Question

The acute angle between two line 2x+y = 3 & 3x-y = 4 is
​

Answer 1

Solution 1:-

First line is,

$\longrightarrow L_1:2x+y-3=0$

whose slope is,

$\longrightarrow m_1=-\dfrac{2}{1}=-2$

Second line,

$\longrightarrow L_2:3x-y-4=0$

whose slope is,

$\longrightarrow m_2=-\dfrac{3}{-1}=3$

Let $\theta$ be the acute angle between the two lines which is given by,

$\longrightarrow\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$

$\longrightarrow\tan\theta=\left|\dfrac{-2-3}{1+(-2\times3)}\right|$

$\longrightarrow\tan\theta=\left|\dfrac{-5}{-5}\right|$

$\longrightarrow\tan\theta=1$

$\longrightarrow\underline{\underline{\theta=45^o}}$

Hence the angle between the lines is 45°.

Solution 2:-

Here we are changing the given lines into vector forms.

Equation of first line is,

$\longrightarrow2x+y-3=0$

$\longrightarrow2x=-y+3$

$\longrightarrow\dfrac{x-0}{\left(\dfrac{1}{2}\right)}=\dfrac{y-3}{-1}$

Hence vector form of first line is,

$\longrightarrow\vec{r_1}=\left<0,\ 3\right>+\lambda_1\left<\dfrac{1}{2},\ -1\right>$

where $\left<x,\ y\right>=x\,\hat i+y\,\hat j.$

Equation of second line is,

$\longrightarrow3x-y-4=0$

$\longrightarrow3x=y+4$

$\longrightarrow\dfrac{x-0}{\left(\dfrac{1}{3}\right)}=\dfrac{y+4}{1}$

Hence vector form of second line is,

$\longrightarrow\vec{r_2}=\left<0,\ -4\right>+\lambda_2\left<\dfrac{1}{3},\ 1\right>$

Let $\theta$ be the acute angle between the two lines, which is given by the dot product of their direction ratios, as,

$\longrightarrow\cos\theta=\dfrac{\left|\left<\dfrac{1}{2},\ -1\right>\cdot\left<\dfrac{1}{3},\ 1\right>\right|}{\sqrt{\left(\dfrac{1}{2}\right)^2+(-1)^2}\cdot\sqrt{\left(\dfrac{1}{3}\right)^2+(1)^2}}$

$\longrightarrow\cos\theta=\dfrac{\left|\dfrac{1}{6}-1\right|}{\dfrac{\sqrt5}{2}\cdot\dfrac{\sqrt{10}}{3}}$

$\longrightarrow\cos\theta=\dfrac{\left(\dfrac{5}{6}\right)}{\left(\dfrac{5}{3\sqrt2}\right)}$

$\longrightarrow\cos\theta=\dfrac{1}{\sqrt2}$

$\longrightarrow\underline{\underline{\theta=45^o}}$

Answer 2

Given:

The equations of the lines are

2x - y + 3 = 0 ... (1)

X + y + 2 = 0... (2)

Let m1 and m2 be the slopes of these

lines.

m1 = 2, m2 = -1

Let 0 be the angle between the lines. Then, by using the formula

tan 8 = [(m1- m2)/(1 + m1m2)]

= [(2 + 1)/(1 + 2)] =

= 3

So,

o = tan-1 (3)

:: The acute angle between the lines is

tan1 (3).