Questions and Answers Section

Example 1: Slope and Inclination

Question: Find the slope of a line which passes through the origin and the mid-point of the line segment joining P(0, -4) and B(8, 0).

Solution:

Step 1: Find the mid-point of the segment PB.

Mid-point M = $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (\frac{0+8}{2}, \frac{-4+0}{2}) = (4, -2)$.

Step 2: Find the slope of the line passing through the origin O(0,0) and M(4, -2).

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 0}{4 - 0} = -\frac{2}{4} = \mathbf{-\frac{1}{2}}$$.

Example 2: Collinearity of Points

Question: Find the value of k for which the points A(k, 1), B(5, 5), and C(10, 7) are collinear.

Solution:

For the points to be collinear, the slope of AB must equal the slope of BC.

Step 1: Set up the slope expressions.

$m_{AB} = \frac{5-1}{5-k} = \frac{4}{5-k}$.

$m_{BC} = \frac{7-5}{10-5} = \frac{2}{5}$.

Step 2: Equate the slopes and solve for k.

$\frac{4}{5-k} = \frac{2}{5} \implies 4(5) = 2(5-k) \implies 20 = 10-2k \implies 10 = -2k \implies \mathbf{k = -5}$.

Example 3: Angle Between Lines

Question: Find the acute angle between two lines whose slopes are $\sqrt{3}$ and $1/\sqrt{3}$.

Solution:

Let $m_1 = \sqrt{3}$ and $m_2 = 1/\sqrt{3}$. The angle $\theta$ is given by the formula:

$$\tan\theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| = \left| \frac{1/\sqrt{3} - \sqrt{3}}{1 + (\sqrt{3})(1/\sqrt{3})} \right|$$

$$= \left| \frac{(1-3)/\sqrt{3}}{1+1} \right| = \left| \frac{-2/\sqrt{3}}{2} \right| = \frac{1}{\sqrt{3}}$$

Since $\tan\theta = 1/\sqrt{3}$, the acute angle $\mathbf{\theta}$ is $30^\circ$.

Example 4: Equation using Point-Slope Form

Question: Find the equation of the line passing through (-4, 3) with a slope of 1/2.

Solution:

Using the point-slope form $y - y_1 = m(x - x_1)$ with $(x_1,y_1)=(-4,3)$ and $m=1/2$.

$y - 3 = \frac{1}{2}(x - (-4)) \implies 2(y-3) = x+4$.

$2y-6=x+4 \implies \mathbf{x - 2y + 10 = 0}$.

Example 5: Equation using Two-Point Form

Question: Find the equation of the line passing through (-1, 1) and (2, -4).

Solution:

Using the two-point form: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.

$y - 1 = \frac{-4 - 1}{2 - (-1)}(x - (-1)) \implies y - 1 = \frac{-5}{3}(x + 1)$.

$3(y-1) = -5(x+1) \implies 3y-3=-5x-5 \implies \mathbf{5x + 3y + 2 = 0}$.

Example 6: Equation using Intercept Form

Question: Find the equation of a line that makes intercepts -3 and 2 on the x- and y-axes respectively.

Solution:

Using the intercept form $\frac{x}{a} + \frac{y}{b} = 1$ with x-intercept $a=-3$ and y-intercept $b=2$.

$\frac{x}{-3} + \frac{y}{2} = 1$.

To clear the fractions, multiply by the LCM (6):

$6(\frac{x}{-3}) + 6(\frac{y}{2}) = 6(1) \implies -2x + 3y = 6$.

In general form, this is $\mathbf{2x - 3y + 6 = 0}$.

Example 7: Equation using Normal Form

Question: Find the equation of a line whose perpendicular distance from the origin is 5 and the angle the normal makes with the positive x-axis is 30°.

Solution:

Using the normal form $x\cos\omega + y\sin\omega = p$ with $p=5$ and $\omega=30^\circ$.

$x\cos(30^\circ) + y\sin(30^\circ) = 5 \implies x(\frac{\sqrt{3}}{2}) + y(\frac{1}{2}) = 5$.

