Question 1

Draw a quadrilateral in the Cartesian plane, whose vertices are  $(-4, 5)$ ,  $(0, 7)$ ,  $(5, -5)$  and  $(-4, -2)$ . Also, find its area.

Accepted Answer

To draw the quadrilateral, plot the points (-4,5), (0,7), (5,-5), and (-4,-2) on the Cartesian plane and join them in order. To find the area, use the shoelace formula:

Vertices in order: A(-4,5), B(0,7), C(5,-5), D(-4,-2)

Area = (1/2) |x1y2 + x2y3 + x3y4 + x4y1 - (y1x2 + y2x3 + y3x4 + y4x1)|
= (1/2) |(-4)(7) + 0(-5) + 5(-2) + (-4)(5) - [5(0) + 7(5) + (-5)(-4) + (-2)(-4)]|
= (1/2) |(-28) + 0 + (-10) + (-20) - [0 + 35 + 20 + 8]|
= (1/2) |(-58) - 63|
= (1/2) |-121|
= 60.5 units²

So, the area of the quadrilateral is 60.5 square units. — Plot the points and join them to form the quadrilateral. Use the shoelace formula for area calculation by substituting the coordinates in order and simplifying.

Question 2

The base of an equilateral triangle with side  $2a$  lies along the  $y$ -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Accepted Answer

Given an equilateral triangle with side length 2a, base lies along y-axis with midpoint at origin.

Base endpoints: Since base length = 2a along y-axis, and midpoint at origin, the base vertices are at (0, a) and (0, -a).

To find the third vertex, let it be at (x, 0) on x-axis (since base is vertical, the apex lies on x-axis).

Distance from apex to each base vertex = side length = 2a.

Using distance formula:
Distance between (x,0) and (0,a) = 2a
\Rightarrow \sqrt{(x-0)^2 + (0 - a)^2} = 2a
\Rightarrow x^2 + a^2 = 4a^2
\Rightarrow x^2 = 3a^2
\Rightarrow x = \pm a\sqrt{3}

So, the third vertex is at (a\sqrt{3}, 0) or (-a\sqrt{3}, 0).

Vertices of the triangle are:
(0, a), (0, -a), and (a\sqrt{3}, 0) (or (-a\sqrt{3}, 0)). — Base vertices are on y-axis at (0,a) and (0,-a). The third vertex lies on x-axis. Using distance formula equating side length, solve for x-coordinate of apex.

Question 3

Find the distance between P(x_1, y_1) and Q(x_2, y_2) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Accepted Answer

(i) If PQ is parallel to y-axis, then x-coordinates of P and Q are the same, i.e., x_1 = x_2.
Distance = |y_2 - y_1|

(ii) If PQ is parallel to x-axis, then y-coordinates of P and Q are the same, i.e., y_1 = y_2.
Distance = |x_2 - x_1| — For line parallel to y-axis, horizontal distance is zero, so distance is vertical difference. For line parallel to x-axis, vertical distance is zero, so distance is horizontal difference.

Question 4

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Accepted Answer

Let the point on x-axis be (x, 0).
Distance from (7,6) to (x,0): \sqrt{(x - 7)^2 + (0 - 6)^2} = \sqrt{(x - 7)^2 + 36}
Distance from (3,4) to (x,0): \sqrt{(x - 3)^2 + (0 - 4)^2} = \sqrt{(x - 3)^2 + 16}

Equate distances:
\sqrt{(x - 7)^2 + 36} = \sqrt{(x - 3)^2 + 16}
Square both sides:
(x - 7)^2 + 36 = (x - 3)^2 + 16
Expand:
(x^2 - 14x + 49) + 36 = (x^2 - 6x + 9) + 16
Simplify:
x^2 - 14x + 85 = x^2 - 6x + 25
Cancel x^2:
-14x + 85 = -6x + 25
Bring variables to one side:
-14x + 6x = 25 - 85
-8x = -60
x = \frac{-60}{-8} = 7.5

Therefore, the point is (7.5, 0). — Let the point be (x,0). Equate distances from given points to this point and solve for x.

Question 5

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

Accepted Answer

Mid-point of P(0, -4) and B(8, 0) is:
\left( \frac{0 + 8}{2}, \frac{-4 + 0}{2} \right) = (4, -2)

Slope of line passing through origin (0,0) and (4,-2) is:
\frac{-2 - 0}{4 - 0} = \frac{-2}{4} = -\frac{1}{2} — Calculate mid-point of given points, then use slope formula between origin and mid-point.

Question 6

Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle.

Accepted Answer

To show the triangle is right angled without Pythagoras theorem, check if the slopes of any two sides are negative reciprocals (indicating perpendicularity).

