Straight Lines
Miscellaneous Exercise Q1-Q15
Q1
1. Find k if $(k-3)x – (4-k^2)y + k^2 – 7k + 6 = 0$ is (a) Parallel to x-axis, (b) Parallel to y-axis, (c) Passing through origin.
(a) || x-axis
Coefficient of x must be 0.
$k-3 = 0 \Rightarrow k = 3$.
$k-3 = 0 \Rightarrow k = 3$.
(b) || y-axis
Coefficient of y must be 0.
$-(4-k^2) = 0 \Rightarrow k^2 = 4 \Rightarrow k = \pm 2$.
$-(4-k^2) = 0 \Rightarrow k^2 = 4 \Rightarrow k = \pm 2$.
(c) Origin
Constant term must be 0.
$k^2 – 7k + 6 = 0 \Rightarrow (k-6)(k-1) = 0 \Rightarrow k = 1, 6$.
$k^2 – 7k + 6 = 0 \Rightarrow (k-6)(k-1) = 0 \Rightarrow k = 1, 6$.
(a) 3, (b) $\pm 2$, (c) 1, 6
Q2
2. Find equation of lines with intercepts sum = 1, product = -6.
Given
$a+b=1$ and $ab=-6$.
Solve
Roots of $x^2 – (sum)x + (prod) = 0 \Rightarrow x^2 – x – 6 = 0$.
$(x-3)(x+2)=0 \Rightarrow 3, -2$.
$(x-3)(x+2)=0 \Rightarrow 3, -2$.
Case 1
$a=3, b=-2$. Eq: $\frac{x}{3} + \frac{y}{-2} = 1 \Rightarrow 2x – 3y – 6 = 0$.
Case 2
$a=-2, b=3$. Eq: $\frac{x}{-2} + \frac{y}{3} = 1 \Rightarrow -3x + 2y – 6 = 0 \Rightarrow 3x – 2y + 6 = 0$.
$2x – 3y – 6 = 0$ or $3x – 2y + 6 = 0$
Q3
3. Points on y-axis whose distance from $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.
[Image of distance of a point from a line]
Line Eq
$4x + 3y – 12 = 0$.
Point
On y-axis: $(0, k)$.
Distance
$\frac{|4(0) + 3(k) – 12|}{\sqrt{4^2 + 3^2}} = 4 \Rightarrow \frac{|3k-12|}{5} = 4$.
Solve
$|3k-12| = 20$.
$3k-12=20 \Rightarrow k=32/3$.
$3k-12=-20 \Rightarrow k=-8/3$.
$3k-12=20 \Rightarrow k=32/3$.
$3k-12=-20 \Rightarrow k=-8/3$.
$(0, 32/3)$ and $(0, -8/3)$
Q4
4. Perpendicular distance from origin to line joining $(\cos\theta, \sin\theta)$ and $(\cos\phi, \sin\phi)$.
Eq of Line
$y – \sin\theta = \frac{\sin\phi – \sin\theta}{\cos\phi – \cos\theta}(x – \cos\theta)$.
Slope Simp
Using trig identities: slope $= -\cot(\frac{\phi+\theta}{2})$.
Normal Form
$x \cos(\frac{\phi+\theta}{2}) + y \sin(\frac{\phi+\theta}{2}) = \cos(\frac{\phi-\theta}{2})$.
Distance
From origin, $d = |\cos(\frac{\phi-\theta}{2})|$.
$|\cos(\frac{\phi-\theta}{2})|$
Q5
5. Line parallel to y-axis drawn through intersection of $x – 7y + 5 = 0$ and $3x + y = 0$.
Intersection
From eq 2: $y = -3x$. Substitute in eq 1:
$x – 7(-3x) + 5 = 0 \Rightarrow 22x = -5 \Rightarrow x = -5/22$.
$x – 7(-3x) + 5 = 0 \Rightarrow 22x = -5 \Rightarrow x = -5/22$.
Parallel Y
Line parallel to y-axis has form $x = c$.
Equation
$x = -5/22 \Rightarrow 22x + 5 = 0$.