Multiply by 2 to simplify: $\mathbf{\sqrt{3}x+y-10=0}$.

Example 8: Converting to Slope-Intercept Form

Question: Reduce $6x+3y-5=0$ to slope-intercept form and find its slope and y-intercept.

Solution:

To convert to the slope-intercept form $y=mx+c$, we isolate the y-term.

$3y = -6x+5$.

Divide by 3: $\mathbf{y = -2x + 5/3}$.

Comparing this to $y=mx+c$, the slope is $\mathbf{m=-2}$ and the y-intercept is $\mathbf{c=5/3}$.

Example 9: Distance of a Point from a Line

Question: Find the distance of the point (3, -5) from the line $3x - 4y - 26 = 0$.

Solution:

Using the distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$:

$d = \frac{|3(3) - 4(-5) - 26|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 + 20 - 26|}{\sqrt{9+16}} = \frac{|3|}{\sqrt{25}} = \mathbf{\frac{3}{5}}$.

Example 10: Distance Between Parallel Lines

Question: Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $6x - 8y - 10 = 0$.

Solution:

Step 1: Normalize the equations so the coefficients of x and y match. Divide the second equation by 2:

$(6x - 8y - 10 = 0) \div 2 \implies 3x-4y-5=0$.

Step 2: Now compare $3x-4y+7=0$ and $3x-4y-5=0$. We have $C_1=7, C_2=-5$.

Step 3: Use the formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$:

$d = \frac{|7 - (-5)|}{\sqrt{3^2 + (-4)^2}} = \frac{|12|}{\sqrt{25}} = \mathbf{\frac{12}{5}}$.

Example 11: Using Family of Lines

Question: Find the equation of the line through the intersection of $2x+y-1=0$ and $x-3y+1=0$ and the point (1,1).

Solution:

Step 1: The equation of any line passing through the intersection is $L_1 + \lambda L_2 = 0$.

$(2x+y-1) + \lambda(x-3y+1) = 0$.

Step 2: Since this line passes through (1,1), substitute these coordinates to find $\lambda$.

$(2(1)+1-1) + \lambda(1-3(1)+1) = 0 \implies 2 + \lambda(-1) = 0 \implies \mathbf{\lambda=2}$.

Step 3: Substitute $\lambda=2$ back into the family equation:

$(2x+y-1) + 2(x-3y+1) = 0 \implies 2x+y-1+2x-6y+2=0 \implies \mathbf{4x-5y+1=0}$.

Example 12: Position of a Point

Question: Are the points (2,1) and (-3,5) on the same side of the line $2x-3y+6=0$?

Solution:

Let the line's expression be $L(x,y) = 2x-3y+6$.

Step 1: Substitute the first point into the expression.

$L(2,1) = 2(2)-3(1)+6 = 4-3+6=7$ (Positive).

Step 2: Substitute the second point into the expression.

$L(-3,5) = 2(-3)-3(5)+6 = -6-15+6 = -15$ (Negative).

Since the signs of the results are opposite, the points lie on opposite sides of the line.

Example 13: Equation of a Perpendicular Bisector

Question: Find the equation of the perpendicular bisector of the segment joining (7, 1) and (3, 5).

Solution:

Step 1: Find the midpoint of the segment: M = $(\frac{7+3}{2}, \frac{1+5}{2})=\mathbf{(5,3)}$.

Step 2: Find the slope of the segment: $m_1 = \frac{5-1}{3-7} = \frac{4}{-4} = \mathbf{-1}$.

Step 3: The slope of the perpendicular bisector is $m_2 = -1/m_1 = \mathbf{1}$.

Step 4: The equation of the bisector is the line through M(5,3) with slope 1:

$y-3=1(x-5) \implies y-3=x-5 \implies \mathbf{x-y-2=0}$.

Example 14: Equation of an Altitude

Question: Find the equation of the altitude from vertex A of a triangle with vertices A(2,-1), B(4,0), and C(3,2).

Solution:

The altitude from A is perpendicular to the opposite side BC.

Step 1: The slope of BC is $m_{BC} = \frac{2-0}{3-4} = \mathbf{-2}$.

