Question

Derive an expression for the acute angle between two lines having slopes m1 and m2 and hence find the acute angle between the lines x+y-6=0 and x-y -5=0.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that line having slope m₁ makes an angle a with positive of x - axis and line having slope m₂ makes an angle b with positive direction of x - axis.

So,

$\rm\implies \:tana = m_1$

and

$\rm\implies \:tanb = m_2$

Let further assume that θ be the acute angle between the lines.

Now, from the figure,

b is exterior angle and a and θ are interior opposite angles of a triangle.

So, using exterior angle triangle property of a triangle, the exterior angle is always equals to sum of interior opposite angles.

So,

$\rm \: b \: = \: a \: + \: \theta \:$

$\rm \: θ \: = \: b \: - \: a$

$\rm \: tanθ \: = tan(\: b \: - \: a \: )$

$\rm \: tan\theta = \dfrac{tanb - tana}{1 + tanb \: tana} \:$

So, on substituting the values of tanb and tana, we get

$\rm \: tan\theta = \dfrac{m_2 - m_1}{1 + m_1 \times m_2}$

As we have to find the acute angle, so

$\rm \: tan\theta = \bigg |\dfrac{m_2 - m_1}{1 + m_1 \times m_2} \bigg |$

Now, given lines are

$\rm \: x + y - 6 = 0$

and

$\rm \: x - y - 5 = 0$

Now, Consider

$\rm \: x + y - 6 = 0$

can be rewritten as

$\rm \: y = - x + 6$

We know,

The line of the form y = mx + c represents equation of line having slope m.

$\rm\implies \:m_1 = - 1$

Now, Consider

$\rm \: x - y - 5 = 0$

can be rewritten as

$\rm \: y = x - 5$

$\rm\implies \:m_2 = 1$

Let assume that θ be the angle between the lines.

We know,

$\rm \: tan\theta = \bigg |\dfrac{m_1 - m_2}{1 + m_1 \times m_2} \bigg |$

On substituting the values, we get

$\rm \: tan\theta = \bigg |\dfrac{ 1 - ( - 1)}{1 + ( - 1) \times (1)} \bigg |$

$\rm \: tan\theta = \bigg |\dfrac{ 1 + 1}{1 - 1} \bigg |$

$\rm \: tan\theta = \bigg |\dfrac{ 2}{0} \bigg |$

$\rm \: tan\theta = \infty$

$\rm\implies \:\theta = \dfrac{\pi}{2}$

It means, lines are perpendicular to each other.

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Additional Information :-

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of line parallel to y - axis passes through the point (a, b) is x = a.

Equation of line parallel to x - axis passes through the point (a, b) is y = b.

2. Point-slope form equation of line

Equation of line passing through the point (a, b) having slope m is y - b = m(x - a)

3. Slope-intercept form equation of line

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

4. Intercept Form of Line

Equation of line which makes an intercept of a and b units on x - axis and y - axis respectively is x/a + y/b = 1.

5. Normal form of Line

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle β with the positive X-axis is x cosβ + y sinβ = p.

Answer 2

only half and is there

angle between two lines