$22x + 5 = 0$
Q6
6. Line perp to $\frac{x}{4} + \frac{y}{6} = 1$ through point where it meets y-axis.
y-intercept
Put $x=0$ in line eq: $y/6 = 1 \Rightarrow y = 6$. Point $(0, 6)$.
Slope 1
$3x + 2y = 12 \Rightarrow m_1 = -3/2$.
Perp Slope
$m_2 = -1/m_1 = 2/3$.
Equation
$y – 6 = \frac{2}{3}(x – 0) \Rightarrow 3y – 18 = 2x$.
$2x – 3y + 18 = 0$
Q7
7. Area of triangle formed by $y – x = 0$, $x + y = 0$ and $x – k = 0$.
Vertices
1. $y=x$ & $y=-x \Rightarrow (0,0)$.
2. $y=x$ & $x=k \Rightarrow (k, k)$.
3. $y=-x$ & $x=k \Rightarrow (k, -k)$.
2. $y=x$ & $x=k \Rightarrow (k, k)$.
3. $y=-x$ & $x=k \Rightarrow (k, -k)$.
Area
$\frac{1}{2} |0(k – (-k)) + k(-k – 0) + k(0 – k)| = \frac{1}{2} |0 – k^2 – k^2| = |-k^2| = k^2$.
$k^2$ sq. units
Q8
8. Value of p for concurrent lines: $3x + y – 2 = 0$, $px + 2y – 3 = 0$, $2x – y – 3 = 0$.
Intersect
Solve line 1 & 3: $3x+y=2$ and $2x-y=3$.
Add: $5x=5 \Rightarrow x=1$. Sub: $2(1)-y=3 \Rightarrow y=-1$. Pt $(1, -1)$.
Add: $5x=5 \Rightarrow x=1$. Sub: $2(1)-y=3 \Rightarrow y=-1$. Pt $(1, -1)$.
Substitute
Put $(1, -1)$ in line 2: $p(1) + 2(-1) – 3 = 0$.
Solve p
$p – 2 – 3 = 0 \Rightarrow p = 5$.
p = 5
Q9
9. Three lines $y = m_ix + c_i$ are concurrent. Show $\sum m_1(c_2 – c_3) = 0$.
Intersection
Intersection of $y=m_1x+c_1$ and $y=m_2x+c_2$:
$x = \frac{c_2-c_1}{m_1-m_2}, y = \frac{m_1c_2-m_2c_1}{m_1-m_2}$.
$x = \frac{c_2-c_1}{m_1-m_2}, y = \frac{m_1c_2-m_2c_1}{m_1-m_2}$.
Substitute
Plug point into $y = m_3x + c_3$.
Simplify
$\frac{m_1c_2-m_2c_1}{m_1-m_2} = m_3(\frac{c_2-c_1}{m_1-m_2}) + c_3$. Rearranging yields result.
$m_1(c_2-c_3) + m_2(c_3-c_1) + m_3(c_1-c_2) = 0$
Q10
10. Lines through $(3, 2)$ at $45^\circ$ with $x – 2y = 3$.
Slope 1
$2y = x-3 \Rightarrow m_1 = 1/2$. Let req slope be $m$.
Formula
$\tan 45^\circ = 1 = |\frac{m – 1/2}{1 + m/2}|$.
Case 1
$\frac{2m-1}{2+m} = 1 \Rightarrow 2m-1=2+m \Rightarrow m=3$.
Case 2
$\frac{2m-1}{2+m} = -1 \Rightarrow 2m-1=-2-m \Rightarrow 3m=-1 \Rightarrow m=-1/3$.
$3x – y – 7 = 0$ and $x + 3y – 9 = 0$
Q11
11. Line passing through intersection of $4x+7y-3=0$ and $2x-3y+1=0$ with equal intercepts.
Intersection
Solve system: Mult eq2 by 2 $\Rightarrow 4x – 6y + 2 = 0$. Subtract from eq1.
$13y – 5 = 0 \Rightarrow y = 5/13$. $x = 1/13$. Point $(1/13, 5/13)$.
$13y – 5 = 0 \Rightarrow y = 5/13$. $x = 1/13$. Point $(1/13, 5/13)$.
Form
Equal intercepts $\Rightarrow x + y = a$.
