Chapter 9: Straight lines

Let ABCD be the given quadrilateral with vertices A (-4,5), B (0,7), C (5.-5) and D (-4,-2).

Now, let us plot the points on the Cartesian plane by joining the points AB, BC, CD, and AD, which give us the required quadrilateral.

To find the area, draw diagonal AC.

So, area (ABCD) = area (∆ABC) + area (∆ADC)

Then, area of triangle with vertices (x₁,y₁) , (x₂, y₂) and (x₃,y₃) is

Are of ∆ ABC = ½ [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]

= ½ [-4 (7 + 5) + 0 (-5 – 5) + 5 (5 – 7)] unit²

= ½ [-4 (12) + 5 (-2)] unit²

= ½ (58) unit²

= 29 unit²

Are of ∆ ACD = ½ [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]

= ½ [-4 (-5 + 2) + 5 (-2 – 5) + (-4) (5 – (-5))] unit²

= ½ [-4 (-3) + 5 (-7) – 4 (10)] unit²

= ½ (-63) unit²

= -63/2 unit²

Since area cannot be negative, area ∆ ACD = 63/2 unit²

Area (ABCD) = 29 + 63/2

= 121/2 unit²

Let us consider ABC, the given equilateral triangle with side 2a.

Where, AB = BC = AC = 2a

In the above figure, assuming that the base BC lies on the x-axis such that the mid-point of BC is at the origin, i.e., BO = OC = a, where O is the origin.

The coordinates of point C are (0, a) and that of B are (0,-a).

The line joining a vertex of an equilateral ∆ with the mid-point of its opposite side is perpendicular.

So, vertex A lies on the y –axis.

By applying Pythagoras’ theorem,

(AC)² = OA² + OC²

(2a)²= a² + OC²

4a² – a² = OC²

3a² = OC²

OC =√3a

Co-ordinates of point C = ± √3a, 0

∴ The vertices of the given equilateral triangle are (0, a), (0, -a), (√3a, 0)

Or (0, a), (0, -a) and (-√3a, 0)

Given:

Points P (x₁, y₁) and Q(x₂, y₂)

(i) When PQ is parallel to the y-axis, then x₁ = x₂

So, the distance between P and Q is given by

= |y₂ – y₁|

(ii) When PQ is parallel to the x-axis, then y₁ = y₂

So, the distance between P and Q is given by =

= 

= |x₂ – x₁|

Let us consider (a, 0) to be the point on the x-axis that is equidistant from the point (7, 6) and (3, 4).

So,

Now, let us square on both sides; we get,

a² – 14a + 85 = a² – 6a + 25

-8a = -60

a = 60/8

= 15/2

∴ The required point is (15/2, 0)

The co-ordinates of the mid-point of the line segment joining the points P (0, – 4) and B (8, 0) are (0+8)/2, (-4+0)/2 = (4, -2)

The slope ‘m’ of the line non-vertical line passing through the point (x₁, y₁) and

(x₂, y₂) is given by m = (y₂ – y₁)/(x₂ – x₁) where, x ≠ x₁

The slope of the line passing through (0, 0) and (4, -2) is (-2-0)/(4-0) = -1/2

∴ The required slope is -1/2.

The vertices of the given triangle are (4, 4), (3, 5) and (–1, –1).

The slope (m) of the line non-vertical line passing through the point (x₁, y₁) and

(x₂, y₂) is given by m = (y₂ – y₁)/(x₂ – x₁) where, x ≠ x₁

So, the slope of the line AB (m₁) = (5-4)/(3-4) = 1/-1 = -1

The slope of the line BC (m₂) = (-1-5)/(-1-3) = -6/-4 = 3/2

The slope of the line CA (m₃) = (4+1)/(4+1) = 5/5 = 1

It is observed that m₁.m₃ = -1.1 = -1

Hence, the lines AB and CA are perpendicular to each other.

∴ given triangle is right-angled at A (4, 4)

And the vertices of the right-angled ∆ are (4, 4), (3, 5) and (-1, -1)

We know that if a line makes an angle of 30° with the positive direction of the y-axis measured anti-clock-wise, then the angle made by the line with the positive direction of the x-axis measured anti-clock-wise is 90° + 30° = 120°

∴ The slope of the given line is tan 120° = tan (180° – 60°)

= – tan 60°

= –√3

If the points (x, – 1), (2, 1) and (4, 5) are collinear, then the Slope of AB = Slope of BC

Then, (1+1)/(2-x) = (5-1)/(4-2)

2/(2-x) = 4/2

2/(2-x) = 2

2 = 2(2-x)

2 = 4 – 2x

2x = 4 – 2

2x = 2

x = 2/2

= 1

∴ The required value of x is 1.

Let the given point be A (-2, -1) , B (4, 0) , C ( 3, 3) and D ( -3, 2)

So now, the slope of AB = (0+1)/(4+2) = 1/6

The slope of CD = (3-2)/(3+3) = 1/6

Hence, the Slope of AB = Slope of CD

∴ AB ∥ CD

Now,

The slope of BC = (3-0)/(3-4) = 3/-1 = -3

The slope of AD = (2+1)/(-3+2) = 3/-1 = -3

Hence, the Slope of BC = Slope of AD

∴ BC ∥ AD

Thus, the pair of opposite sides are quadrilateral are parallel, so we can say that ABCD is a parallelogram.

Hence, the given vertices, A (-2, -1), B (4, 0), C(3, 3) and D(-3, 2) are vertices of a parallelogram.

The Slope of the line joining the points (3, -1) and (4, -2) is given by

m = (y₂ – y₁)/(x₂ – x₁) where, x ≠ x₁

m = (-2 –(-1))/(4-3)

= (-2+1)/(4-3)

= -1/1

= -1

The angle of inclination of the line joining the points (3, -1) and (4, -2) is given by

tan θ = -1

θ = (90° + 45°) = 135°

∴ The angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.