This exercise reviews all key concepts of the Straight Lines chapter, including different forms of lines, distances, angles, concurrency, and geometric applications.Get detailed, step-by-step solutions for the NCERT Class 11 Maths Chapter 9 Miscellaneous Exercise
1. Line Parallel to Axes and Passing Through Origin
The given line is $L: (k-3) x – (4 – k^2) y + k^2 -7k + 6 = 0$.
The general form is $Ax + By + C = 0$, where $A=k-3$, $B=-(4-k^2)=k^2-4$, and $C=k^2-7k+6$.
(a) Parallel to the $x$-axis
A line is parallel to the $x$-axis if its slope is $0$. Slope $m = -A/B = 0$.
This requires $A=0$ and $B \ne 0$.
- $A = k – 3 = 0 \implies k = 3$.
- $B = k^2 – 4$. If $k=3$, $B = 3^2 – 4 = 5 \ne 0$. (Valid)The value is $\mathbf{k=3}$.
(b) Parallel to the $y$-axis
A line is parallel to the $y$-axis if its slope is undefined.
This requires $B=0$ and $A \ne 0$.
- $B = k^2 – 4 = 0 \implies k = \pm 2$.
- $A = k – 3$.
- If $k=2$, $A = 2 – 3 = -1 \ne 0$. (Valid)
- If $k=-2$, $A = -2 – 3 = -5 \ne 0$. (Valid)The values are $\mathbf{k=2}$ and $\mathbf{k=-2}$.
(c) Passing through the origin
A line passes through the origin $(0, 0)$ if substituting $(0, 0)$ into the equation results in $0$.
$$(k-3)(0) – (4 – k^2)(0) + k^2 -7k + 6 = 0$$
$$k^2 – 7k + 6 = 0$$
Factor the quadratic:
$$(k – 1)(k – 6) = 0$$
The values are $\mathbf{k=1}$ and $\mathbf{k=6}$.
2. Line with Given Intercept Sum and Product
Let the $x$-intercept be $a$ and the $y$-intercept be $b$.
Given:
- Sum: $a + b = 1 \implies b = 1 – a$
- Product: $ab = -6$
Substitute $b = 1 – a$ into the product equation:
$$a(1 – a) = -6$$
$$a – a^2 = -6$$
$$a^2 – a – 6 = 0$$
Factor the quadratic:
$$(a – 3)(a + 2) = 0$$
$$a = 3 \quad \text{or} \quad a = -2$$
This gives two possible lines:
- Case 1: $a = 3$. Then $b = 1 – 3 = -2$.Using the intercept form $\frac{x}{a} + \frac{y}{b} = 1$:$$\frac{x}{3} + \frac{y}{-2} = 1$$Multiply by $6$: $2x – 3y = 6$.$$\mathbf{2x – 3y – 6 = 0}$$
- Case 2: $a = -2$. Then $b = 1 – (-2) = 3$.$$\frac{x}{-2} + \frac{y}{3} = 1$$Multiply by $6$: $-3x + 2y = 6$.$$\mathbf{3x – 2y + 6 = 0}$$
3. Points on $y$-axis Equidistant from a Line
Find the points on the $y$-axis, $P(0, y_1)$, whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.
1. Convert the line equation to the General Form $Ax + By + C = 0$:
$$\frac{x}{3} + \frac{y}{4} = 1$$
Multiply by $12$: $4x + 3y – 12 = 0$.
So, $A=4$, $B=3$, $C=-12$. The point is $(0, y_1)$. Distance $d=4$.
2. Use the Distance Formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$:
$$4 = \frac{|4(0) + 3(y_1) – 12|}{\sqrt{4^2 + 3^2}}$$
$$4 = \frac{|3y_1 – 12|}{5}$$
$$20 = |3y_1 – 12|$$
$$4(5) = |3y_1 – 12|$$
This absolute value equation leads to two cases:
- Case 1: $3y_1 – 12 = 20$$$3y_1 = 32$$$$y_1 = \frac{32}{3}$$The point is $\mathbf{\left(0, \frac{32}{3}\right)}$.
- Case 2: $3y_1 – 12 = -20$$$3y_1 = -8$$$$y_1 = -\frac{8}{3}$$The point is $\mathbf{\left(0, -\frac{8}{3}\right)}$.
4. Perpendicular Distance from Origin to Two-Point Line
Find the perpendicular distance from the origin $O(0, 0)$ to the line joining $P(\cos\theta, \sin\theta)$ and $Q(\cos\phi, \sin\phi)$.
1. Find the equation of the line using the two-point form:
$y – y_1 = \frac{y_2 – y_1}{x_2 – x_1} (x – x_1)$
$$y – \sin\theta = \frac{\sin\phi – \sin\theta}{\cos\phi – \cos\theta} (x – \cos\theta)$$
Use sum-to-product identities: $\sin A – \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$ and $\cos A – \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$.
