# ICSE Solutions for Selina Concise Chapter 14 Equation of a Line Class 10 Maths

### Exercise 14(A)

**1. Find, which of the following points lie on the line x – 2y + 5 = 0:**

**(i) (1, 3)**

**(ii) (0, 5)**

**(iii) (-5, 0)**

**(iv) (5, 5)**

**(v) (2, -1.5)**

**(vi) (-2, -1.5)**

**Solution**

Given line equation is x – 2y + 5 = 0.

**(i)** On substituting x = 1 and y = 3 in the given line equation, we have

L.H.S. = 1 – 2(3) + 5 = 1 – 6 + 5 = 6 – 6 = 0 = R.H.S.

Hence, the point (1, 3) lies on the given line.

**(ii)** On substituting x = 0 and y = 5 in the given line equation, we have

L.H.S. = 0 – 2(5) + 5 = -10 + 5 = -5 ≠ R.H.S.

Hence, the point (0, 5) does not lie on the given line.

**(iii)** On substituting x = -5 and y = 0 in the given line equation, we have

L.H.S. = -5 – 2(0) + 5 = -5 – 0 + 5 = 5 – 5 = 0 = R.H.S.

Hence, the point (-5, 0) lies on the given line.

**(iv)** On substituting x = 5 and y = 5 in the given line equation, we have

L.H.S. = 5 – 2(5) + 5 = 5 – 10 + 5 = 10 – 10 = 0 = R.H.S.

Hence, the point (5, 5) lies on the given line.

**(v)** On substituting x = 2 and y = -1.5 in the given line equation, we have

L.H.S. = 2 – 2(-1.5) + 5 = 2 + 3 + 5 = 10 ≠ R.H.S.

Hence, the point (2, -1.5) does not lie on the given line.

**(vi)** On substituting x = -2 and y = -1.5 in the given line equation, we have

L.H.S. = -2 – 2(-1.5) + 5 = -2 + 3 + 5 = 6 ≠ R.H.S.

Hence, the point (-2, -1.5) does not lie on the given line.

**2. State, true or false:**

**(i) the line x/2 + y/3 = 0 passes through the point (2, 3).**

**(ii) the line x/2 + y/3 = 0 passes through the point (4, -6).**

**(iii) the point (8, 7) lies on the line y – 7 = 0.**

**(iv) the point (-3, 0) lies on the line x + 3 = 0.**

**(v) if the point (2, a) lies on the line 2x – y = 3, then a = 5.**

**Solution**

**(i)** The given line is x/2 + y/3 = 0

Substituting x = 2 and y = 3 in the given equation,

L.H.S = 2/2 + 3/3 = 1 + 1 = 2 ≠ R.H.S

Hence, the given statement is false.

**(ii) **The given line is x/2 + y/3 = 0

Substituting x = 4 and y = -6 in the given equation,

L.H.S = 4/2 + (-6)/3 = 2 – 2 = R.H.S

Hence, the given statement is true.

**(iii)** The given line is y – 7 = 0

Substituting y = 7 in the given equation,

L.H.S = y – 7 = 7 – 7 = 0 = R.H.S.

Hence, the given statement is true.

**(iv)** The given line is x + 3 = 0

Substituting x = -3 in the given equation,

L.H.S. = x + 3 = -3 + 3 = 0 = R.H.S

Hence, the given statement is true.

**(v)** The point (2, a) lies on the line 2x – y = 3.

Substituting x = 2 and y = a in the given equation, we get

So, 2(2) – a = 3

⇒ 4 – a = 3

⇒ a = 4 – 3 = 1

Hence, the given statement is false.

**3. The line given by the equation 2x – y/3 = 7 passes through the point (k, 6); calculate the value of k.**

**Solution**

Given line equation is 2x – y/3 = 7 passes through the point (k, 6).

So, on substituting x = k and y = 6 in the given equation, we have

2k – 6/3 = 7

⇒ 6k – 6 = 21

⇒ 6k = 27

⇒ k = 27/6 = 9/2

⇒ k = 4.5

**4. For what value of k will the point (3, -k) lie on the line 9x + 4y = 3?**

**Solution**

The given line equation is 9x + 4y = 3.

