Angle between two lines

Short Questions

Write down the formula to find the angle between the lines y = m₁x + c₁ and y = m₂x + c₂. Also write the condition of perpendicularity of these lines.

The formula to find the angle between the lines y = m₁x + c₁ and y = m₂x + c₂ is:
tanθ = ±

(\frac{m_1 \, - \, m_2}{1 \, + \, m_{1}m_{2}})

Also, the condition of perpendicularity of these lines is m₁ × m₂ = -1.

Prove that the straight line passing through the points (3, -4) and (-2, 6) and the straight line having equation 2x + y + 3 = 0 are parallel.

Given equation of straight line,
2x + y + 3 = 0 --- (i)

Slope of equation (i),
m₁ = - $\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = - $\frac 21$ = -2 --- (a)

Again, slope of another straight line that pass through the point (3, -4) = (x₁, y₁) and (-2, 6) = (x₂, y₂),
m₂ = $\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}$ = $\frac{\text{6 + 4}}{\text{-2 - 3}}$ = -2 --- (b)

From (a) and (b),
m₁ = m₂ = -2

Hence, given straight lines are parallel.

If the lines 3x + my = 5 and $\frac x2$ + $\frac y3$ = 1 are parallel to each other, find the value of m.

Here, Slope of 3x + my = 5 is,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

= -

\frac 3m

Again, Slope of $\frac x2$ + $\frac y3$ = 1 is,
m₂ = - $\frac{\frac 12}{\frac 13}$ = - $\frac 3m$ = - $\frac 32$

Since, the lines are parallel, m₁ = m₂ or, - $\frac 3m$ = - $\frac 32$
∴ m = 2.

The straight line having equation 3x + 4y = 12 and 2x - py = 5 are perpendicular to each other, find the value of p.

Here, Slope of straight line having equation 3x + 4y = 12 is,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

= -

\frac 34

Again, slope of 2x - py = 5 is, m₂ = $\frac 2p$

Since, the lines are perpendicular to each other,
m₁ × m₂ = -1 or, - $\frac {6}{4p}$ = -1
∴ p = $\frac 32$

Find the slope of the line perpendicular to the straight line 3x - 4y = 10.

Here, Slope of straight line having equation 3x - 4y = 10 is,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

\frac 34

Let, Slope of another line = m₂

Since, the lines are perpendicular to each other, m₁ × m₂ = -1 or, $\frac 34$ × m₂ = -1
∴ m₂ = - $\frac 43$

If the line passing through points (3, -4) and (-2, k) is perpendicular to the line having equation 5x + y = -3, find the value of k.

Here, slope of line passing through points (3, -4) and (-2, k) is,
m₁ =

\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}

\frac{\text{k + 4}}{\text{-2 - 3}}

Also, slope of line having equation 5x + y = -3,
m₂ = - $\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = - $\frac 51$

Since, the lines are perpendicular, m₁ × m₂ = -1 or, $\frac{\text{k + 4}}{\text{-5}}$ × -5 = -1
∴ k = -5

Find the acute angle between the lines having slope $-\sqrt3$ and $\sqrt3$ respectively.

Given slopes,
m₁ =

-\sqrt3

and m₂ =

\sqrt3

Now, for acute angle,
tanθ = + $\frac{m_1\text{ - }m_2}{1 \text{ + }m_1 m_2}$ = $\frac{-\sqrt3 \text{ - } \sqrt3}{\text{1 - 3}}$
= $\frac{-2\sqrt3}{-2}$
= tan60° ∴ θ = 60°

If the slopes of the lines AB and CD are 3 and -2 respectively, find the acute angle between them.

Given slopes,
Slope of line AB (m₁) = 3
Slope of line CD (m₂) = -2

Let, θ be the angle between given lines,
tanθ = ± $\frac{m_1\text{ - }m_2}{1 \text{ + }m_1 m_2}$ = ± $\frac{3 \text{ + } 2}{\text{1 - 6}}$
= ± $\frac 55$
= ±1

Taking positive sign,
or, tanθ = 45°
∴ θ = 45°

Note: Acute angle means an angle less than 90° while obtuse angle is an angle greater than 90° but less than 180°. Example for obtuse angle

If the straight lines Px + 3y - 12 = 0 and 4y - 3x + 7 = 0 are parallel to each other, find the value of P.

Here, Slope of line Px + 3y - 12 = 0,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

= -

\frac P3

Again, slope of line 4y - 3x + 7 = 0,
m₂ = $\frac 34$

Since, the lines are parallel,
m₁ = m₂ or, - $\frac P3$ = $\frac 34$
∴ P = - $\frac 94$

If the lines 2x - 8y + 6 = 0 and px - 12y - 4 = 3 are parallel to each other, find the value of p.