Let $A=\phi$, $B=\theta$.
$$\frac{\sin\phi – \sin\theta}{\cos\phi – \cos\theta} = \frac{2\cos\left(\frac{\phi+\theta}{2}\right)\sin\left(\frac{\phi-\theta}{2}\right)}{-2\sin\left(\frac{\phi+\theta}{2}\right)\sin\left(\frac{\phi-\theta}{2}\right)} = -\frac{\cos\left(\frac{\phi+\theta}{2}\right)}{\sin\left(\frac{\phi+\theta}{2}\right)} = -\cot\left(\frac{\theta+\phi}{2}\right)$$
Let $m = -\cot\left(\frac{\theta+\phi}{2}\right)$. Let $\alpha = \frac{\theta+\phi}{2}$. $m = -\cot\alpha$.
The equation is: $y – \sin\theta = (-\cot\alpha) (x – \cos\theta)$
$$y – \sin\theta = -\frac{\cos\alpha}{\sin\alpha} (x – \cos\theta)$$
Multiply by $\sin\alpha$:
$$y\sin\alpha – \sin\theta\sin\alpha = -x\cos\alpha + \cos\theta\cos\alpha$$
$$x\cos\alpha + y\sin\alpha – (\cos\theta\cos\alpha + \sin\theta\sin\alpha) = 0$$
Using the cosine difference formula, $\cos(\theta – \alpha) = \cos\theta\cos\alpha + \sin\theta\sin\alpha$:
$$x\cos\alpha + y\sin\alpha – \cos(\theta – \alpha) = 0$$
Substitute $\alpha = \frac{\theta+\phi}{2}$:
$$x\cos\left(\frac{\theta+\phi}{2}\right) + y\sin\left(\frac{\theta+\phi}{2}\right) – \cos\left(\theta – \frac{\theta+\phi}{2}\right) = 0$$
$$x\cos\left(\frac{\theta+\phi}{2}\right) + y\sin\left(\frac{\theta+\phi}{2}\right) – \cos\left(\frac{\theta-\phi}{2}\right) = 0$$
2. Find the distance $d$ from the origin $(0, 0)$:
This is now in the normal form $x\cos\alpha + y\sin\alpha = p$. The length of the perpendicular $p$ is $\cos\left(\frac{\theta-\phi}{2}\right)$.
The General Form is $Ax + By + C = 0$:
$$A = \cos\left(\frac{\theta+\phi}{2}\right), \quad B = \sin\left(\frac{\theta+\phi}{2}\right), \quad C = -\cos\left(\frac{\theta-\phi}{2}\right)$$
$$d = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{|-\cos\left(\frac{\theta-\phi}{2}\right)|}{\sqrt{\cos^2\left(\frac{\theta+\phi}{2}\right) + \sin^2\left(\frac{\theta+\phi}{2}\right)}} = \frac{|\cos\left(\frac{\theta-\phi}{2}\right)|}{1}$$
The perpendicular distance is $\mathbf{|\cos\left(\frac{\theta-\phi}{2}\right)|}$.
5. Line Parallel to $y$-axis Through Intersection
Find the equation of the line parallel to the $y$-axis and passing through the intersection of $L_1: x – 7y + 5 = 0$ and $L_2: 3x + y = 0$.
1. Find the point of intersection $(x, y)$:
From $L_2$: $y = -3x$.
Substitute into $L_1$:
$$x – 7(-3x) + 5 = 0$$
$$x + 21x + 5 = 0$$
$$22x = -5 \implies x = -\frac{5}{22}$$
Find $y$:
$$y = -3x = -3\left(-\frac{5}{22}\right) = \frac{15}{22}$$
The intersection point is $\left(-\frac{5}{22}, \frac{15}{22}\right)$.
2. Find the equation of the line.
A line parallel to the $y$-axis has the equation $x = k$, where $k$ is the $x$-coordinate of the point it passes through.
The line passes through $\left(-\frac{5}{22}, \frac{15}{22}\right)$, so $k = -\frac{5}{22}$.
$$x = -\frac{5}{22}$$
$$\mathbf{22x + 5 = 0}$$
6. Perpendicular Line Through $y$-intercept
Find the equation of a line drawn perpendicular to the line $\frac{x}{4} + \frac{y}{6} = 1$ through the point where it meets the $y$-axis.
1. Find the $y$-intercept.
The equation is in intercept form $\frac{x}{a} + \frac{y}{b} = 1$, so the $y$-intercept is $b=6$.
The line meets the $y$-axis at the point $\mathbf{(0, 6)}$.
2. Find the slope $m_1$ of the given line.
Convert to slope-intercept form $y = mx + c$:
$$\frac{y}{6} = 1 – \frac{x}{4}$$
$$y = 6 – \frac{6}{4}x \implies y = -\frac{3}{2}x + 6$$
The slope is $m_1 = -\frac{3}{2}$.
