# Angle between two lines

## Short Questions

Write down the formula to find the angle between the lines y = m_{1}x + c_{1} and y = m_{2}x + c_{2}. Also write the condition of perpendicularity of these lines.

_{1}x + c

_{1}and y = m

_{2}x + c

_{2}is:

tanθ = ±$(\frac{m_1 \, - \, m_2}{1 \, + \, m_{1}m_{2}})$

Also, the condition of perpendicularity of these lines is m_{1} × m_{2} = -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.

2x + y + 3 = 0 --- (i)

Slope of equation (i),

m_{1} = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -$\frac 21$ = -2 --- (a)

Again, slope of another straight line that pass through the point (3, -4) = (x_{1}, y_{1}) and (-2, 6) = (x_{2}, y_{2}),

m_{2} = $\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}$ = $\frac{\text{6 + 4}}{\text{-2 - 3}}$ = -2 --- (b)

From (a) and (b),

m_{1} = m_{2} = -2

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

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -$\frac 3m$

Again, Slope of $\frac x2$ + $\frac y3$ = 1 is,

m_{2} = -$\frac{\frac 12}{\frac 13}$ = -$\frac 3m$ = -$\frac 32$

Since, the lines are parallel,
m_{1} = m_{2}
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.

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -$\frac 34$

Again, slope of 2x - py = 5 is,
m_{2} = $\frac 2p$

Since, the lines are perpendicular to each other,

m_{1} × m_{2} = -1
or, -$\frac {6}{4p}$ = -1

∴ p = $\frac 32$

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

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac 34$

Let, Slope of another line = m_{2}

Since, the lines are perpendicular to each other,
m_{1} × m_{2} = -1
or, $\frac 34$ × m_{2} = -1

∴ m_{2} = -$\frac 43$

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.

m

_{1}= $\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}$ = $\frac{\text{k + 4}}{\text{-2 - 3}}$

Also, slope of line having equation 5x + y = -3,

m_{2} = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -$\frac 51$

Since, the lines are perpendicular,
m_{1} × m_{2} = -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.

m

_{1}= $-\sqrt3$ and m

_{2}= $\sqrt3$

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.

Slope of line AB (m

_{1}) = 3

Slope of line CD (m

_{2}) = -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.

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -$\frac P3$

Again, slope of line 4y - 3x + 7 = 0,

m_{2} = $\frac 34$

Since, the lines are parallel,

m_{1} = m_{2}
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.

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac 28$ = $\frac 14$

Also, slope of the line px - 12y - 4 = 3 is,

m_{2} = $\frac{p}{12}$

Since, the lines are parallel,
m_{1} = m_{2}
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.

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -$\frac pq$

Also, slope of the line lx - my + n = 0 is,

m_{2} = $\frac lm$

Since, the lines are perpendicular,
m_{1} × m_{2} = -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.

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -$\frac p3$

Also, slope of the line 4y - 3x + 7 = 0 is,

m_{2} = $\frac 34$

Since, the lines are perpendicular,
m_{1} × m_{2} = -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.

m

_{1}= $\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}$ = $\frac{\text{6 + p}}{\text{-2 - 4}}$ = $\frac{\text{6 + p}}{\text{-6}}$

Also, slope of the line 2y + 3x = -3 is,

m_{2} = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac{-3}{2}$

Since, the lines are parallel,
m_{1} = m_{2}
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.

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac{-\sqrt3}{1}$

Again, slope of the line x - $\sqrt3$y = 5,

m_{2} = $\frac{1}{\sqrt3}$

Now, for the case of perpendicular,
m_{1} × m_{2} = -1
or, $\frac{-\sqrt3}{1}$ × $\frac{1}{\sqrt3}$ = -1

∴ -1 = -1

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

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -$\frac ab$

Again, slope of the line bx - ay + c = 0,

m_{2} = $\frac ba$

Now, for the case of perpendicular,
m_{1} × m_{2} = -1
or, -$\frac ab$ × $\frac ba$ = -1

∴ -1 = -1

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

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -$\frac 23$

Again, slope of the line ax - 5y + 20 = 0,

m_{2} = $\frac a5$

Since, the lines are perpendicular,

m_{1} × m_{2} = -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.

m

_{1}= -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = 2

Slope of line 3x + y + 3 = 0,

m

_{2}= -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°).