Here, slope of the line 2x - 8y + 6 = 0 is,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

\frac 28

\frac 14

Also, slope of the line px - 12y - 4 = 3 is,
m₂ = $\frac{p}{12}$

Since, the lines are parallel, m₁ = m₂ or, - $\frac 14$ = $\frac{p}{12}$
∴ p = 3

If two straight lines px + qy + r = 0 and lx - my + n = 0 are perpendicular to each other, show pl = qm.

Here, slope of the line px + qy + r = 0 is,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

= -

\frac pq

Also, slope of the line lx - my + n = 0 is,
m₂ = $\frac lm$

Since, the lines are perpendicular, m₁ × m₂ = -1 or, - $\frac pq$ × $\frac lm$ = -1
∴ pl = qm

Two straight lines px + 3y - 12 = 0 and 4y - 3x + 7 = 0 are perpendicular to each other, find the value of p.

Here, slope of the line px + 3y - 12 = 0 is,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

= -

\frac p3

Also, slope of the line 4y - 3x + 7 = 0 is,
m₂ = $\frac 34$

Since, the lines are perpendicular, m₁ × m₂ = -1 or, - $\frac p3$ × $\frac 34$ = -1
∴ p = 4

If a line passing through the points (4, -p) and (-2, 6) is parallel to the line having equation 2y + 3x = -3, find the value of p.

Here, slope of the line passing through the points (4, -p) and (-2, 6) is,
m₁ =

\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}

\frac{\text{6 + p}}{\text{-2 - 4}}

\frac{\text{6 + p}}{\text{-6}}

Also, slope of the line 2y + 3x = -3 is,
m₂ = - $\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac{-3}{2}$

Since, the lines are parallel, m₁ = m₂ or, $\frac{\text{6 + p}}{\text{-6}}$ = $\frac{-3}{2}$
or, 6 + p = 9
∴ p = 3

Show that the lines y + $\sqrt3$ x + 4 = 0 and x - $\sqrt3$ y = 5 are perpendicular to each other.

Here, slope of the line y +

\sqrt3

x + 4 = 0,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

\frac{-\sqrt3}{1}

Again, slope of the line x - $\sqrt3$ y = 5,
m₂ = $\frac{1}{\sqrt3}$

Now, for the case of perpendicular, m₁ × m₂ = -1 or, $\frac{-\sqrt3}{1}$ × $\frac{1}{\sqrt3}$ = -1
∴ -1 = -1

Hence, the given lines are perpendicular to each other.

Show that the lines ax + by + c = 0 and bx - ay + c = 0 are perpendicular to each other.

Here, slope of the line ax + by + c = 0,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

= -

\frac ab

Again, slope of the line bx - ay + c = 0,
m₂ = $\frac ba$

Now, for the case of perpendicular, m₁ × m₂ = -1 or, - $\frac ab$ × $\frac ba$ = -1
∴ -1 = -1

Hence, the given lines are perpendicular to each other.

If the straight lines 2x + 3y + 6 = 0 and ax - 5y + 20 = 0 are perpendicular to each other, find the value of a.

Here, slope of the line 2x + 3y + 6 = 0,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

= -

\frac 23

Again, slope of the line ax - 5y + 20 = 0,
m₂ = $\frac a5$

Since, the lines are perpendicular,
m₁ × m₂ = -1 or, - $\frac 23$ × $\frac a5$ = -1
∴ a = $\frac{15}{2}$

Find the obtuse angle between two lines 2x - y + 4 = 0 and 3x + y + 3 = 0.

Slope of line 2x - y + 4 = 0,
m₁ = -

\frac{\text{Coefficient of x}}{\text{Coefficient of y}}

= 2
Slope of line 3x + y + 3 = 0,
m₂ = -3

Let, θ be the angle between given lines,
tanθ = ± $\frac{m_1\text{ - }m_2}{1 \text{ + }m_1 m_2}$ = ± $\frac{\text{2 - (-3)}}{\text{1 - 6}}$
= ± $\frac {5}{-5}$ = ± (-1)
= ± 1

Now, for obtuse angle, taking negative sign,
so, tanθ = -tan45°
or, tanθ = tan(180° - 45°)
∴ θ = 135°

Note: Instead of using '±' × '-' = '±' technique you can also go with 'taking positive sign' from '= ± (-1)' step. The main goal is to find the obtuse angle (which means angle between 90° to 180°).