3. Find the slope $m_{req}$ of the perpendicular line.
$$m_{req} = -\frac{1}{m_1} = -\frac{1}{-3/2} = \frac{2}{3}$$
4. Find the equation of the line with $m_{req} = \frac{2}{3}$ and passing through $(0, 6)$.
Use the slope-intercept form $y = mx + c$, where $c=6$:
$$y = \frac{2}{3}x + 6$$
Multiply by 3:
$$3y = 2x + 18$$
$$\mathbf{2x – 3y + 18 = 0}$$
7. Area of the Triangle
Find the area of the triangle formed by the lines $L_1: y – x = 0$, $L_2: x + y = 0$, and $L_3: x – k = 0$.
1. Find the vertices (points of intersection).
- Vertex A ($L_1$ and $L_2$):$$y = x \text{ and } x + y = 0$$$$x + x = 0 \implies 2x = 0 \implies x = 0, y = 0$$$A = (0, 0)$.
- Vertex B ($L_1$ and $L_3$):$$y = x \text{ and } x = k$$$$y = k$$$B = (k, k)$.
- Vertex C ($L_2$ and $L_3$):$$x + y = 0 \text{ and } x = k$$$$k + y = 0 \implies y = -k$$$C = (k, -k)$.
2. Find the area of $\triangle ABC$.
Use the area formula with $A(0, 0)$, $B(k, k)$, $C(k, -k)$:
$$\text{Area} = \frac{1}{2} |x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)|$$
Since $A=(0, 0)$, the formula simplifies:
$$\text{Area} = \frac{1}{2} |x_2 y_3 – x_3 y_2|$$
$$\text{Area} = \frac{1}{2} |k(-k) – k(k)|$$
$$\text{Area} = \frac{1}{2} |-k^2 – k^2| = \frac{1}{2} |-2k^2| = \frac{1}{2} (2|k^2|)$$
The area is $\mathbf{k^2 \text{ square units}}$.
8. Concurrent Lines
Find the value of $p$ so that the three lines $L_1: 3x + y – 2 = 0$, $L_2: px + 2y – 3 = 0$, and $L_3: 2x – y – 3 = 0$ may intersect at one point (be concurrent).
If the lines are concurrent, the intersection point of any two lines must satisfy the third line’s equation.
1. Find the intersection of $L_1$ and $L_3$:
$$L_1: 3x + y = 2$$
$$L_3: 2x – y = 3$$
Add the equations:
$$(3x + y) + (2x – y) = 2 + 3$$
$$5x = 5 \implies x = 1$$
Substitute $x=1$ into $L_1$:
$$3(1) + y = 2 \implies y = 2 – 3 = -1$$
The intersection point is $(1, -1)$.
2. Substitute the intersection point $(1, -1)$ into $L_2$:
$$px + 2y – 3 = 0$$
$$p(1) + 2(-1) – 3 = 0$$
$$p – 2 – 3 = 0$$
$$p – 5 = 0$$
$$\mathbf{p = 5}$$
9. Condition for Concurrent Lines (Proof)
If $L_1: y = m_1x + c_1$, $L_2: y = m_2x + c_2$ and $L_3: y = m_3x + c_3$ are concurrent, show that $m_1(c_2 – c_3) + m_2(c_3 – c_1) + m_3(c_1 – c_2) = 0$.
If the lines are concurrent, the area of the triangle formed by them must be zero. Alternatively, the determinant of the coefficients in the general form $Ax+By+C=0$ must be zero.
General forms:
- $L_1: m_1x – y + c_1 = 0$
- $L_2: m_2x – y + c_2 = 0$
- $L_3: m_3x – y + c_3 = 0$
The condition for concurrency is:
$$\begin{vmatrix} m_1 & -1 & c_1 \\ m_2 & -1 & c_2 \\ m_3 & -1 & c_3 \end{vmatrix} = 0$$
Expand the determinant along the second column (or the first row):
$$-(-1) \begin{vmatrix} m_2 & c_2 \\ m_3 & c_3 \end{vmatrix} + (-1) \begin{vmatrix} m_1 & c_1 \\ m_3 & c_3 \end{vmatrix} – (-1) \begin{vmatrix} m_1 & c_1 \\ m_2 & c_2 \end{vmatrix} = 0$$
$$1(m_2 c_3 – m_3 c_2) – 1(m_1 c_3 – m_3 c_1) + 1(m_1 c_2 – m_2 c_1) = 0$$
$$m_2 c_3 – m_3 c_2 – m_1 c_3 + m_3 c_1 + m_1 c_2 – m_2 c_1 = 0$$
Rearrange the terms:
$$m_1(c_2 – c_3) + m_2(c_3 – c_1) + m_3(c_1 – c_2) = 0$$
(Proved)
10. Lines Making a $45^\circ$ Angle
Find the equation of the lines through $(3, 2)$ which make an angle of $45^\circ$ with the line $L_1: x – 2y = 3$.
