line QT is

y – 1 = -\(\frac{1}{3}\)(x - 6)                        

or,  3y – 3 = - x + 6 

Or,  x + 3y = 9 ………………(2)

Now, solving equations (1) and (2) we get, x = 3, y = 2.

Therefore, the co-ordinates of the point of intersection of the lines (1) and (2) are (3, 2).

Therefore, the co-ordinates of the ortho-centre of the ∆PQR = the co-ordinates of the point of intersection of the straight lines PS and QT = (3, 2).

● The Straight Line

11 and 12 Grade Math

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Let $l$ be a line with slope $\frac{1}{0}$, and $m$ a line with slope $0$. Now, line $l$ is perpendicular to line $l$, therefore the product of the slope of line $l$ and the slope of line $m$ equals $-1$. This implies that $\frac{1}{0} \times 0 = -1$!

Perpendicular lines are lines that intersect at right angles.

If you multiply the slopes of two perpendicular lines in the plane, you get − 1 . That is, the slopes of perpendicular lines are opposite reciprocals .

(Exception: Horizontal and vertical lines are perpendicular, though you can't multiply their slopes, since the slope of a vertical line is undefined.)

We can write the equation of a line perpendicular to a given line if we know a point on the line and the equation of the given line.

Example :

Write the equation of a line that passes through the point ( 1 , 3 ) and is perpendicular to the line y = 3 x + 2 .

Perpendicular lines are lines that intersect at right angles.

The slope of the line with equation y = 3 x + 2 is 3 . If you multiply the slopes of two perpendicular lines, you get − 1 .

3 ⋅ ( − 1 3 ) = − 1

So, the line perpendicular to y = 3 x + 2 has the slope − 1 3 .

Now use the point-slope form to find the equation.

y − y 1 = m ( x − x 1 )

We have to find the equation of the line which has the slope − 1 3 and passes through the point ( 1 , 3 ) . So, replace m with − 1 3 , x 1 with 1 , and y 1 with 3 .

y − 3 = − 1 3 ( x − 1 )

Use the distributive property .

y − 3 = − 1 3 x + 1 3

Add 3 to each side.

y − 3 + 3 = − 1 3 x + 1 3 + 3                         y = − 1 3 x + 10 3

Therefore, the line y = − 1 3 x + 10 3 is perpendicular to the line y = 3 x + 2 and passes through the point ( 1 , 3 ) .