Points: A(4,4), B(3,5), C(-1,-1)

Slope AB = (5 - 4)/(3 - 4) = 1 / (-1) = -1
Slope BC = (-1 - 5)/(-1 - 3) = (-6)/(-4) = 3/2
Slope AC = (-1 - 4)/(-1 - 4) = (-5)/(-5) = 1

Check if any two slopes multiply to -1:
Slope AB * Slope AC = -1 * 1 = -1

Therefore, AB is perpendicular to AC.

Hence, triangle ABC is right angled at A. — Calculate slopes of sides and check for perpendicularity by verifying product of slopes is -1.

Question 7

Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Accepted Answer

The slope m of a line making an angle θ with the positive x-axis is m = tan θ.

Here, the line makes 30° with positive y-axis anticlockwise.

Angle with positive x-axis = 90° - 30° = 60°

Therefore, slope m = tan 60° = \sqrt{3}. — Since slope is tan of angle with x-axis, convert angle from y-axis to x-axis by subtracting from 90°, then find tan.

Question 8

Without using distance formula, show that points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.

Accepted Answer

To show the quadrilateral is a parallelogram without distance formula, show that opposite sides are parallel by comparing slopes.

Points: A(-2,-1), B(4,0), C(3,3), D(-3,2)

Slope AB = (0 +1)/(4 + 2) = 1/6
Slope DC = (3 - 2)/(3 + 3) = 1/6

Slope BC = (3 - 0)/(3 - 4) = 3 / (-1) = -3
Slope AD = (2 + 1)/(-3 + 2) = 3 / (-1) = -3

Since AB || DC and BC || AD, the quadrilateral ABCD is a parallelogram. — Calculate slopes of opposite sides and verify they are equal, confirming parallelism and hence parallelogram.

Question 9

Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).

Accepted Answer

Slope of line joining (3, -1) and (4, -2):
\frac{-2 + 1}{4 - 3} = \frac{-1}{1} = -1

Angle θ with x-axis is given by:
	an 	heta = |m| = 1
\Rightarrow 	heta = 45^{\circ}

Since slope is negative, line makes 135° with positive x-axis, but angle between line and x-axis is 45°. — Calculate slope, then find arctangent of absolute slope value to get angle with x-axis.

Question 10

The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.

Accepted Answer

Let slopes be m and 2m.

Angle between lines θ satisfies:
	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \frac{1}{3}

Substitute m_1 = m, m_2 = 2m:
\frac{|m - 2m|}{|1 + 2m^2|} = \frac{1}{3}
\Rightarrow \frac{| -m |}{|1 + 2m^2|} = \frac{1}{3}
\Rightarrow \frac{|m|}{|1 + 2m^2|} = \frac{1}{3}

Cross multiply:
3|m| = |1 + 2m^2|

Consider positive values (since slopes can be positive or negative):
3m = 1 + 2m^2

Rearranged:
2m^2 - 3m + 1 = 0

Solve quadratic:
Discriminant D = 9 - 8 = 1

m = [3 ± 1]/(2*2) = (3 ± 1)/4

So,
m = (3 + 1)/4 = 4/4 = 1
or
m = (3 - 1)/4 = 2/4 = 0.5

Therefore, slopes are:
If m = 1, then other slope = 2
If m = 0.5, then other slope = 1

Hence, slopes are (1, 2) or (0.5, 1). — Use formula for tangent of angle between two lines and substitute given relation between slopes. Solve quadratic to find slopes.

Question 11

A line passes through (x_1, y_1) and (h, k). If slope of the line is m, show that k - y_1 = m(h - x_1).

Accepted Answer

Slope m of line passing through points (x_1, y_1) and (h, k) is given by:

m = \frac{k - y_1}{h - x_1}

Multiply both sides by (h - x_1):

m(h - x_1) = k - y_1

Hence proved. — Use definition of slope and rearrange the equation to get the required relation.

Question 12

In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:

1. Write the equations for the  $x$ -and  $y$ -axes.
2. Passing through the point  $(-4, 3)$  with slope  $\frac{1}{2}$ .
3. Passing through  $(0, 0)$  with slope  $m$ .
4. Passing through  $\left(2, 2\sqrt{3}\right)$  and inclined with the  $x$ -axis at an angle of  $75^{\circ}$ .
5. Intersecting the  $x$ -axis at a distance of 3 units to the left of origin with slope  $-2$ .
6. Intersecting the  $y$ -axis at a distance of 2 units above the origin and making an angle of  $30^{\circ}$  with positive direction of the  $x$ -axis.
7. Passing through the points $(-1, 1)$ and $(2, -4)$.
8. The vertices of $\Delta$ PQR are P (2, 1), Q (-2, 3) and R (4, 5). Find equation of the median through the vertex R.