1. Find the slope $m_1$ of the given line $L_1$:
$$2y = x – 3$$
$$y = \frac{1}{2}x – \frac{3}{2}$$
$$m_1 = \frac{1}{2}$$
2. Use the Angle between Lines Formula $\tan \theta = \left|\frac{m_2 – m_1}{1 + m_1 m_2}\right|$:
Here $\theta = 45^\circ$, $\tan 45^\circ = 1$, and $m_1 = 1/2$. Let $m_2$ be the slope of the required line.
$$1 = \left|\frac{m_2 – 1/2}{1 + (1/2)m_2}\right|$$
$$1 = \left|\frac{2m_2 – 1}{2 + m_2}\right|$$
This leads to two cases:
- Case 1: Positive value$$1 = \frac{2m_2 – 1}{2 + m_2}$$$$2 + m_2 = 2m_2 – 1$$$$3 = m_2 \implies m_2 = 3$$
- Case 2: Negative value$$1 = -\left(\frac{2m_2 – 1}{2 + m_2}\right) = \frac{1 – 2m_2}{2 + m_2}$$$$2 + m_2 = 1 – 2m_2$$$$3m_2 = -1 \implies m_2 = -\frac{1}{3}$$
3. Find the equation of the two possible lines through $(3, 2)$:
- Line $L_a$ ($m_2 = 3$):$$y – 2 = 3(x – 3)$$$$y – 2 = 3x – 9$$$$\mathbf{3x – y – 7 = 0}$$
- Line $L_b$ ($m_2 = -1/3$):$$y – 2 = -\frac{1}{3}(x – 3)$$$$3(y – 2) = -1(x – 3)$$$$3y – 6 = -x + 3$$$$\mathbf{x + 3y – 9 = 0}$$
11. Intersection and Equal Intercepts
Find the equation of the line passing through the intersection of $L_1: 4x + 7y – 3 = 0$ and $L_2: 2x – 3y + 1 = 0$ that has equal intercepts on the axes.
1. Find the point of intersection $(x, y)$:
Multiply $L_2$ by 2: $4x – 6y + 2 = 0 \quad (L_2′)$
Subtract $L_2’$ from $L_1$:
$$(4x + 7y – 3) – (4x – 6y + 2) = 0 – 0$$
$$13y – 5 = 0 \implies y = \frac{5}{13}$$
Substitute $y$ into $L_2$:
$$2x – 3\left(\frac{5}{13}\right) + 1 = 0$$
$$2x – \frac{15}{13} + 1 = 0$$
$$2x = \frac{15}{13} – 1 = \frac{15 – 13}{13} = \frac{2}{13}$$
$$x = \frac{1}{13}$$
The intersection point is $\left(\frac{1}{13}, \frac{5}{13}\right)$.
2. Find the line with equal intercepts $a=b$.
The equation is $\frac{x}{a} + \frac{y}{a} = 1 \implies x + y = a$.
The line passes through $\left(\frac{1}{13}, \frac{5}{13}\right)$:
$$\frac{1}{13} + \frac{5}{13} = a$$
$$a = \frac{6}{13}$$
The equation is $x + y = \frac{6}{13}$.
$$\mathbf{13x + 13y – 6 = 0}$$
12. Line through Origin with Angle to another Line (Proof)
Show that the equation of the line passing through the origin and making an angle $\theta$ with the line $y = mx + c$ is $y/x = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}$.
The required line passes through the origin, so its equation is $y = m_2 x$, and its slope is $m_2 = y/x$.
The angle $\theta$ between $L_1: y = mx + c$ (slope $m_1=m$) and $L_2: y = m_2 x$ (slope $m_2$) is given by:
$$\tan\theta = \left|\frac{m_2 – m_1}{1 + m_1 m_2}\right|$$
Remove the absolute value, leading to $\tan\theta = \pm \frac{m_2 – m}{1 + m m_2}$:
$$m_2 – m = \pm \tan\theta (1 + m m_2)$$
$$m_2 – m = \pm \tan\theta \pm m m_2 \tan\theta$$
Group terms with $m_2$:
$$m_2 (1 \mp m\tan\theta) = m \pm \tan\theta$$
$$m_2 = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}$$
Since $m_2 = y/x$:
$$\mathbf{\frac{y}{x} = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}}$$
(Proved)
13. Ratio of Division by a Line
In what ratio, the line $L: x + y = 4$ divides the line segment joining $A(-1, 1)$ and $B(5, 7)$?
Let the line $L$ divide $AB$ at point $P(x, y)$ in the ratio $k:1$.
1. Find the coordinates of $P$ using the section formula:
$$x = \frac{k(5) + 1(-1)}{k + 1} = \frac{5k – 1}{k + 1}$$
$$y = \frac{k(7) + 1(1)}{k + 1} = \frac{7k + 1}{k + 1}$$
2. Substitute $(x, y)$ into the line equation $x + y = 4$:
$$\frac{5k – 1}{k + 1} + \frac{7k + 1}{k + 1} = 4$$
$$\frac{(5k – 1) + (7k + 1)}{k + 1} = 4$$
$$\frac{12k}{k + 1} = 4$$
$$12k = 4(k + 1)$$
$$12k = 4k + 4$$
$$8k = 4$$
$$k = \frac{4}{8} = \frac{1}{2}$$
The ratio $k:1$ is $\frac{1}{2}:1$, or $\mathbf{1:2}$.
