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NCERT Exemplar · Q26

Q.The tangent of angle between the lines whose intercepts on the axes are a,−ba,-b and b,−ab,-a, respectively, is
(A) a2−b2ab\dfrac{a^2-b^2}{ab}
(B) b2−a22\dfrac{b^2-a^2}{2}
(C) b2−a22ab\dfrac{b^2-a^2}{2ab}
(D) None of these

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The key is to find the slopes of the two lines from their intercept form equations, then use the tangent formula for the angle between lines. The tangent of the angle is a2−b22ab\dfrac{a^2 - b^2}{2ab}, which corresponds to option (C).

Concept and Intuition

When two lines are given by their intercepts on the axes, the quickest way to find the angle between them is to first write each line in intercept form, then convert to slope-intercept form to extract the slopes. Once you have the slopes m1m_1 and m2m_2, the tangent of the angle θ\theta between them is given by:

tan⁡θ=∣m1−m21+m1m2∣\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

The absolute value gives the acute angle; the problem asks for "the tangent of angle between the lines", which typically means the acute angle's tangent.

Watch out

A common mistake is to confuse the intercepts: for a line with intercepts pp on the x-axis and qq on the y-axis, the equation is xp+yq=1\frac{x}{p} + \frac{y}{q} = 1, not xp−yq=1\frac{x}{p} - \frac{y}{q} = 1 unless the intercepts have signs built in. Here, the intercepts are given as a,−ba, -b — that means the x-intercept is aa and the y-intercept is −b-b.

Step-by-Step Solution

1. Write the equations of the two lines in intercept form.

For the first line, intercepts are aa on the x-axis and −b-b on the y-axis. Its equation is:

xa+y−b=1⇒xa−yb=1\frac{x}{a} + \frac{y}{-b} = 1 \quad \Rightarrow \quad \frac{x}{a} - \frac{y}{b} = 1

For the second line, intercepts are bb on the x-axis and −a-a on the y-axis. Its equation is:

xb+y−a=1⇒xb−ya=1\frac{x}{b} + \frac{y}{-a} = 1 \quad \Rightarrow \quad \frac{x}{b} - \frac{y}{a} = 1

2. Convert each equation to slope-intercept form (y=mx+cy = mx + c).

For the first line:

xa−yb=1\frac{x}{a} - \frac{y}{b} = 1

Multiply through by abab:

bx−ay=abbx - ay = ab

Solve for yy:

−ay=ab−bx⇒ay=bx−ab⇒y=bax−b-ay = ab - bx \quad \Rightarrow \quad ay = bx - ab \quad \Rightarrow \quad y = \frac{b}{a}x - b

So the slope of the first line is m1=bam_1 = \dfrac{b}{a}.

For the second line:

xb−ya=1\frac{x}{b} - \frac{y}{a} = 1

Multiply through by abab:

ax−by=abax - by = ab

Solve for yy:

−by=ab−ax⇒by=ax−ab⇒y=abx−a-by = ab - ax \quad \Rightarrow \quad by = ax - ab \quad \Rightarrow \quad y = \frac{a}{b}x - a

So the slope of the second line is m2=abm_2 = \dfrac{a}{b}.

Tip

Notice that m1⋅m2=ba⋅ab=1m_1 \cdot m_2 = \frac{b}{a} \cdot \frac{a}{b} = 1. This means the lines are not perpendicular (which would require m1m2=−1m_1 m_2 = -1), but they are symmetric in a certain way — one slope is the reciprocal of the other.

3. Apply the formula for tan⁡θ\tan \theta between two lines.

tan⁡θ=∣m1−m21+m1m2∣\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

Substitute m1=bam_1 = \frac{b}{a} and m2=abm_2 = \frac{a}{b}:

tan⁡θ=∣ba−ab1+(ba)(ab)∣\tan \theta = \left| \frac{\frac{b}{a} - \frac{a}{b}}{1 + \left(\frac{b}{a}\right)\left(\frac{a}{b}\right)} \right|

4. Simplify the numerator. …

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