Class 11 Straight Lines Exemplar 9 3 Exemplar Solutions

Long Answer Type Questions

Find the coordinates of the orthocentre of the triangle whose vertices are $A(-1, 3)$, $B(2, -1)$, and $C(0, 0)$.

Orthocentre is the point of intersection of altitudes of the triangle.

Slope of $BC$: $m_{BC} = \frac{-1 - 0}{2 - 0} = -\frac{1}{2}$.

Altitude from $A(-1, 3)$ is perpendicular to $BC$, so its slope is $m_A = 2$.

Equation of altitude from $A$: $y - 3 = 2(x + 1) \implies 2x - y + 5 = 0$.

Slope of $AC$: $m_{AC} = \frac{3 - 0}{-1 - 0} = -3$.

Altitude from $B(2, -1)$ is perpendicular to $AC$, so its slope is $m_B = \frac{1}{3}$.

Equation of altitude from $B$: $y - (-1) = \frac{1}{3}(x - 2) \implies x - 3y - 5 = 0$.

Solve $2x - y + 5 = 0$ and $x - 3y - 5 = 0$: from first equation $y = 2x + 5$.

Substitute into second: $x - 3(2x + 5) - 5 = 0 \implies -5x - 20 = 0 \implies x = -4$.

Then $y = 2(-4) + 5 = -3$.

Orthocentre is $(-4, -3)$.

$(-4, -3)$

If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x\cos\theta - y\sin\theta = k\cos 2\theta$ and $x\sec\theta + y\csc\theta = k$, respectively, prove that $p^2 + 4q^2 = k^2$.

Perpendicular distance $p$ from $(0, 0)$ to $x\cos\theta - y\sin\theta - k\cos 2\theta = 0$ is:

$p = \frac{|-k\cos 2\theta|}{\sqrt{\cos^2\theta + \sin^2\theta}} = k|\cos 2\theta| \implies p^2 = k^2\cos^2 2\theta$.

The second line is $\frac{x}{\cos\theta} + \frac{y}{\sin\theta} = k \implies x\sin\theta + y\cos\theta - k\sin\theta\cos\theta = 0$.

Perpendicular distance $q$ from origin is:

$q = \frac{|-k\sin\theta\cos\theta|}{\sqrt{\sin^2\theta + \cos^2\theta}} = k\sin\theta\cos\theta = \frac{k}{2}\sin 2\theta$.

$2q = k\sin 2\theta \implies 4q^2 = k^2\sin^2 2\theta$.

Adding $p^2 + 4q^2 = k^2\cos^2 2\theta + k^2\sin^2 2\theta = k^2(\cos^2 2\theta + \sin^2 2\theta) = k^2$.

$p^2 + 4q^2 = k^2$ (Hence proved)

A ray of light passing through the point $(1, 2)$ reflects on the $x$-axis at point $A$ and the reflected ray passes through the point $(5, 3)$. Find the coordinates of $A$.

Let $P = (1, 2)$ and $Q = (5, 3)$. The point of reflection is $A(x, 0)$ on the $x$-axis.

By the laws of reflection, the angle of incidence equals the angle of reflection.

The image of $P(1, 2)$ in the $x$-axis (line $y = 0$) is $P'(1, -2)$.

The reflected ray appears to come from $P'$, so $P'$, $A$, and $Q$ are collinear.

Slope of line $P'Q = \frac{3 - (-2)}{5 - 1} = \frac{5}{4}$.

Equation of line $P'Q$: $y - 3 = \frac{5}{4}(x - 5) \implies 5x - 4y - 13 = 0$.

Point $A$ is the $x$-intercept of line $P'Q$, so put $y = 0$:

$5x - 13 = 0 \implies x = \frac{13}{5}$.

Therefore, $A = \left(\frac{13}{5}, 0\right)$.

$\left(\frac{13}{5}, 0\right)$

Find the equation of the line passing through the intersection of the lines $x + 2y - 3 = 0$ and $4x - y + 7 = 0$ and which is parallel to $5x + 4y - 20 = 0$.

Solve the system $x + 2y = 3$ and $4x - y = -7$:

Multiply second equation by $2$: $8x - 2y = -14$.

Add to first: $9x = -11 \implies x = -\frac{11}{9}$.

Substitute into first: $-\frac{11}{9} + 2y = 3 \implies 2y = \frac{38}{9} \implies y = \frac{19}{9}$.

Intersection point is $\left(-\frac{11}{9}, \frac{19}{9}\right)$.

The required line is parallel to $5x + 4y - 20 = 0$, so its equation is $5x + 4y + k = 0$.

Substitute the intersection point: $5\left(-\frac{11}{9}\right) + 4\left(\frac{19}{9}\right) + k = 0$.

$\frac{-55 + 76}{9} + k = 0 \implies \frac{21}{9} + k = 0 \implies k = -\frac{7}{3}$.

Equation: $5x + 4y - \frac{7}{3} = 0 \implies 15x + 12y - 7 = 0$.

$15x + 12y - 7 = 0$

Find the coordinates of the circumcentre of the triangle whose vertices are $A(1, 3)$, $B(-3, 5)$, and $C(5, -1)$.

