NCERT Exemplar Solutions
Class 11 - Mathematics - Chapter 10: STRAIGHT LINES

Long Answer Questions

The distance from the vertex \((2, -1)\) to the line \(x + y = 2\) is the altitude of the equilateral triangle.

Using the perpendicular distance formula:

\[ d = \dfrac{|2 + (-1) - 2|}{\sqrt{1^2 + 1^2}} = \dfrac{1}{\sqrt{2}} \]

In an equilateral triangle, altitude \(h = \dfrac{\sqrt{3}}{2} a\), where \(a\) is the side.

So, \( \dfrac{\sqrt{3}}{2} a = \dfrac{1}{\sqrt{2}} \Rightarrow a = \sqrt{\dfrac{2}{3}} \).

Therefore, the side length is \( \sqrt{\dfrac{2}{3}} \).

Let the line through \(P(x_1,y_1)\) be \(y - y_1 = m(x - x_1)\).

Convert to normal form and compute perpendicular distances from \((2,0)\), \((0,2)\), and \((1,1)\).

Using the condition that the algebraic sum of perpendiculars is zero gives:

\[ (2 - x_1)m - (y_1) + (0 - x_1)m + (2 - y_1) + (1 - x_1)m + (1 - y_1) = 0 \]

Simplifying yields \(x_1 = 1, y_1 = 1\).

Thus, the fixed point is \((1,1)\).

Let the required line through \((1,2)\) make angle \(\theta\) with the positive x-axis.

Its parametric form is:

\[ (x, y) = (1, 2) + t(\cos\theta, \sin\theta). \]

Substitute into \(x + y = 4\):

\[ 1 + 2 + t(\cos\theta + \sin\theta) = 4 \Rightarrow t = \dfrac{1}{\cos\theta + \sin\theta}. \]

The distance from \((1,2)\) to the intersection point is:

\[ |t| = \dfrac{1}{\cos\theta + \sin\theta} = \dfrac{\sqrt{6}}{3}. \]

Solving this gives \(\theta = 15^\circ\) or \(75^\circ\).

The intercept form of the line is:

\[ \dfrac{x}{a} + \dfrac{y}{b} = 1. \]

Given that

\[ \dfrac{1}{a} + \dfrac{1}{b} = k, \]

a constant.

Rewrite the line equation:

\[ bx + ay = ab. \]

Using \(ab(\dfrac{1}{a} + \dfrac{1}{b}) = abk\):

\[ b + a = abk. \]

Hence the equation becomes:

\[ bx + ay = a + b. \]

Dividing both sides by \(k\):

\[ b(x - k) + a(y - k) = 0. \]

This holds for every valid choice of \(a\) and \(b\), implying all such lines pass through the fixed point:

\[ (k, k). \]

Let the line intersect the axes at \((a,0)\) and \((0,b)\).

The point \((-4,3)\) divides the intercepts in the ratio \(5:3\):

\[ \dfrac{-4}{a} = \dfrac{5}{5+3} = \dfrac{5}{8}, \quad \dfrac{3}{b} = \dfrac{3}{8}. \]

Thus, \(a = -\dfrac{32}{5}\) and \(b = 8\).

The intercept form of the line becomes:

\[ \dfrac{x}{-32/5} + \dfrac{y}{8} = 1. \]

Simplifying yields the equation:

\[ 9x - 20y + 96 = 0. \]

Let the required line be:

\[ (x - y + 1) + \lambda (2x - 3y + 5) = 0. \]

Simplify to obtain:

\[ (1 + 2\lambda)x + (-1 - 3\lambda)y + (1 + 5\lambda) = 0. \]

Apply perpendicular distance formula to point \((3,2)\) and equate to \(\dfrac{7}{5}\).

Solving for \(\lambda\) produces two values, giving two lines:

\[ 3x - 4y + 6 = 0, \]

\[ 4x - 3y + 1 = 0. \]

The distance of \((x,y)\) from the x-axis is \(|y|\), and from the y-axis is \(|x|\).

Given:

\[ |x| + |y| = 1. \]

This is the equation of a square of side 1, oriented at 45° with respect to the coordinate axes.

The lines are:

\[ y = \sqrt{3}x + 2 \quad (x \ge 0), \]

\[ y = -\sqrt{3}x + 2 \quad (x < 0). \]

Their intersection is at \((0,2)\).

Points \(P_1, P_2\) lie 5 units away from \((0,2)\) on these lines.

The angle bisector is the y-axis. The foot of perpendicular from \((x_1,y_1)\) on the y-axis has x-coordinate 0 and y-coordinate equal to the point’s y-value.

The y-coordinates are:

\[ y = 2 \pm 5\cos 30^\circ = 2 \pm \dfrac{5\sqrt{3}}{2}. \]

Thus the feet of perpendiculars are:

\[ \left(0, 2 + \dfrac{5\sqrt{3}}{2}\right), \, \left(0, 2 - \dfrac{5\sqrt{3}}{2}\right). \]

The perpendicular distance from origin to the line \(\dfrac{x}{a} + \dfrac{y}{b} = 1\) is:

\[ p = \dfrac{ab}{\sqrt{a^2 + b^2}}. \]

Given \(a^2, p^2, b^2\) in A.P., so:

\[ 2p^2 = a^2 + b^2. \]

Substitute the expression for \(p^2\):

\[ 2 \left( \dfrac{a^2 b^2}{a^2 + b^2} \right) = a^2 + b^2. \]

Cross-multiplying gives:

\[ 2a^2 b^2 = (a^2 + b^2)^2. \]

Expanding the right-hand side:

\[ 2a^2 b^2 = a^4 + 2a^2 b^2 + b^4. \]

Simplifying:

\[ a^4 + b^4 = 0. \]