On putting x = 3 and y = -k, we have

9(3) + 4(-k) = 3

⇒ 27 – 4k = 3

⇒ 4k = 27 – 3 = 24

⇒ k = 6

**5. The line 3x/5 – 2y/3 + 1 = 0 contains the point (m, 2m – 1); calculate the value of m.**

**Solution**

The equation of the given line is 3x/5 – 2y/3 + 1 = 0

On putting x = m, y = 2m – 1, we have

⇒ 9m – 20m + 10 = -15

⇒ -11m = -25

⇒ m = 25/11

**6. Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2)?**

**Solution**

It’s known that the given line will bisect the join of A (5, -2) and B (-1, 2), if the co-ordinates of the mid-point of AB satisfy the line equation.

The co-ordinates of the mid-point of AB are

(5-1/2, -2+2/2) = (2, 0)

On substituting x = 2 and y = 0 in the given line equation, we have

L.H.S. = 3x – 5y = 3(2) – 5(0) = 6 – 0 = 6 = R.H.S.

∴ the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2).

### Exercise 14(B)

**1. Find the slope of the line whose inclination is:**

**(i) 0**^{o}

**(ii) 30**^{o}

**(iii) 72**^{o}** 30′**

**(iv) 46**^{o}

**Solution**

We know that, the slope of a line is given by the tan of its inclination.

(i) Slope = tan 0^{o} = 0

(ii) Slope = tan 30^{o} = 1/ √3

(iii) Slope = tan 72^{o} 30′ = 3.1716

(iv) Slope = tan 46^{o} = 1.0355

**2. Find the inclination of the line whose slope is:**

**(i) 0**

**(ii) √3**

**(iii) 0.7646**

**(iv) 1.0875**

**Solution**

(i) Slope = tan Î¸ = 0

⇒ Î¸ = 0^{o}

(ii) Slope = tan Î¸ = √3

⇒ Î¸ = 60^{o}

(iii) Slope = tan Î¸ = 0.7646

⇒ Î¸ = 37^{o} 24′

(iv) Slope = tan Î¸ = 1.0875

⇒ Î¸ = 47^{o} 24′

**3. Find the slope of the line passing through the following pairs of points:**

**(i) (-2, -3) and (1, 2)**

**(ii) (-4, 0) and origin**

**(iii) (a, -b) and (b, -a)**

**Solution**

We know that,

Slope = (y_{2} – y_{1})/ (x_{2} – x_{1})

(i) Slope = (2 + 3)/ (1 + 2) = 5/3

(ii) Slope = (0 – 0)/ (0 + 4) = 0

(iii) Slope = (-a + b)/ (b – a) = 1

**4. Find the slope of the line parallel to AB if:**

**(i) A = (-2, 4) and B = (0, 6)**

**(ii) A = (0, -3) and B = (-2, 5)**

**Solution**

(i) Slope of AB = (6 – 4)/ (0 + 2) = 2/2 = 1

Hence, slope of the line parallel to AB = Slope of AB = 1

(ii) Slope of AB = (5 + 3)/ (-2 – 0) = 8/-2 = -4

Hence, slope of the line parallel to AB = Slope of AB = -4

**5. Find the slope of the line perpendicular to AB if:**

**(i) A = (0, -5) and B = (-2, 4)**

**(ii) A = (3, -2) and B = (-1, 2)**

**Solution**

(i) Slope of AB = (4 + 5)/ (-2 – 0) = -9/2

Slope of the line perpendicular to AB = -1/Slope of AB = -1/(-9/2 ) = 2/9

(ii) Slope of AB = (2 + 2)/ (-1 – 3) = 4/-4 = -1

Slope of the line perpendicular to AB = -1/slope of AB = -1/-1 = 1

**6. The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.**

**Solution**

Slope of the line passing through (0, 2) and (-3, -1) = (-1 – 2)/ (-3 – 0) = -3/-3 = 1

Slope of the line passing through (-1, 5) and (4, a) = (a – 5)/ (4 + 1) = (a -5)/ 5

As, the lines are parallel.

The slopes must be equal.

1 = (a – 5)/ 5

⇒ a – 5 = 5

⇒ a = 10

**7. The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.**

**Solution**

Slope of the line passing through (-4, -2) and (2, -3) = (-3 + 2)/ (2 + 4) = -1/6

Slope of the line passing through (a, 5) and (2, -1) = (-1 – 5)/ (2 – a) = -6/ (2 –a)

As, the lines are perpendicular we have

-1/6 = -1/(-6/ 2 – a)

⇒ -1/6 = (2 – a)/6

⇒ 2 – a = -1

⇒ a = 3

**8. Without using the distance formula, show that the points A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.**

**Solution**

Given points are A (4, -2), B (-4, 4) and C (10, 6).