Step 2: The slope of the altitude is the negative reciprocal, $m_{alt} = -1/(-2) = \mathbf{1/2}$.

Step 3: The altitude passes through A(2,-1). Its equation is:

$y-(-1) = \frac{1}{2}(x-2) \implies 2(y+1)=x-2 \implies \mathbf{x-2y-4=0}$.

Example 15: Foot of the Perpendicular

Question: Find the coordinates of the foot of the perpendicular from (2,3) on the line x+y-11=0.

Solution:

Let the point be $P(x_1, y_1)=(2,3)$ and the line be $ax+by+c=0$. The foot $(x,y)$ is given by the formula:

$\frac{x-x_1}{a} = \frac{y-y_1}{b} = -\frac{ax_1+by_1+c}{a^2+b^2}$.

Here $a=1, b=1, c=-11$.

$\frac{x-2}{1} = \frac{y-3}{1} = -\frac{1(2)+1(3)-11}{1^2+1^2} = -\frac{-6}{2} = 3$.

• $x-2=3 \implies \mathbf{x=5}$

• $y-3=3 \implies \mathbf{y=6}$

The foot is (5,6).

Example 16: Image of a Point

Question: Find the image of the point (3,8) in the line x+3y=7.

Solution:

Let the point be $P(x_1, y_1)=(3,8)$ and the line be $x+3y-7=0$. The image $(x,y)$ is given by the formula:

$\frac{x-x_1}{a} = \frac{y-y_1}{b} = -2\frac{ax_1+by_1+c}{a^2+b^2}$.

$\frac{x-3}{1} = \frac{y-8}{3} = -2\frac{1(3)+3(8)-7}{1^2+3^2} = -2\frac{20}{10} = -4$.

• $x-3=-4 \implies \mathbf{x=-1}$

• $y-8=-12 \implies \mathbf{y=-4}$

The image is (-1,-4).

Example 17: Finding the Orthocenter

Question: Find the orthocenter of the triangle with vertices (0,0), (3,4) and (4,0).

Solution:

Let A=(0,0), B=(3,4), C=(4,0). The orthocenter is the intersection of the altitudes.

Altitude 1: The side AC lies on the x-axis. The altitude from B to AC is the vertical line passing through B, which is $\mathbf{x=3}$.

Altitude 2: The slope of AB is $m_{AB}=4/3$. The altitude from C to AB has a slope of $m_{\perp}=-3/4$ and passes through C(4,0). Its equation is $\mathbf{y-0 = -3/4(x-4)}$.

Intersection: Substitute $x=3$ into the second equation: $y = -3/4(3-4) = 3/4$.

The orthocenter is (3, 3/4).

Example 18: Finding the Circumcenter

Question: Find the circumcenter of the triangle with vertices A(1,1), B(2,3), and C(4,1).

Solution:

The circumcenter is the intersection of the perpendicular bisectors.

Bisector 1 (of AC): Midpoint is $(2.5, 1)$. Side AC is horizontal, so the bisector is the vertical line $\mathbf{x = 2.5}$.

Bisector 2 (of BC): Midpoint is $(3,2)$. Slope of BC is $\frac{1-3}{4-2}=-1$. The slope of the bisector is 1. Its equation is $y-2=1(x-3) \implies \mathbf{y=x-1}$.

Intersection: Substitute $x=2.5$ into the second equation: $y=2.5-1=1.5$.

The circumcenter is (2.5, 1.5).

Example 19: Area of a Triangle

Question: Find the area of the triangle formed by the lines $y=x$, $x+y=2$ and the x-axis.

Solution:

Step 1: Find the vertices of the triangle.

• Intersection of $y=x$ and $x+y=2$ is (1,1).

• Intersection of $y=x$ and x-axis (y=0) is (0,0).

• Intersection of $x+y=2$ and x-axis is (2,0).

Step 2: The vertices are (0,0), (2,0), and (1,1). The base of the triangle lies on the x-axis and has length 2. The height is the y-coordinate of the third vertex, which is 1.