Substitute
$1/13 + 5/13 = a \Rightarrow a = 6/13$.
$13x + 13y = 6$
Q12
12. Show equation of line through origin making angle $\theta$ with $y=mx+c$ is $\frac{y}{x} = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}$.
Slope
Let required slope be $m_1 = y/x$. Slope of given line is $m$.
Formula
$\tan\theta = |\frac{m_1 – m}{1 + m m_1}|$.
Expand
$\frac{m_1 – m}{1 + mm_1} = \pm \tan\theta$. Solve for $m_1$.
Result
$m_1 = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}$. Since line passes through origin, $y = m_1 x \Rightarrow \frac{y}{x} = m_1$.
Proved.
Q13
13. In what ratio is line joining $(-1, 1)$ and $(5, 7)$ divided by $x + y = 4$?
Ratio k:1
Point $P = (\frac{5k-1}{k+1}, \frac{7k+1}{k+1})$.
Substitute
Put P in $x+y=4$: $\frac{5k-1 + 7k+1}{k+1} = 4$.
Solve
$12k = 4(k+1) \Rightarrow 12k = 4k + 4 \Rightarrow 8k = 4 \Rightarrow k = 1/2$.
Ratio 1:2
Q14
14. Find distance of $4x + 7y + 5 = 0$ from $(1, 2)$ along $2x – y = 0$.
Direction
Line $2x – y = 0 \Rightarrow y=2x$. Slope = 2.
Eq
Eq of line through $(1, 2)$ with slope 2: $y – 2 = 2(x – 1) \Rightarrow y = 2x$.
Intersection
Solve $4x + 7y + 5 = 0$ and $y=2x$.
$4x + 7(2x) + 5 = 0 \Rightarrow 18x = -5 \Rightarrow x = -5/18, y = -5/9$.
$4x + 7(2x) + 5 = 0 \Rightarrow 18x = -5 \Rightarrow x = -5/18, y = -5/9$.
Distance
Dist between $(1, 2)$ and $(-5/18, -5/9)$.
$d = \sqrt{(1 – (-5/18))^2 + (2 – (-5/9))^2} = \sqrt{(23/18)^2 + (23/9)^2}$.
$d = \sqrt{(1 – (-5/18))^2 + (2 – (-5/9))^2} = \sqrt{(23/18)^2 + (23/9)^2}$.
Calc
$\frac{23}{18} \sqrt{1 + 4} = \frac{23\sqrt{5}}{18}$.
$\frac{23\sqrt{5}}{18}$ units
Q15
15. Direction of line through $(-1, 2)$ so intersection with $x+y=4$ is at distance 3.
Parametric
Point on line at dist 3: $x = -1 + 3\cos\theta, y = 2 + 3\sin\theta$.
Satisfy
This point lies on $x+y=4$.
$(-1 + 3\cos\theta) + (2 + 3\sin\theta) = 4$.
$(-1 + 3\cos\theta) + (2 + 3\sin\theta) = 4$.
Solve
$1 + 3(\cos\theta + \sin\theta) = 4 \Rightarrow \cos\theta + \sin\theta = 1$.
Angle
Squaring: $1 + \sin 2\theta = 1 \Rightarrow \sin 2\theta = 0$.
$\theta = 0^\circ$ (Parallel to x-axis) or $90^\circ$ (Parallel to y-axis).
$\theta = 0^\circ$ (Parallel to x-axis) or $90^\circ$ (Parallel to y-axis).
Parallel to x-axis or y-axis
Straight Lines
Miscellaneous Exercise Solutions (Q16 – Q23)
Q16
16. Hypotenuse ends at $(1, 3)$ and $(-4, 1)$. Legs parallel to axes. Find legs equations.
Concept
Since legs are parallel to axes, the third vertex must share x-coord with one point and y-coord with the other.
Case 1
Vertex C is at $(1, 1)$.
Leg 1 (Vertical): passes through $(1,1)$ and $(1,3) \Rightarrow x = 1$.
Leg 2 (Horizontal): passes through $(1,1)$ and $(-4,1) \Rightarrow y = 1$.
Leg 1 (Vertical): passes through $(1,1)$ and $(1,3) \Rightarrow x = 1$.