Calculating the slopes, we have

Slope of AB = (4 + 2)/ (-4 – 4) = 6/ -8 = -3/4

Slope of BC = (6 – 4)/ (10 + 4) = 2/ 14 = 1/7

Slope of AC = (6 + 2)/ (10 – 4) = 8/ 6 = 4/3

It’s clearly seen that,

Slope of AB = -1/Slope of AC

Thus, AB ⊥ AC.

∴ the given points are the vertices of a right-angled triangle.

**9. Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.**

**Solution**

Given points are A (4, 5), B (1, 2), C (4, 3) and D (7, 6).

Slope of AB = (2 – 5)/ (1 – 4) = -3/ -3 = 1

Slope of CD = (6 – 3)/ (7 – 4) = 3/3 = 1

As the slope of AB = slope of CD

So, we can conclude AB || CD

Now,

Slope of BC = (3 – 2)/ (4 – 1) = 1/3

Slope of DA = (6 – 5)/ (7 – 4) = 1/3

As, slope of BC = slope of DA

Hence, we can say BC || DA

∴ ABCD is a parallelogram.

**10. (-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.**

**Solution**

Let the given points be A (-2, 4), B (4, 8), C (10, 7) and D (11, -5).

And, let P, Q, R and S be the mid-points of AB, BC, CD and DA respectively.

So,

As, slope of PQ = Slope of RS, we can say PQ || RS.

Now,

As, slope of QR = Slope of SP, we can say QR || SP.

∴ PQRS is a parallelogram.

**11. Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.**

**Solution**

We know that,

The points P, Q, R will be collinear if the slopes of PQ and QR are the same.

Calculating for the slopes, we get

Slope of PQ = (c + a – b – c)/(b – a) = (a – b)/(b – a) = -1

Slope of QR = (a + b – c – a)/(c – b) = (b – c)/(c – b) = -1

∴ the points P, Q, and R are collinear.

### Exercise 14(C)

**1. Find the equation of a line whose:**

**y – intercept = 2 and slope = 3.**

**Solution**

Given,

y – intercept = c = 2 and slope = m = 3.

The line equation is given by: y = mx + c

On substituting the values of c and m, we get

y = 3x + 2

The above is the required line equation.

**2. Find the equation of a line whose:**

**y – intercept = -1 and inclination = 45 ^{o}.**

**Solution**

Given,

y – intercept = c = -1 and inclination = 45^{o}.

So, slope = m = tan 45^{o} = 1

Hence, on substituting the values of c and m in the line equation y = mx + c, we get

y = x – 1

The above is the required line equation.

**3. Find the equation of the line whose slope is -4/3 and which passes through (-3, 4).**

**Solution**

Given, slope = -4/3

The line passes through the point (-3, 4) = (x_{1}, y_{1})

Now, on substituting the values in y – y_{1} = m(x – x_{1}), we have

y – 4 = -4/3 (x + 3)

⇒ 3y – 12 = -4x – 12

⇒ 4x + 3y = 0

Hence, the above is the required line equation.

**4. Find the equation of a line which passes through (5, 4) and makes an angle of 60 ^{o} with the positive direction of the x-axis.**

**Solution**

The slope of the line, m = tan 60^{o} = √3

And, the line passes through the point (5, 4) = (x_{1}, y_{1})

Hence, on substituting the values in y – y_{1} = m(x – x_{1}), we have

y – 4 = √3 (x – 5)

⇒ y – 4 = √3x – 5√3

⇒ y = √3x + 4 – 5√3,

which is the required line equation.

