Find the angle between the straight lines $3x + 4y - 7 = 0$ and $4x - 3y + 5 = 0$.

Hint: The slope of a line is defined as the change in y-coordinates with respect to change in x-coordinates. The angle between two lines $3x + 4y - 7 = 0$ and $4x - 3y + 5 = 0$ of given slope $m_1$ and $m_2$ is given by the formula-
$tan\theta = \left| {\dfrac{{{{\text{m}}_2} - {{\text{m}}_1}}}{{1 + {{\text{m}}_1}{{\text{m}}_2}}}} \right|$

Complete step-by-step answer:

To find the slope of a line, we will convert it into the slope-intercept form, which is given by-
$y = mx + c$ where m is the slope of the line.
We can compare the coefficient of x to find the slope of the line.

For the line $3x + 4y - 7 = 0$,
$3x + 4y - 7 = 0$
$4y = - 3x + 7$
${\text{y}} = - \dfrac{3}{4}{\text{x}} + \dfrac{7}{4}$
By comparison we can see that the slope $m_1= \dfrac{-3}{4}$.

For the line $4x - 3y + 5 = 0$,
$4x - 3y + 5 = 0$
$3y = 4x + 5$
$y = \dfrac{4}{3}x + \dfrac{5}{3}$
By comparison we can see that the slope $m_2 = \dfrac{4}{3}$

Now, using the given formula we can find the angle between the two lines as-
$tan\theta = \left| {\dfrac{{{{\text{m}}_2} - {{\text{m}}_1}}}{{1 + {{\text{m}}_1}{{\text{m}}_2}}}} \right|$

$tan\theta = \left| {\dfrac{{\dfrac{4}{3} - \left( { - \dfrac{3}{4}} \right)}}{{1 + \left( {\dfrac{4}{3}} \right)\left( { - \dfrac{3}{4}} \right)}}} \right| \to \infty $
${{\theta }} = {90^{\text{o}}}$
Therefore the angle between the lines is $90^o$. This is the required answer.

Note: Students often forget to consider both the cases while finding the angle. We should remember that whenever we eliminate the modulus sign, then we need to replace it by the plus-minus sign, hence we get two cases and two answers. Here there is only one angle because the angle itself is $90^o$, so both the acute and obtuse angles are the same that is right angle