Leg 2 (Horizontal): passes through $(1,1)$ and $(-4,1) \Rightarrow y = 1$.
Case 2
Vertex C is at $(-4, 3)$.
Leg 1 (Vertical): passes through $(-4,3)$ and $(-4,1) \Rightarrow x = -4$.
Leg 2 (Horizontal): passes through $(-4,3)$ and $(1,3) \Rightarrow y = 3$.
Leg 1 (Vertical): passes through $(-4,3)$ and $(-4,1) \Rightarrow x = -4$.
Leg 2 (Horizontal): passes through $(-4,3)$ and $(1,3) \Rightarrow y = 3$.
Set 1: $x=1, y=1$
Set 2: $x=-4, y=3$
Set 2: $x=-4, y=3$
Q17
17. Find image of point $(3, 8)$ in line $x + 3y = 7$.
Formula
$\frac{h-x_1}{a} = \frac{k-y_1}{b} = -2 \frac{ax_1+by_1+c}{a^2+b^2}$
Substitute
Line: $x + 3y – 7 = 0$. Point: $(3, 8)$.
$\frac{h-3}{1} = \frac{k-8}{3} = -2 \frac{1(3)+3(8)-7}{1^2+3^2}$.
$\frac{h-3}{1} = \frac{k-8}{3} = -2 \frac{1(3)+3(8)-7}{1^2+3^2}$.
Calc RHS
$-2 \frac{3+24-7}{10} = -2 \frac{20}{10} = -4$.
Solve
$h – 3 = -4 \Rightarrow h = -1$.
$\frac{k-8}{3} = -4 \Rightarrow k – 8 = -12 \Rightarrow k = -4$.
$\frac{k-8}{3} = -4 \Rightarrow k – 8 = -12 \Rightarrow k = -4$.
Image: $(-1, -4)$
Q18
18. $y = 3x + 1$ and $2y = x + 3$ are equally inclined to $y = mx + 4$. Find m.
Slopes
$m_1 = 3$. $m_2 = 1/2$. Line L slope = $m$.
Condition
Angle between $L$ & $L_1$ = Angle between $L$ & $L_2$.
$|\frac{m – 3}{1 + 3m}| = |\frac{m – 1/2}{1 + m/2}| \Rightarrow |\frac{m – 3}{1 + 3m}| = |\frac{2m – 1}{2 + m}|$.
$|\frac{m – 3}{1 + 3m}| = |\frac{m – 1/2}{1 + m/2}| \Rightarrow |\frac{m – 3}{1 + 3m}| = |\frac{2m – 1}{2 + m}|$.
Case 1
$\frac{m-3}{1+3m} = \frac{2m-1}{2+m} \Rightarrow (m-3)(2+m) = (2m-1)(1+3m)$.
$m^2 – m – 6 = 6m^2 – m – 1 \Rightarrow 5m^2 + 5 = 0$ (No real solution).
$m^2 – m – 6 = 6m^2 – m – 1 \Rightarrow 5m^2 + 5 = 0$ (No real solution).
Case 2
$\frac{m-3}{1+3m} = -\frac{2m-1}{2+m} \Rightarrow m^2 – m – 6 = -(6m^2 – m – 1)$.
$m^2 – m – 6 = -6m^2 + m + 1 \Rightarrow 7m^2 – 2m – 7 = 0$.
$m^2 – m – 6 = -6m^2 + m + 1 \Rightarrow 7m^2 – 2m – 7 = 0$.
Quadratic
$m = \frac{2 \pm \sqrt{4 – 4(7)(-7)}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14}$.
$m = \frac{1 \pm 5\sqrt{2}}{7}$
Q19
19. Sum of perp distances from $P(x,y)$ to $x+y-5=0$ and $3x-2y+7=0$ is 10. Show P moves on a line.
Distances
$d_1 = \frac{|x+y-5|}{\sqrt{1^2+1^2}}$, $d_2 = \frac{|3x-2y+7|}{\sqrt{3^2+(-2)^2}}$.
Equation
$\frac{|x+y-5|}{\sqrt{2}} + \frac{|3x-2y+7|}{\sqrt{13}} = 10$.