Step 3: Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = \mathbf{1}$ sq. unit.

Example 20: Locus Problem

Question: A point moves such that its distance from the line $3x-4y+1=0$ is always 2 units. Find its locus.

Solution:

Let the point be $(x,y)$. The distance from $(x,y)$ to the line is given by the formula $d = \frac{|3x-4y+1|}{\sqrt{3^2+(-4)^2}}$.

We are given that $d=2$.

$\frac{|3x-4y+1|}{5} = 2 \implies |3x-4y+1|=10$.

This absolute value equation gives two possible linear equations:

1. $3x-4y+1=10 \implies \mathbf{3x-4y-9=0}$

2. $3x-4y+1=-10 \implies \mathbf{3x-4y+11=0}$

The locus is a pair of parallel lines.

Example 21: Family of Lines with Intercept Condition

Question: Find the equation of the line passing through the intersection of $2x+3y=1$ and $3x-4y=1$ which cuts off equal intercepts on the axes.

Solution:

Step 1: Find the intersection point. Solving the system $2x+3y=1$ and $3x-4y=1$ gives the point (7/17, 1/17).

Step 2: Consider the two cases for 'equal intercepts'.

Case A: Intercepts are equal and non-zero. The line has the form $x+y=a$. Since it passes through $(7/17, 1/17)$, we have $7/17+1/17=a \implies a=8/17$. The line is $\mathbf{x+y=8/17}$.

Case B: Both intercepts are zero. The line passes through the origin $(0,0)$ and $(7/17, 1/17)$. Its slope is $m = \frac{1/17}{7/17} = 1/7$. The line is $y=\frac{1}{7}x$, or $\mathbf{x-7y=0}$.

Example 22: Finding the Incenter

Question: Find the incenter of the triangle with vertices (3,0), (0,4), and (0,0).

Solution:

Let the vertices be A(3,0), B(0,4), C(0,0). This is a right-angled triangle.

Step 1: Find the side lengths.

• $a$ (side opposite A) = length of BC = 4.

• $b$ (side opposite B) = length of AC = 3.

• $c$ (side opposite C) = length of AB = $\sqrt{(3-0)^2+(0-4)^2}=5$.

Step 2: Use the incenter formula: $I = (\frac{ax_A+bx_B+cx_C}{a+b+c}, \frac{ay_A+by_B+cy_C}{a+b+c})$.

$I = (\frac{4(3)+3(0)+5(0)}{4+3+5}, \frac{4(0)+3(4)+5(0)}{4+3+5}) = (\frac{12}{12}, \frac{12}{12}) = \mathbf{(1,1)}$.

Example 23: Line parallel to axes

Question: Find the equation of the line through (-2,3) and parallel to the x-axis.

Solution:

A line parallel to the x-axis is a horizontal line. All points on a horizontal line have the same y-coordinate. Since the line passes through (-2,3), its y-coordinate must always be 3. Therefore, the equation is $\mathbf{y=3}$.

Example 24: Concurrent Lines

Question: For what value of k are the lines $2x-3y+k=0$, $3x-4y-13=0$ and $8x-11y-33=0$ concurrent?

Solution:

For lines to be concurrent, they must all intersect at a single point.

Step 1: Find the intersection of the two known lines: $3x-4y=13$ and $8x-11y=33$. Solving this system gives $x=11, y=5$. The point of intersection is (11,5).

Step 2: For the lines to be concurrent, this point must also lie on the first line. Substitute (11,5) into $2x-3y+k=0$.

$2(11)-3(5)+k=0 \implies 22-15+k=0 \implies 7+k=0 \implies \mathbf{k=-7}$.

Example 25: Angle Bisectors

Question: Find the equation of the bisectors of the angles between the lines $3x-4y+7=0$ and $12x-5y-8=0$.