14. Distance Along Another Line
Find the distance of the line $L_1: 4x + 7y + 5 = 0$ from the point $P(1, 2)$ along the line $L_2: 2x – y = 0$.
This requires finding the intersection point $N$ of $L_1$ and $L_2$, and then finding the distance $PN$.
1. Find the intersection point $N(x, y)$ of $L_1$ and $L_2$:
From $L_2$: $y = 2x$.
Substitute into $L_1$:
$$4x + 7(2x) + 5 = 0$$
$$4x + 14x + 5 = 0$$
$$18x = -5 \implies x = -\frac{5}{18}$$
Find $y$:
$$y = 2x = 2\left(-\frac{5}{18}\right) = -\frac{5}{9} = -\frac{10}{18}$$
The intersection point is $N\left(-\frac{5}{18}, -\frac{10}{18}\right)$.
2. Find the distance $PN$ between $P(1, 2)$ and $N(-\frac{5}{18}, -\frac{10}{18})$.
$P = \left(\frac{18}{18}, \frac{36}{18}\right)$
$$\text{Distance } PN = \sqrt{\left(\frac{18}{18} – \left(-\frac{5}{18}\right)\right)^2 + \left(\frac{36}{18} – \left(-\frac{10}{18}\right)\right)^2}$$
$$PN = \sqrt{\left(\frac{23}{18}\right)^2 + \left(\frac{46}{18}\right)^2}$$
$$PN = \sqrt{\left(\frac{23}{18}\right)^2 + \left(2 \cdot \frac{23}{18}\right)^2}$$
$$PN = \sqrt{\left(\frac{23}{18}\right)^2 (1 + 4)} = \frac{23}{18} \sqrt{5}$$
The distance is $\mathbf{\frac{23\sqrt{5}}{18} \text{ units}}$.
15. Direction and Distance
Find the direction (slope $m$) in which a straight line must be drawn through $P(-1, 2)$ so that its point of intersection $Q$ with $L: x + y = 4$ may be at a distance of 3 units from $P$.
Let $m$ be the slope of the required line $PQ$. The equation of $PQ$ is:
$$y – 2 = m(x – (-1)) \implies y – 2 = m(x + 1)$$
$$mx – y + m + 2 = 0$$
The distance $PQ$ is 3 units. We can use the formula for distance in terms of the angle $\theta$ (where $\tan\theta = m$).
Let $Q(x_Q, y_Q)$. The parametric equation of the line $PQ$ is:
$$x_Q = x_P + r\cos\theta = -1 + 3\cos\theta$$
$$y_Q = y_P + r\sin\theta = 2 + 3\sin\theta$$
where $r=3$.
Since $Q$ lies on $x + y = 4$:
$$(-1 + 3\cos\theta) + (2 + 3\sin\theta) = 4$$
$$1 + 3(\cos\theta + \sin\theta) = 4$$
$$3(\cos\theta + \sin\theta) = 3$$
$$\cos\theta + \sin\theta = 1$$
To solve this, divide by $\sqrt{1^2 + 1^2} = \sqrt{2}$:
$$\frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta = \frac{1}{\sqrt{2}}$$
$$\cos\left(\frac{\pi}{4}\right)\cos\theta + \sin\left(\frac{\pi}{4}\right)\sin\theta = \frac{1}{\sqrt{2}}$$
$$\cos\left(\theta – \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)$$
$$\theta – \frac{\pi}{4} = \pm \frac{\pi}{4} + 2n\pi$$
- Case 1: $\theta – \frac{\pi}{4} = \frac{\pi}{4} \implies \theta = \frac{\pi}{2}$ (Line is vertical)$$m = \tan(\frac{\pi}{2}) \text{ (undefined)}$$
- Case 2: $\theta – \frac{\pi}{4} = -\frac{\pi}{4} \implies \theta = 0$ (Line is horizontal)$$m = \tan(0) = 0$$
The slope $m$ (direction) is 0 or undefined.
16. Legs of a Right Angled Triangle
The hypotenuse of a right angled triangle has its ends at $A(1, 3)$ and $B(-4, 1)$. Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.
If the legs are parallel to the axes, the right angle must occur at a point $C$ where $AC$ is horizontal and $BC$ is vertical, or vice-versa.
- Case 1: Right angle at $C_1$$AC_1$ is parallel to $x$-axis (Horizontal), $BC_1$ is parallel to $y$-axis (Vertical).$C_1$ has $y$-coordinate of $A$ and $x$-coordinate of $B$: $C_1(-4, 3)$.
- Leg 1 ($AC_1$): Horizontal line through $A(1, 3)$, $y = 3$.
- Leg 2 ($BC_1$): Vertical line through $B(-4, 1)$, $x = -4$.The equations are $\mathbf{y = 3}$ and $\mathbf{x = -4}$.
- Case 2: Right angle at $C_2$$BC_2$ is parallel to $x$-axis (Horizontal), $AC_2$ is parallel to $y$-axis (Vertical).$C_2$ has $y$-coordinate of $B$ and $x$-coordinate of $A$: $C_2(1, 1)$.