Analysis
The expression involves linear terms $|Ax+By+C|$. Removing modulus signs creates linear equations of form $ax + by + c = 0$.
Conclusion
Since the resulting equation is linear in x and y, the locus of P is a straight line (or set of lines).
P moves on a line
Q20
20. Equation of line equidistant from $9x + 6y – 7 = 0$ and $3x + 2y + 6 = 0$.
Step 1
Make coefficients same. Multiply eq2 by 3:
Eq1: $9x + 6y – 7 = 0$
Eq2: $9x + 6y + 18 = 0$.
Eq1: $9x + 6y – 7 = 0$
Eq2: $9x + 6y + 18 = 0$.
Logic
The equidistant line is the parallel mid-line. Constant is average of $C_1$ and $C_2$.
Average C
$C_{mid} = \frac{-7 + 18}{2} = \frac{11}{2}$.
Equation
$9x + 6y + \frac{11}{2} = 0 \Rightarrow 18x + 12y + 11 = 0$.
$18x + 12y + 11 = 0$
Q21
21. Ray through $(1, 2)$ reflects on x-axis at A, passes through $(5, 3)$. Find A.
Reflection
Image of point $P(1, 2)$ in x-axis is $P'(1, -2)$.
Property
Point A lies on the line joining $P'(1, -2)$ and $Q(5, 3)$.
Line Eq
Slope $m = \frac{3 – (-2)}{5 – 1} = \frac{5}{4}$.
Eq: $y – 3 = \frac{5}{4}(x – 5)$.
Eq: $y – 3 = \frac{5}{4}(x – 5)$.
x-intercept
Put $y=0$ to find A. $-3 = \frac{5}{4}(x – 5) \Rightarrow -12 = 5x – 25 \Rightarrow 5x = 13 \Rightarrow x = 13/5$.
A is $(13/5, 0)$
Q22
22. Product of perpendiculars from $(\pm\sqrt{a^2-b^2}, 0)$ to $\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$ is $b^2$.
Line
$bx\cos\theta + ay\sin\theta – ab = 0$. Let $c = \sqrt{a^2-b^2}$. Points $(\pm c, 0)$.
Dist 1
$p_1 = \frac{|bc\cos\theta – ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}$.
Dist 2
$p_2 = \frac{|-bc\cos\theta – ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}$.
Product
$p_1 p_2 = \frac{|b^2c^2\cos^2\theta – a^2b^2|}{b^2\cos^2\theta + a^2\sin^2\theta}$. Numerator $= b^2|c^2\cos^2\theta – a^2|$.
Simplify
Using $c^2 = a^2 – b^2$, num becomes $b^4$. Denom simplifies to $b^2$. Total $b^2$.
Proved product is $b^2$.
Q23
23. From junction of $2x-3y+4=0$ and $3x+4y-5=0$, reach $6x-7y+8=0$ in least time.
Intersection
Solve system: Mult eq1 by 4, eq2 by 3.
$8x – 12y + 16 = 0$ and $9x + 12y – 15 = 0$.
Add: $17x + 1 = 0 \Rightarrow x = -1/17$.
Sub x: $2(-1/17) – 3y + 4 = 0 \Rightarrow 3y = 66/17 \Rightarrow y = 22/17$.
$8x – 12y + 16 = 0$ and $9x + 12y – 15 = 0$.
Add: $17x + 1 = 0 \Rightarrow x = -1/17$.
Sub x: $2(-1/17) – 3y + 4 = 0 \Rightarrow 3y = 66/17 \Rightarrow y = 22/17$.
Slope
Least time $\Rightarrow$ shortest path $\Rightarrow$ perpendicular.
Target line $6x – 7y + 8 = 0$ has slope $6/7$.
Required slope $m = -7/6$.
Target line $6x – 7y + 8 = 0$ has slope $6/7$.
Required slope $m = -7/6$.
Equation
$y – \frac{22}{17} = -\frac{7}{6}(x + \frac{1}{17})$.
Simplify
$6(17y – 22) = -7(17x + 1)$
$102y – 132 = -119x – 7$.
$102y – 132 = -119x – 7$.
$119x + 102y – 125 = 0$