Accepted Answer

1. Equations of the axes:
   - x-axis: y = 0
   - y-axis: x = 0

2. Passing through (-4, 3) with slope 1/2:
   Equation: y - 3 = (1/2)(x + 4) => y = (1/2)x + 5

3. Passing through (0,0) with slope m:
   Equation: y = mx

4. Passing through (2, 2√3) and inclined at 75° to x-axis:
   Slope m = tan 75° = 3.732
   Equation: y - 2√3 = 3.732(x - 2)

5. Intersecting x-axis 3 units left of origin means x-intercept = -3, slope = -2
   Equation in intercept form: y = m(x + 3) => y = -2(x + 3) => y = -2x - 6

6. Intersecting y-axis 2 units above origin means y-intercept = 2, angle with x-axis = 30°
   Slope m = tan 30° = 1/√3
   Equation: y - 2 = (1/√3)(x - 0) => y = (1/√3)x + 2

7. Passing through (-1,1) and (2,-4):
   Slope m = (-4 - 1)/(2 + 1) = -5/3
   Equation: y - 1 = (-5/3)(x + 1) => y = (-5/3)x - 2/3

8. Median through vertex R(4,5):
   Midpoint of PQ:
   P(2,1), Q(-2,3)
   Midpoint M = ((2 - 2)/2, (1 + 3)/2) = (0, 2)
   Equation of median RM:
   Slope m = (2 - 5)/(0 - 4) = (-3)/(-4) = 3/4
   Equation: y - 5 = (3/4)(x - 4) => 4y - 20 = 3x - 12 => 3x - 4y + 8 = 0 — Step-by-step solutions:

1. The x-axis is the line where y=0 and the y-axis is where x=0.

2. Using point-slope form y - y1 = m(x - x1), substitute point (-4,3) and slope 1/2.

3. Line through origin with slope m is y = mx.

4. Slope is tan 75°, calculate slope and use point-slope form.

5. x-intercept at -3 means line passes through (-3,0), slope -2, use point-slope form.

6. y-intercept 2 means line passes through (0,2), slope tan 30°, use point-slope form.

7. Calculate slope between two points, then use point-slope form.

8. Median passes through vertex and midpoint of opposite side, find midpoint and then equation of line through vertex and midpoint.

Question 13

Find the equation of the line passing through $(-3, 5)$ and perpendicular to the line through the points (2, 5) and $(-3, 6)$.

Accepted Answer

Step 1: Find slope of line through (2,5) and (-3,6):
Slope m = (6 - 5)/(-3 - 2) = 1 / (-5) = -1/5

Step 2: Slope of line perpendicular to this is the negative reciprocal:
m_perp = 5

Step 3: Equation of line passing through (-3,5) with slope 5:
Using point-slope form:
y - 5 = 5(x + 3)
=> y - 5 = 5x + 15
=> y = 5x + 20 — Calculate slope of given line, then find negative reciprocal for perpendicular slope. Use point-slope form with given point and perpendicular slope to find equation.

Question 14

A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: $n$. Find the equation of the line.

Accepted Answer

Step 1: Find coordinates of point dividing segment (1,0) to (2,3) in ratio 1:n:
Using section formula:
(x, y) = \left( \frac{1 	imes 2 + n 	imes 1}{1 + n}, \frac{1 	imes 3 + n 	imes 0}{1 + n} ight) = \left( \frac{2 + n}{1 + n}, \frac{3}{1 + n} ight)

Step 2: Find slope of segment:
m = (3 - 0)/(2 - 1) = 3

Step 3: Slope of line perpendicular to segment:
m_{perp} = -1/3

Step 4: Equation of line passing through point dividing segment:
Using point-slope form:
y - \frac{3}{1 + n} = -\frac{1}{3} \left( x - \frac{2 + n}{1 + n} ight)

This is the required equation. — Use section formula to find point dividing segment in ratio 1:n. Find slope of segment and then perpendicular slope. Use point-slope form with dividing point and perpendicular slope to get equation.

Question 15

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Accepted Answer

Let the intercepts on x and y axes be a.
Equation of line with equal intercepts:
\frac{x}{a} + \frac{y}{a} = 1 \implies x + y = a

Since line passes through (2,3):
2 + 3 = a \implies a = 5

Therefore, equation is:
x + y = 5 — For equal intercepts, intercept form is x/a + y/a = 1, which simplifies to x + y = a. Substitute given point to find a.

Straight Lines

Straight Lines — Study Notes

9.1 Introduction

9.2 Slope of a Line

9.2.2 Conditions for parallelism and perpendicularity of lines in terms of their slopes

Practice Questions — Straight Lines

All 14 Chapters in Mathematics