Question

Find the tangent of the angle between the lines whose intercepts on the axes are respectively a, -b and b, -a.

Answer 1

- Hint:- With the given intercepts find the points through which the 2 lines pass. Find the slope of the 2 lines, i.e. find \[{{m}_{1}}\] and \[{{m}_{2}}\] using the formula to find slope. Thus find the tangent of angle, \[\tan \theta \].

Complete step-by-step answer: -
We have been given the intercepts of the axes as a, -b and b, -a.
The intercepts of the first line are a and –b.
Thus the line passes through the point, (a, 0) and (0, -b). We can find the slope of line m, using the formula,
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
Thus the slope of the line formed by point (a, 0)and (0, -b) is \[{{m}_{1}}\].
Take \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( a,0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( 0,-b \right)\].
\[\therefore \]Slope of line, \[{{m}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{-b-0}{0-a}=\dfrac{-b}{-a}=\dfrac{b}{a}\].
Thus we got the slope of the 1st line as, \[{{m}_{1}}={}^{b}/{}_{a}\].
Now let us find the slope of the other line. Given to us are the intercepts of the lines as –b, a. Thus the points are (b, 0) and (0, -a).
Thus the line passes through the points (b, 0) and (0, -a).
The slope of the line formed by these points (b, 0) and (0, -a) is \[{{m}_{2}}\].
\[\therefore \]Slope of line, \[{{m}_{2}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\].
Take \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( b,0 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)=\left( 0,-a \right)\].
\[\therefore \]Slope of line, \[{{m}_{2}}=\dfrac{-a-0}{0-b}=\dfrac{-a}{-b}=\dfrac{a}{b}\].
Thus we got the slope of the 2nd line as, \[{{m}_{2}}={}^{a}/{}_{b}\].
Now we have 2 slopes of two lines \[{{m}_{1}}\] and \[{{m}_{2}}\]. The angle formed between them is given by the formula,
\[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\].
Thus substitute the value of \[{{m}_{1}}\] and \[{{m}_{2}}\] in the above expression.
\[\tan \theta =\left| \dfrac{\dfrac{b}{a}-\dfrac{a}{b}}{1+\left( \dfrac{b}{a}\times \dfrac{a}{b} \right)} \right|=\left| \dfrac{\dfrac{{{b}^{2}}-{{a}^{2}}}{ab}}{1+1} \right|=\left| \dfrac{{{b}^{2}}-{{a}^{2}}}{2ab} \right|\]
\[\therefore \tan \theta =\left| \dfrac{{{b}^{2}}-{{a}^{2}}}{2ab} \right|\].
Thus we got the tangent of the angle between the line as \[\left( \dfrac{{{b}^{2}}-{{a}^{2}}}{2ab} \right)\].

Note:- The formula for tangent of angle \[\left( \tan \theta \right)\] works only on acute angles, i.e. acute angles formed by two lines. This formula doesn’t work for perpendicular lines. From the question tangent of angle means \[\tan \theta \].