- Leg 1 ($BC_2$): Horizontal line through $B(-4, 1)$, $y = 1$.
- Leg 2 ($AC_2$): Vertical line through $A(1, 3)$, $x = 1$.The equations are $\mathbf{y = 1}$ and $\mathbf{x = 1}$.
Since the problem asks for “an equation of the legs,” either pair of equations is a valid answer.
We’ll provide the first pair derived: $\mathbf{x = -4}$ and $\mathbf{y = 3}$.
17. Image of a Point in a Line
Find the image $Q(h, k)$ of the point $P(3, 8)$ with respect to the line $L: x + 3y = 7$.
The line $L$ is the perpendicular bisector of $PQ$.
1. The line $PQ$ is perpendicular to $L$.
Slope of $L$: $3y = -x + 7 \implies m_L = -\frac{1}{3}$.
Slope of $PQ$: $m_{PQ} = -\frac{1}{m_L} = 3$.
Also, $m_{PQ} = \frac{k – 8}{h – 3} = 3$.
$$k – 8 = 3(h – 3)$$
$$k – 8 = 3h – 9 \implies \mathbf{3h – k = 1} \quad \text{(i)}$$
2. The mid-point $M$ of $PQ$ lies on $L$.
$M = \left(\frac{h + 3}{2}, \frac{k + 8}{2}\right)$.
Substitute $M$ into $x + 3y = 7$:
$$\frac{h + 3}{2} + 3\left(\frac{k + 8}{2}\right) = 7$$
$$h + 3 + 3k + 24 = 14$$
$$h + 3k + 27 = 14 \implies \mathbf{h + 3k = -13} \quad \text{(ii)}$$
3. Solve the system of equations (i) and (ii):
(i) $3h – k = 1 \implies k = 3h – 1$.
Substitute into (ii):
$$h + 3(3h – 1) = -13$$
$$h + 9h – 3 = -13$$
$$10h = -10 \implies h = -1$$
Substitute $h=-1$ into (i) $k = 3h – 1$:
$$k = 3(-1) – 1 = -4$$
The image of the point is $Q(h, k) = \mathbf{(-1, -4)}$.
18. Equally Inclined Lines
If the lines $L_1: y = 3x + 1$ and $L_2: 2y = x + 3$ are equally inclined to the line $L_3: y = mx + 4$, find the value of $m$.
$L_1$ has slope $m_1 = 3$.
$L_2$: $y = \frac{1}{2}x + \frac{3}{2}$, so $m_2 = \frac{1}{2}$.
$L_3$ has slope $m_3 = m$.
The angle $\alpha$ between $L_1$ and $L_3$ must equal the angle $\beta$ between $L_2$ and $L_3$.
$\tan\alpha = \tan\beta$.
$$\left|\frac{m_3 – m_1}{1 + m_1 m_3}\right| = \left|\frac{m_3 – m_2}{1 + m_2 m_3}\right|$$
$$\left|\frac{m – 3}{1 + 3m}\right| = \left|\frac{m – 1/2}{1 + (1/2)m}\right| = \left|\frac{2m – 1}{2 + m}\right|$$
This implies $\frac{m – 3}{1 + 3m} = \pm \frac{2m – 1}{2 + m}$.
- Case 1: Positive value (same sign)$$\frac{m – 3}{1 + 3m} = \frac{2m – 1}{2 + m}$$$$(m – 3)(2 + m) = (2m – 1)(1 + 3m)$$$$2m + m^2 – 6 – 3m = 2m + 6m^2 – 1 – 3m$$$$m^2 – m – 6 = 6m^2 – m – 1$$$$5m^2 = -5 \implies m^2 = -1$$(No real solution for $m$).
- Case 2: Negative value (opposite sign)$$\frac{m – 3}{1 + 3m} = -\left(\frac{2m – 1}{2 + m}\right) = \frac{1 – 2m}{2 + m}$$$$(m – 3)(2 + m) = (1 – 2m)(1 + 3m)$$$$m^2 – m – 6 = 1 + 3m – 2m – 6m^2$$$$m^2 – m – 6 = 1 + m – 6m^2$$$$7m^2 – 2m – 7 = 0$$Use the quadratic formula $m = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$:$$m = \frac{-(-2) \pm \sqrt{(-2)^2 – 4(7)(-7)}}{2(7)}$$$$m = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14}$$$$m = \mathbf{\frac{1 \pm 5\sqrt{2}}{7}}$$
19. Sum of Perpendicular Distances
If the sum of the perpendicular distances of a variable point $P(x, y)$ from the lines $L_1: x + y – 5 = 0$ and $L_2: 3x – 2y + 7 = 0$ is always 10. Show that $P$ must move on a line.