**5. Find the equation of the line passing through:**

**(i) (0, 1) and (1, 2)**

**(ii) (-1, -4) and (3, 0)**

**Solution**

**(i)** Let (0, 1) = (x_{1}, y_{1}) and (1, 2) = (x_{2}, y_{2})

So, Slope of the line = (2 – 1)/ (1 – 0) = 1

Now,

The required line equation is given by,

y – y_{1} = m(x – x_{1})

⇒ y – 1 = 1(x – 0)

⇒ y – 1 = x

⇒ y = x + 1

**(ii)** Let (-1, -4) = (x_{1}, y_{1}) and (3, 0) = (x_{2}, y_{2})

So, Slope of the line = (0 + 4)/ (3 + 1) = 4/4 = 1

The required line equation is given by,

y – y_{1} = m(x – x_{1})

⇒ y + 4 = 1(x + 1)

⇒ y + 4 = x + 1

⇒ y = x – 3

**6. The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find:**

**(i) the gradient of PQ;**

**(ii) the equation of PQ;**

**(iii) the co-ordinates of the point where PQ intersects the x-axis.**

**Solution**

Given,

The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively.

**(i)** Gradient of PQ = (5 – 6)/ (-3 – 2) = -1/-5 = 1/5

**(ii)** The line equation of PQ is given by

y – y_{1} = m(x – x_{1})

⇒ y – 6 = 1/5 (x – 2)

⇒ 5y – 30 = x – 2

⇒ 5y = x + 28

**(iii)** Let the line PQ intersect the x-axis at point A (x, 0).

So, on putting y = 0 in the line equation of PQ, we get

0 = x + 28

⇒ x = -28

Hence, the co-ordinates of the point where PQ intersects the x-axis are A (-28, 0).

**7. The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find:**

**(i) the equation of AB;**

**(ii) the co-ordinates of the point where the line AB intersects the y-axis.**

**Solution**

(i) Given, co-ordinates of two points A and B are (-3, 4) and (2, -1).

Slope = (-1 – 4)/ (2 + 3) = -5/5 = -1

The equation of the line AB is given by:

y – y_{1} = m(x – x_{1})

⇒ y + 1 = -1(x – 2)

⇒ y + 1 = -x + 2

⇒ x + y = 1

(ii) Let’s consider the line AB intersect the y-axis at point (0, y).

On putting x = 0 in the line equation, we get

0 + y = 1

⇒ y = 1

Thus, the co-ordinates of the point where the line AB intersects the y-axis are (0, 1).

**8. The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equation of AB and CD.**

**Solution**

The slope of line AB = tan 45^{o} = 1

And, the line AB passes through P (3, 4).

Hence, the equation of the line AB is given by

y – y_{1} = m(x – x_{1})

⇒ y – 4 = 1(x – 3)

⇒ y – 4 = x – 3

⇒ y = x + 1

The slope of line CD = tan 60^{o} = √3

And, the line CD passes through P (3, 4).

Hence, the equation of the line CD is given by

y – y_{1} = m(x – x_{1})

⇒ y – 4 = √3 (x – 3)

⇒ y – 4 = √3x – 3√3

⇒ y = √3x + 4 – 3√3

**9. In Î”ABC, A = (3, 5), B = (7, 8) and C = (1, -10). Find the equation of the median through A.**

**Solution**

Given,

Vertices of Î”ABC, A = (3, 5), B = (7, 8) and C = (1, -10).

Coordinates of the mid-point D of BC = (x_{1} + x_{2}) / 2, (y_{1} + y_{2}) / 2

The slope of AD = (y_{2 }– y_{1})/ (x_{2 }– x_{1})

= (-1 – 5)/ (4 – 3)

= -6/1

= -6

Hence, the equation of the median AD through A is given by

y – y_{1 }= m (x – x_{1})

⇒ y – 5 = -6(x – 3)

⇒ y – 5 = -6x + 18

⇒ 6x + y = 23

**10. The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis,** ∠**A = 60 ^{o} and vertex C = (7, 5). Find the equations of BC and CD.**

**Solution**

Given, ∠A = 60^{o} and vertex C = (7, 5)

As, ABCD is a parallelogram, we have

∠A + ∠B = 180^{o} **[corresponding angles]**

⇒ ∠B = 180^{o} – 60^{o} = 120^{o}

Slope of BC = tan 120^{o} = tan (90^{o} + 30^{o}) = cot 30^{o} = √3

So, the equation of line BC is given by

y – y_{1} = m(x – x_{1})

⇒ y – 5 = √3 (x – 7)

⇒ y – 5 = √3x – 7√3

⇒ y = √3x + 5 – 7√3

As, CD || AB and AB || x-axis

Slope of CD = Slope of AB = 0 [As slope of x-axis is zero]

So, the equation of the line CD is given by

y – y_{1} = m(x – x_{1})

⇒ y – 5 = 0(x – 7)

⇒ y = 5

**11. Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x – y = 4.**

**Solution**

The given line equations are:

x + 2y = 7 **….(1)**

x – y = 4 **….(2)**

On solving the above line equations, we can find the point of intersection of the two lines.

