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Exercise 10.1

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Q1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

Solution:

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

Now, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can be drawn as joining Point A to C forms a diagonal AC.

Accordingly, area (ABCD) = area(∆ABC) + area (∆ACD)

Using area of triangle formula. 

Therefore, Area of ∆ACD where A (-4, 5), C (5, -5), D (-4, -2)

Q2. 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.

Solution: 

Q3. Find the dis tance between P (x₁, y₁) and Q (x₂, y₂) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Solution:

Given points are : P(x₁, y₁) and Q(x₂, y₂)

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

Using Distance formula for distance between P and Q 

Q4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Solution:

Let the point on x-axis be P (x, 0) which is equidistance from points A (7, 6) and B (3, 4).

Accordingly,

AP = BP 

Or    AP² = BP²  [Squaring both sides]

⇒   (x – 7)² + (0 – 6)² = (x – 3)² + (0 – 4)²

⇒   x² – 14x + 49 + 36 = x² – 6x + 9 + 16

⇒   x² – 14x + 85 = x² – 6x + 25

⇒   85 – 25 = 14x – 6x

Q5. 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).

Solution:

The coordinates of the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Q6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Solution:

The vertices of the given triangle are

A (4, 4), B (3, 5), and C (–1, –1).

If given vertices are of a right angle triangle.

m₁m₃ = -1

Slope of AB (m₁) × Slope of AC (m₃) = -1

It means side AB and AC are perpendicular to each other.

Here, given triangle is right-angled at point A (4, 4).

Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

Q7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Q8. Find the value of x for which the points (x, – 1), (2,1) and (4, 5) are collinear.

Solution:

Let point be A (x, –1), B (2, 1), and C (4, 5).

If points A (x, –1), B (2, 1), and C (4, 5) are collinear, then

Slope of AB = Slope of BC

Q9. Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

Solution:

To be the points A(– 2, – 1), B(4, 0), C(3, 3) and D(–3, 2) of the vertices of a parallelogram.

There must be AB || CD or BC || AD

Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.

Q10. Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2).

Solution:

Let be points A(3,–1) and B(4,–2) are given for a line. 

 

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

Solution:

Q12. A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).

Solution:

Line passes through points (x₁, y₁) and (h, k). 

Q14. Consider the following population and year graph, find the slope of the line AB and using it, find what will be the population in the year 2010?

Solution:

Line AB passes through points A(1985, 92) and B(1995, 97).

Let y be the population in the year 2010. Then, according to the given graph, line AB must pass through point C (2010, y). 

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Exercise 10.3 

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Q1. Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

(i) x + 7y = 0,

(ii) 6x + 3y – 5 = 0,

(iii) y = 0.

Solution: 

Solution: 

Solution: 

(iii) y = 0

Q2. Reduce the following equations into intercept form and find their intercepts on the axes.

(i) 3x + 2y – 12 = 0

(ii) 4x – 3y = 6

(iii) 3y + 2 = 0.

Solution: (i) 3x + 2y – 12 = 0

Reducing into intercept form

3x + 2y = 12

Dividing by 12 

Solution: (ii) 4x – 3y = 6

Reducing equation into intercept form

4x – 3y = 6

Dividing by 6 

Solution: (iii) 3y + 2 = 0

Reducing equation into intercept form

0.x  + 3y  = - 2

Dividing by -2 

Q3. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

Required normal form of the line.

X cos 120° + y sin 120° = 4

Perpendicular distance (p) = 4

And angle between perpendicular and the positive x-axis = 120°

Solution: (ii) y – 2 = 0

On reducing the equation we have 

Here, comparing with general normal form x cos ω + y sin ω  = p

cos ω =  , sin ω =  and p = 2

Hence point lies on y-axis and θ is in I quadrant. 

θ = 90°

∴ ω = 90°

Required normal form of the line.

X cos 90° + y sin 90° = 2

Perpendicular distance (p) = 2

And angle between perpendicular and the positive x-axis = 90°

∴ ω θ

Solution: (iii) x – y = 4

On reducing the equation we have

x - y = 2    …. (1)

A = 1 and B = -1

Hence ω lies in VI quadrant.

θ = 45°   [ θ is value of angle between 0 - 90°]

∴ ω = 360° - θ = 360° - 45° = 315°

Required normal form of the line.

x cos 315° + y sin 315° = 2

Perpendicular distance (p) = 2  

And angle between perpendicular and the positive x-axis = 315°

Q4.  Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Solution:

Given line is 12(x + 6) = 5(y – 2) which gives

⇒ 12(x + 6) = 5(y – 2)

⇒ 12x + 72 = 5y – 10

⇒ 12x – 5y + 72 + 10 = 0

⇒ 12x – 5y + 82 = 0     ….. (1)

On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain

A = 12, B = - 5 and C = 82

Now distance from given point (-1, 1) to line 12x – 5y + 82 = 0  given by 

Solution: 

⇒  4x + 3y = 12

⇒  4x + 3y - 12 = 0   …..(1)

Comparing equation (1) with general equation of line Ax + By + C = 0

We obtain, A = 4, B = 3 C = -12

Let the point on x-axis be (a, 0) whose distance from given line is 4 units.

Using perpendicular distance formula;

⇒  ±(4a – 12) = 20

Here we take both +ve and –ve signs

⇒  4a – 12 = 20   Or  – 4a + 12 = 20

⇒  4a = 32   Or  – 4a = 20 – 12

⇒  4a = 32   Or  – 4a = 20 – 12 = 8

⇒  a = 8 or  a =  – 2 

Thus the required point on x-axis are (8, 0) and (-2, 0) 

Q6.  Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

(ii) l (x + y) + p = 0 and l (x + y) – r = 0.

Solution: 

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

 

Q12.  Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.

Solution:

The slope of one line m₁ = 2,

Let the slope of other line be m₂.

And the angle between two lines is 60°

⇒ θ = 60°

Now equation of the given live which passes through (2, 3)

Now equation of the given live which passes through (2, 3)

 

Q13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Solution:

Right bisector means perpendicular bisector of given line segment are A(3, 4) and B(-1, 2).

∵ Line bisects AB ​  

The equation of the line passing through (1, 3) and having a slope of –2 is

(y – 3) = –2 (x – 1) y – 3 = –2x + 2

2x + y = 5

Thus, the required equation of the line is 2x + y = 5.

Q14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Solution:

Equation of given line is

3x – 4y – 16 = 0     …………. (1) 

Q15. The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.

Solution:

Slope of perpendicular line from origin (0, 0) and (–1, 2).

Q16. If p and q are the lengths of perpendiculars from the origin to the  lines x cosθ - ysin θ = k cos2θ and x sec θ + y cosec θ  = k, respectively,  prove that p² + 4q² = k².

Solution:

x cos θ – y sinθ = k cos 2θ ………………. (1)

x secθ + y cosec θ = k ………………….… (2)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

Q17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

Solution:

ABC is the triangle which vertices are A (2, 3), B (4, –1) and C (1, 2).

AD is altitude on side BC from vertex A.

Equation for Altitude AD

⇒ (y – 3) = 1 (x – 2)

⇒ y – 3 = x – 2

⇒ x – y – 2 + 3 = 0

 ⇒ x – y + 1 = 0

The required equation of Altitude is x – y + 1 = 0.

Now equation for line BC where slope is – 1.

⇒ (y + 1) = –1 (x – 4)

⇒  y + 1  = – x + 4  

⇒  x + y + 1 – 4 = 0

⇒  x + y – 3 = 0

Length of AD = Length of the perpendicular from A (2, 3) to BC The equation of BC is x + y – 3 = 0

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