The perpendicular distance $d_1$ from $P(x, y)$ to $L_1$:
$$d_1 = \frac{|x + y – 5|}{\sqrt{1^2 + 1^2}} = \frac{|x + y – 5|}{\sqrt{2}}$$
The perpendicular distance $d_2$ from $P(x, y)$ to $L_2$:
$$d_2 = \frac{|3x – 2y + 7|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x – 2y + 7|}{\sqrt{13}}$$
Given $d_1 + d_2 = 10$:
$$\frac{|x + y – 5|}{\sqrt{2}} + \frac{|3x – 2y + 7|}{\sqrt{13}} = 10$$
For $P$ to move on a line, the absolute value signs must yield a single linear equation. This happens if the expressions inside the absolute signs maintain their signs (or change signs simultaneously, which means they are multiples of each other, which is not the case here).
The two possible linear paths are formed when we consider the four sign combinations. However, if the sum of two distances is a constant, the locus is a line segment (or a single line if the signs never change).
Consider the case where both expressions are positive:
$$\frac{(x + y – 5)}{\sqrt{2}} + \frac{(3x – 2y + 7)}{\sqrt{13}} = 10$$
Multiply by $\sqrt{26}$:
$$\sqrt{13}(x + y – 5) + \sqrt{2}(3x – 2y + 7) = 10\sqrt{26}$$
$$x(\sqrt{13} + 3\sqrt{2}) + y(\sqrt{13} – 2\sqrt{2}) + (-5\sqrt{13} + 7\sqrt{2} – 10\sqrt{26}) = 0$$
Since this resulting equation is of the form $Ax + By + C = 0$, it represents a straight line.
Since the locus is given by $d_1 + d_2 = \text{constant}$, and the lines are not parallel, the locus of $P$ is a line segment (a straight line, where the region is restricted by the absolute value conditions).
The equation derived above $\mathbf{x(\sqrt{13} + 3\sqrt{2}) + y(\sqrt{13} – 2\sqrt{2}) + (-5\sqrt{13} + 7\sqrt{2} – 10\sqrt{26}) = 0}$ is the equation of a line, proving the statement.
20. Equidistant Line from Parallel Lines
Find equation of the line which is equidistant from parallel lines $L_1: 9x + 6y – 7 = 0$ and $L_2: 3x + 2y + 6 = 0$.
1. Convert $L_2$ to have the same coefficients as $L_1$:
Multiply $L_2$ by 3:
$$L_2′: 3(3x + 2y + 6) = 0 \implies 9x + 6y + 18 = 0$$
Now the two parallel lines are:
$$L_1: 9x + 6y – 7 = 0 \quad (C_1 = -7)$$
$$L_2′: 9x + 6y + 18 = 0 \quad (C_2 = 18)$$
2. The equidistant line $L_m$ (mid-line) has the equation $Ax + By + C_m = 0$, where $C_m$ is the average of $C_1$ and $C_2$:
$$C_m = \frac{C_1 + C_2}{2} = \frac{-7 + 18}{2} = \frac{11}{2}$$
The equation of the line is $9x + 6y + \frac{11}{2} = 0$.
Multiply by 2 to clear the fraction:
$$\mathbf{18x + 12y + 11 = 0}$$
21. Reflection of a Ray of Light
A ray of light passing through $P(1, 2)$ reflects on the $x$-axis at point $A$ and the reflected ray passes through $Q(5, 3)$. Find the coordinates of $A$.
Let $A = (a, 0)$ on the $x$-axis.
The angle of incidence equals the angle of reflection. This means the slope of the incident ray $PA$ and the reflected ray $AQ$ are related by the reflection rule (if $m_{PA}$ and $m_{AQ}$ are slopes, they are equal in magnitude but opposite in sign with respect to the normal, which is the $y$-axis, or $m_{PA} = -m_{AQ}$ is simpler to check).
If $Q'(5, -3)$ is the image of $Q$ in the $x$-axis, then $P, A, Q’$ are collinear.
1. Use the condition for collinearity of $P(1, 2)$, $A(a, 0)$, and $Q'(5, -3)$:
The slope of $PA$ must equal the slope of $AQ’$.
$$m_{PA} = \frac{0 – 2}{a – 1} = \frac{-2}{a – 1}$$
$$m_{AQ’} = \frac{-3 – 0}{5 – a} = \frac{-3}{5 – a}$$
Set the slopes equal:
$$\frac{-2}{a – 1} = \frac{-3}{5 – a}$$
$$2(5 – a) = 3(a – 1)$$
$$10 – 2a = 3a – 3$$
$$13 = 5a$$
$$a = \frac{13}{5}$$
The coordinates of $A$ are $\mathbf{\left(\frac{13}{5}, 0\right)}$.
22. Product of Perpendicular Lengths (Proof)
Prove that the product of the lengths of the perpendiculars drawn from the points $P_1(\sqrt{a^2 – b^2}, 0)$ and $P_2(-\sqrt{a^2 – b^2}, 0)$ to the line $\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$ is $b^2$.
The line is $L: \frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta – 1 = 0$.
Multiply by $ab$: $L: bx\cos\theta + ay\sin\theta – ab = 0$.
$A = b\cos\theta, B = a\sin\theta, C = -ab$.