Solution:

The equation of the angle bisectors is given by $\frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}} = \pm \frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}$.

$$\frac{3x-4y+7}{\sqrt{3^2+(-4)^2}} = \pm \frac{12x-5y-8}{\sqrt{12^2+(-5)^2}}$$

$$\frac{3x-4y+7}{5} = \pm \frac{12x-5y-8}{13}$$

This gives two equations:

1. $13(3x-4y+7) = 5(12x-5y-8) \implies 39x-52y+91 = 60x-25y-40 \implies \mathbf{21x+27y-131=0}$

2. $13(3x-4y+7) = -5(12x-5y-8) \implies 39x-52y+91 = -60x+25y+40 \implies \mathbf{99x-77y+51=0}$.

Example 26: Area of a Parallelogram

Question: Find the area of the parallelogram whose sides are $x+2y+3=0$, $x+2y-1=0$, $2x-y+1=0$ and $2x-y-2=0$.

Solution:

The area of a parallelogram formed by lines $a_1x+b_1y+c_1=0, a_1x+b_1y+d_1=0, a_2x+b_2y+c_2=0, a_2x+b_2y+d_2=0$ is given by $|\frac{(c_1-d_1)(c_2-d_2)}{a_1b_2-a_2b_1}|$.

• Pair 1 ($a_1=1, b_1=2$): $c_1=3, d_1=-1$.

• Pair 2 ($a_2=2, b_2=-1$): $c_2=1, d_2=-2$.

Area = $|\frac{(3-(-1))(1-(-2))}{(1)(-1)-(2)(2)}| = |\frac{(4)(3)}{-1-4}| = |\frac{12}{-5}| = \mathbf{12/5}$ sq. units.

Example 27: Ratio of Division

Question: Find the ratio in which the line $3x+4y-7=0$ divides the line segment joining the points (1,2) and (-2,1).

Solution:

A quick method is to use the formula: Ratio = $-\frac{Ax_1+By_1+C}{Ax_2+By_2+C}$.

Ratio = $-\frac{3(1)+4(2)-7}{3(-2)+4(1)-7} = -\frac{3+8-7}{-6+4-7} = -\frac{4}{-9} = \mathbf{4/9}$.

Since the ratio is positive, the division is internal in the ratio 4:9.

Example 28: Condition for three lines to be concurrent

Question: Show that the lines $x-y=6$, $4x-3y=20$ and $6x+5y+8=0$ are concurrent.

Solution:

Step 1: Find the intersection of the first two lines.

From $x-y=6$, we have $x=y+6$. Substituting into the second: $4(y+6)-3y=20 \implies 4y+24-3y=20 \implies y=-4$.

Then $x=(-4)+6=2$. The intersection point is (2,-4).

Step 2: Check if this point lies on the third line.

Substitute (2,-4) into $6x+5y+8=0$: $6(2)+5(-4)+8 = 12-20+8 = 0$.

Since the point satisfies the third equation, the lines are concurrent.

Example 29: Parametric Form

Question: Find the coordinates of a point on the line $x+y=4$ that is at a unit distance from the line $4x+3y=10$.

Solution:

Let the point be $(h,k)$.

Condition 1: $h+k=4 \implies k=4-h$.

Condition 2: The distance from $(h,k)$ to $4x+3y-10=0$ is 1.

$\frac{|4h+3k-10|}{\sqrt{4^2+3^2}}=1 \implies |4h+3k-10|=5$.

Substitute $k=4-h$ into the second condition:

$|4h+3(4-h)-10|=5 \implies |4h+12-3h-10|=5 \implies |h+2|=5$.

This gives two possibilities:

• $h+2=5 \implies h=3$. Then $k=4-3=1$. Point is (3,1).

• $h+2=-5 \implies h=-7$. Then $k=4-(-7)=11$. Point is (-7,11).

Example 30: Shift of Origin

Question: Find the new coordinates of the point (3,-4) if the origin is shifted to (1,2).

Solution:

Let the original coordinates be $(x, y)=(3,-4)$ and the new origin be $(h,k)=(1,2)$. Let the new coordinates be $(x', y')$.

The transformation equations are $x = x'+h$ and $y=y'+k$.

• For x: $3=x'+1 \implies \mathbf{x'=2}$.

• For y: $-4=y'+2 \implies \mathbf{y'=-6}$.

The new coordinates are (2,-6).