So, subtracting (2) from (1), we get

3y = 3

⇒ y = 1

Now,

x = 4 + y = 4 + 1 = 5 [From (2)]

It’s given that,

The required line passes through (0, 0) and (5, 1).

The slope of the line = (1 – 0)/ (5 – 0) = 1/5

Hence, the required equation of the line is given by

y – y_{1} = m(x – x_{1})

⇒ y – 0 = 1/5(x – 0)

⇒ 5y = x

⇒ x – 5y = 0

### Exercise 14(D)

**1. Find the slope and y-intercept of the line:**

**(i) y = 4**

**(ii) ax – by = 0**

**(iii) 3x – 4y = 5**

**Solution**

**(i)** y = 4

On comparing the given equation with y = mx + c, we get

Slope = m = 0

y – intercept = c = 4

**(ii)** ax – by = 0 ⇒ by = ax ⇒ y = (a/b)x

On comparing the above equation with y = mx + c, we get

Slope = m = a/b

y – intercept = c = 0

**(iii)** 3x – 4y = 5 ⇒ 4y = 3x – 5 ⇒ y = ¾ x – 5/4

On comparing the above equation with y = mx + c, we get

Slope = m = 3/4

y-intercept = c = -5/4

**2. The equation of a line x – y = 4. Find its slope and y-intercept. Also, find its inclination.**

**Solution**

Given equation of a line: x – y = 4

⇒ y = x – 4

Comparing the above equation with y = mx + c, we get

Slope = m = 1

y – intercept = c = -4

Let the inclination be Î¸.

Slope = 1 = tan Î¸ = tan 45^{o}

Î¸ = 45^{o}

^{}

**3. (i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0?**

**(ii) Is the line x – 3y = 4 perpendicular to the line 3x – y = 7?**

**(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1?**

**(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.**

**Solution**

**(i) **Given,

3x + 4y + 7 = 0

⇒ 4y = -3x – 7

⇒ y = (-3/4)x – 7/4

Slope of this line = -3/4

And, for

28x – 21y + 50 = 0

⇒ 21y = 28x + 50

⇒ y = (28/21)x + 50/21

⇒ y = (4/3)x + 50/21

Slope of this line = 4/3

As, the product of slopes of the two lines = 4/3 x -3/4 = -1

∴ the lines are perpendicular to each other.

**(ii)** Given,

x – 3y = 4

⇒ 3y = x – 4

⇒ y = (1/3)x – 4/3

So, the slope of this line = 1/3

And, for

3x – y = 7

⇒ y = 3x – 7

So, the slope of this line = 3

Now, the slopes of the two lines = 1/3 x 3 = 1 and not equal to -1.

Hence, the lines are not perpendicular to each other.

**(iii) **Given,

3x + 2y = 5

⇒ 2y = -3x + 5

⇒ y = (-3/2) x + 5/2

So, the slope of this line = -3/2

And, for

x + 2y = 1

⇒ 2y = -x + 1

⇒ y = -1/2x + 1/2

So, the slope of this line = -1/2

The slopes of the two lines are not equal.

Hence, the lines are not parallel to each other.

**(iv)** Given, the slope of the line through (1, 4) and (x, 2) is 2.

So,

⇒ -1 = x – 1

⇒ x = 0

**4. Find the slope of the line which is parallel to:**

**(i) x + 2y + 3 = 0**

**(ii) x/2 – y/3 – 1 = 0 **

**Solution**

**(i)** x + 2y + 3 = 0

⇒ 2y = -x – 3

⇒ y = -1/2x – 3/2

Slope of this line = -1/2

Thus, slope of the line which is parallel to the given line = slope of the given line = -1/2

**(ii)** x/2 – y/3 – 1 = 0

⇒ y/3 = x/2 – 1

⇒ y = (3/2)x – 3

So, the slope of this line = 3/2

Thus, slope of the line which is parallel to the given line = Slope of the given line = 3/2