Let $d_1$ be the perpendicular distance from $P_1(x_1, 0)$ and $d_2$ from $P_2(x_2, 0)$, where $x_1 = -x_2 = \sqrt{a^2 – b^2}$.
$$d_1 = \frac{|b\cos\theta(x_1) + a\sin\theta(0) – ab|}{\sqrt{(b\cos\theta)^2 + (a\sin\theta)^2}} = \frac{|bx_1\cos\theta – ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}$$
$$d_2 = \frac{|b\cos\theta(x_2) + a\sin\theta(0) – ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} = \frac{|bx_2\cos\theta – ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}$$
Since $x_2 = -x_1$:
$$d_1 d_2 = \frac{|bx_1\cos\theta – ab| \cdot |-bx_1\cos\theta – ab|}{b^2\cos^2\theta + a^2\sin^2\theta}$$
$$d_1 d_2 = \frac{|(ab – bx_1\cos\theta)(ab + bx_1\cos\theta)|}{b^2\cos^2\theta + a^2\sin^2\theta}$$
Using $|(A-B)(A+B)| = |A^2 – B^2|$:
$$d_1 d_2 = \frac{|(ab)^2 – (bx_1\cos\theta)^2|}{b^2\cos^2\theta + a^2\sin^2\theta} = \frac{|a^2 b^2 – b^2 x_1^2 \cos^2\theta|}{b^2\cos^2\theta + a^2\sin^2\theta}$$
Substitute $x_1^2 = a^2 – b^2$:
$$d_1 d_2 = \frac{|a^2 b^2 – b^2 (a^2 – b^2) \cos^2\theta|}{b^2\cos^2\theta + a^2\sin^2\theta}$$
$$d_1 d_2 = \frac{|b^2 [a^2 – (a^2 – b^2) \cos^2\theta]|}{b^2\cos^2\theta + a^2\sin^2\theta}$$
$$d_1 d_2 = \frac{b^2 |a^2 – a^2 \cos^2\theta + b^2 \cos^2\theta|}{b^2\cos^2\theta + a^2\sin^2\theta}$$
$$d_1 d_2 = \frac{b^2 |a^2 (1 – \cos^2\theta) + b^2 \cos^2\theta|}{b^2\cos^2\theta + a^2\sin^2\theta}$$
Using $1 – \cos^2\theta = \sin^2\theta$:
$$d_1 d_2 = \frac{b^2 |a^2 \sin^2\theta + b^2 \cos^2\theta|}{b^2\cos^2\theta + a^2\sin^2\theta}$$
Since the denominator and the absolute value term in the numerator are equal:
$$d_1 d_2 = b^2$$
(Proved)
23. Shortest Path
A person standing at the junction $J$ of $L_1: 2x – 3y + 4 = 0$ and $L_2: 3x + 4y – 5 = 0$ wants to reach the path $L_3: 6x – 7y + 8 = 0$ in the least time. Find the equation of the path he should follow.
The path of least time is the shortest path, which is the line segment perpendicular to $L_3$. The required line $L_{req}$ must pass through $J$ and be perpendicular to $L_3$.
1. Find the junction point $J$ (intersection of $L_1$ and $L_2$).
$$L_1: 2x – 3y = -4 \quad \text{(i)}$$
$$L_2: 3x + 4y = 5 \quad \text{(ii)}$$
Multiply (i) by 4 and (ii) by 3:
$$8x – 12y = -16$$
$$9x + 12y = 15$$
Add the equations:
$$17x = -1 \implies x = -\frac{1}{17}$$
Substitute $x$ into $L_1$:
$$2\left(-\frac{1}{17}\right) – 3y = -4$$
$$-\frac{2}{17} + 4 = 3y$$
$$3y = \frac{-2 + 68}{17} = \frac{66}{17}$$
$$y = \frac{22}{17}$$
The junction is $J\left(-\frac{1}{17}, \frac{22}{17}\right)$.
2. Find the slope $m_3$ of the target path $L_3$.
$$L_3: 6x – 7y + 8 = 0$$
$$7y = 6x + 8 \implies m_3 = \frac{6}{7}$$
3. Find the slope $m_{req}$ of the required perpendicular path.
$$m_{req} = -\frac{1}{m_3} = -\frac{7}{6}$$
4. Find the equation of the line with $m_{req} = -\frac{7}{6}$ passing through $J\left(-\frac{1}{17}, \frac{22}{17}\right)$.
$$y – y_J = m_{req}(x – x_J)$$
$$y – \frac{22}{17} = -\frac{7}{6}\left(x – \left(-\frac{1}{17}\right)\right)$$
$$y – \frac{22}{17} = -\frac{7}{6}\left(x + \frac{1}{17}\right)$$
Multiply by $17 \times 6 = 102$:
$$6(17y – 22) = -7(17x + 1)$$
$$102y – 132 = -119x – 7$$
$$119x + 102y – 132 + 7 = 0$$
$$\mathbf{119x + 102y – 125 = 0}$$