Let the circumcentre be $O(x, y)$. Since $O$ is equidistant from all vertices: $OA^2 = OB^2 = OC^2$.

$OA^2 = (x - 1)^2 + (y - 3)^2 = x^2 + y^2 - 2x - 6y + 10$.

$OB^2 = (x + 3)^2 + (y - 5)^2 = x^2 + y^2 + 6x - 10y + 34$.

$OC^2 = (x - 5)^2 + (y + 1)^2 = x^2 + y^2 - 10x + 2y + 26$.

From $OA^2 = OB^2$: $-2x - 6y + 10 = 6x - 10y + 34 \implies 8x - 4y + 24 = 0 \implies 2x - y + 6 = 0$.

From $OA^2 = OC^2$: $-2x - 6y + 10 = -10x + 2y + 26 \implies 8x - 8y - 16 = 0 \implies x - y - 2 = 0$.

Subtracting $(x - y - 2 = 0)$ from $(2x - y + 6 = 0)$: $x + 8 = 0 \implies x = -8$.

Then $-8 - y - 2 = 0 \implies y = -10$.

Circumcentre is $(-8, -10)$.

$(-8, -10)$

If one diagonal of a square is along the line $8x - 15y = 0$ and one of its vertices is at $(1, 2)$, find the equations of the sides of the square passing through this vertex.

The diagonal line $8x - 15y = 0$ does not pass through $(1, 2)$ because $8(1) - 15(2) = -22 \ne 0$.

In a square, the sides make an angle of $45^\circ$ with either diagonal.

Slope of the diagonal $8x - 15y = 0$ is $m_1 = \frac{8}{15}$.

Let $m$ be the slope of a side: $\tan 45^\circ = \left|\frac{m - 8/15}{1 + 8m/15}\right| = 1$.

$\frac{15m - 8}{15 + 8m} = \pm 1$.

Case 1: $15m - 8 = 15 + 8m \implies 7m = 23 \implies m = \frac{23}{7}$.

Equation of side through $(1, 2)$: $y - 2 = \frac{23}{7}(x - 1) \implies 23x - 7y - 9 = 0$.

Case 2: $15m - 8 = -(15 + 8m) \implies 23m = -7 \implies m = -\frac{7}{23}$.

Equation of side through $(1, 2)$: $y - 2 = -\frac{7}{23}(x - 1) \implies 7x + 23y - 53 = 0$.

$23x - 7y - 9 = 0$ and $7x + 23y - 53 = 0$

Find the equations of the lines which pass through the point $(4, 5)$ and make equal angles with the lines $5x - 12y + 6 = 0$ and $3x = 4y - 7$.

Rewrite lines: $L_1: 5x - 12y + 6 = 0$ (slope $m_1 = \frac{5}{12}$) and $L_2: 3x - 4y + 7 = 0$ (slope $m_2 = \frac{3}{4}$).

A line making equal angles with $L_1$ and $L_2$ is parallel to one of the angle bisectors of $L_1$ and $L_2$.

Equations of angle bisectors: $\frac{5x - 12y + 6}{\sqrt{25 + 144}} = \pm \frac{3x - 4y + 7}{\sqrt{9 + 16}}$.

$\frac{5x - 12y + 6}{13} = \pm \frac{3x - 4y + 7}{5}$.

Taking $(+)$: $25x - 60y + 30 = 39x - 52y + 91 \implies 14x + 8y + 61 = 0$ (slope $M_1 = -\frac{7}{4}$).

Taking $(-)$: $25x - 60y + 30 = -39x + 52y - 91 \implies 64x - 112y + 121 = 0$ (slope $M_2 = \frac{4}{7}$).

Line with slope $-\frac{7}{4}$ through $(4, 5)$: $y - 5 = -\frac{7}{4}(x - 4) \implies 7x + 4y - 48 = 0$.

Line with slope $\frac{4}{7}$ through $(4, 5)$: $y - 5 = \frac{4}{7}(x - 4) \implies 4x - 7y + 19 = 0$.

$7x + 4y - 48 = 0$ and $4x - 7y + 19 = 0$

A variable line passes through the fixed point $(a, b)$ and meets the coordinate axes in $A$ and $B$. Show that the locus of the point of intersection of lines drawn through $A$ parallel to $y$-axis and through $B$ parallel to $x$-axis is $\frac{a}{x} + \frac{b}{y} = 1$.

Let the equation of the variable line in intercept form be $\frac{x}{h_1} + \frac{y}{k_1} = 1$.

It meets the $x$-axis at $A(h_1, 0)$ and the $y$-axis at $B(0, k_1)$.

Line through $A$ parallel to $y$-axis is $x = h_1$.

Line through $B$ parallel to $x$-axis is $y = k_1$.

Let their point of intersection be $P(x, y)$, then $x = h_1$ and $y = k_1$.

Since the variable line passes through the fixed point $(a, b)$:

$\frac{a}{h_1} + \frac{b}{k_1} = 1$.

Substituting $h_1 = x$ and $k_1 = y$, we obtain the required locus:

$\frac{a}{x} + \frac{b}{y} = 1$.

$\frac{a}{x} + \frac{b}{y} = 1$ (Hence proved)

Chapter 9 - Straight Lines

Overview

Long Answer Type Questions