**5. Find the slope of the line which is perpendicular to:**

**(i) x – y/2 + 3 = 0**

**(ii) x/3 – 2y = 4**

**Solution**

**(i)** x – y/2 + 3 = 0

⇒ y/2 = x + 3

⇒ y = 2x + 6

So, the slope of this line = 2

We know that,

Slope of the line which is perpendicular to the given line = -1/(Slope of the given line) = -1/2

**(ii)** x/3 – 2y = 4

⇒ 2y = x/3 – 4

⇒ y = x/6 – 2

So, the slope of this line = 1/6

We know that,

Slope of the line which is perpendicular to the given line = -1/(Slope of the given line) = -1/(1/6) = -6

**6. (i) Lines 2x – by + 3 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.**

**(ii) Lines mx + 3y + 7 = 0 and 5x – ny – 3 = 0 are perpendicular to each other. Find the relation connecting m and n.**

**Solution**

**(i)** We know that, if two lines are parallel then, the slopes of the two lines must be equal.

For, 2x – by + 3 = 0

⇒ by = 2x + 3

⇒ y = (2/b)x + 3/b

So, the slope of this line = 2/b

And, ax + 3y = 2

⇒ 3y = -ax + 2

⇒ y = (-a/3)x + 2/3

So, the slope of this line = -a/3

Now, equating the slopes we get

2/b = -a/3

⇒ ab = -6

**(ii)** We know that, if two lines are perpendicular to each other then, the product of their slopes = -1.

For, mx + 3y + 7 = 0

⇒ 3y = -mx – 7

⇒ y = -m/3x – 7/3

Slope of this line = -m/3

And, 5x – ny – 3 = 0

⇒ ny = 5x – 3

⇒ y = (5/n)x – 3/n

Slope of this line = 5/n

Products of slopes is

(-m/3) × (5/n) = -1

⇒ 5m = 3n

**7. Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.**

**Solution**

2x – y + 5 = 0 **…. (1)**

px + 3y = 4 **….. (2)**

⇒ y = 2x + 5

Now,

Slope of line = 2

px + 3y = 4

The above equation can be rewritten as

3y = -px + 4

⇒ y = (-p/3)x + 4/3

So, the slope of this line = -p/3

For 2 lines to be perpendicular to each other, the product of their slopes must be -1.

So,

(2) x (-p/3) = -1

⇒ 2p/3 = 1

⇒ p = 3/2

**8. The equation of a line AB is 2x – 2y + 3 = 0.**

**(i) Find the slope of the line AB.**

**(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.**

**Solution**

**(i)** Given, equation of the line

2x – 2y + 3 = 0

⇒ 2y = 2x + 3

⇒ y = x + (3/2)

So, the slope of the line AB = 1

**(ii) **Required to find the angle of the line AB = Î¸

We have,

Slope = tan Î¸ = 1

And, tan 45° = 1

Hence, Î¸ = 45°

**9. The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.**

**Solution**

4x + 3y = 9

⇒ 3y = -4x + 9

⇒ y = (-4/3)x + 3

Slope of this line = -4/3

And,

px – 6y + 3 = 0

⇒ 6y = px + 3

⇒ y = (p/6)x + ½

Slope of this line = p/6

For two lines to be parallel, their slopes must be equal.

-4/3 = p/6

⇒ -4 = p/2

⇒ p = -8

**10. If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.**

**Solution**

y = 3x + 7

Slope of this line = 3

And,

2y + px = 3

⇒ 2y = -px + 3

⇒ y = (-p/2)x + 3

So, the slope of this line = -p/2

If two lines are perpendicular to each other, then the product of their slopes is -1.

(3) × (-p/2) = -1

⇒ 3p/2 = 1

⇒ p = 2/3

**11. The line through A(-2,3) and B(4,b) is perpendicular to the line 2x – 4y =5. Find the value of b.**

**Solution**

Given,

Points A (-2, 3) and B (4, b)

And, line equation: 2x – 4y = 5

4y = 2x – 5

⇒ y = (1/2)x – 5/4

So, the slope of this line = ½

From the question, it’s said that

The line through A and B is perpendicular to above given line.

We know that, when two lines are perpendicular their product of slopes is -1

Hence, the slope of the line through A and B must be -2.

Now,

The slope of the line through A and B is given by,

Slope of AB = (b – 3)/ (4 + 2)

= (b – 3)/6

Thus,

(b – 3)/6 = -2

⇒ b – 3 = -12

⇒ b = -9

**12. Find the equation of the line through (-5, 7) and parallel to:**

**(i) x-axis**

**(ii) y-axis**

**Solution**

**(i)** We know that, the slope of a line parallel to x-axis is 0.

Here, (x_{1}, y_{1}) = (-5, 7) and m = 0

So, the required line equation is

y – y_{1} = m(x – x_{1})

⇒ y – 7 = 0(x + 5)

⇒ y = 7

**(ii) **We know that, the slope of a line parallel to y-axis is not defined. (tan90^{o})

So, the given line is parallel to y-axis.

Here, (x_{1}, y_{1}) = (-5, 7)

So, the required equation of the line is

x – x_{1 }= 0

⇒ x + 5 = 0

### Exercise 14(E)

**1. Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5. Find its co-ordinates of point P.**

**Also, find the equation of the line through P and parallel to 3x + 5y = 7.**

**Solution**

Given points, A (8, 0) and B (16, -8)

By section formula, the co-ordinates of the point P which divides AB in the ratio 3: 5 is given by

= (11, -3) = (x_{1}, y_{1})

Given line equation is,

3x + 5y = 7

⇒ 5y = -3x + 7

⇒ y = (-3/5)x + 7/5

So, the slope of this line = -3/5

The line parallel to the line 3x + 5y = 7 will have the same slope,

Hence, the slope of the required line = Slope of the given line = -3/5

Thus,

The equation of the required line is

y – y_{1} = m(x – x_{1})

⇒ y + 3 = (-3/5)(x – 11)

⇒ 5y + 15 = -3x + 33

⇒ 3x + 5y = 18

**2. The line segment joining the points A(3, -4) and B (-2, 1) is divided in the ratio 1: 3 at point P in it. Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y + 4 = 0.**

**Solution**

Given points, A (3, -4) and B (-2, 1)

By section formula, the co-ordinates of the point P which divides AB in the ratio 1: 3 is given by

= (7/4, -11/4) = (x_{1}, y_{1})

Given line equation is,

5x – 3y + 4 = 0

⇒ 3y = 5x + 4

⇒ y = (5/3) x + 4/3

So, the slope of this line = 5/3

The line perpendicular to the given line will have slope

Slope of the required line = -1/(5/3) = -3/5

Hence,

The equation of the required line is given by

y – y_{1} = m(x – x_{1})

⇒ 20y + 55 = -12x + 21

⇒ 12x + 20y + 34 = 0

⇒ 6x + 10y + 17 = 0

**3. A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.**

**Solution**

As the point P lies on y-axis,

Putting x = 0 in the equation 5x + 3y + 15 = 0, we get

5(0) + 3y + 15 = 0

⇒ y = -5

Hence, the co-ordinates of the point P are (0, -5).

Given line equation,

x – 3y + 4 = 0

⇒ 3y = x + 4

⇒ y = (1/3)x + 4/3

Slope of this line = 1/3

From the question, the required line equation is perpendicular to the given equation: x – 3y + 4 = 0.

So, the product of their slopes is -1.

Slope of the required line = -1/(1/3) = -3

And,

(x_{1}, y_{1}) = (0, -5)

∴ The required line equation is

y – y1 = m(x – x1)

⇒ y + 5 = -3(x – 0)

⇒ 3x + y + 5 = 0

**4. Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other.**

**Solution**

Given,

kx – 5y + 4 = 0

⇒ 5y = kx + 4

⇒ y = (k/5)x + 4/5

So, the slope of this line = m1 = k/5

And, for 5x – 2y + 5 = 0

⇒ 2y = 5x + 5

⇒ y = (5/2) x + 5/2

Slope of this line = m2 = 5/2

As, the lines are perpendicular to each other

m_{1 }× m_{2} = -1

⇒ (k/5) × (5/2) = -1

⇒ k = -2

**5. A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid-point of the segment AB. Find:**

**(i) the equation of the line.**

**(ii) the co-ordinates of A and B.**

**(iii) the co-ordinates of M.**

**Solution**

**(i)** Given points, P (-1, 4) and Q (5, -2)

Slope of PQ = (-2 – 4)/ (5 + 1) = -6/6 = -1

Equation of the line PQ is given by,

y – y_{1} = m(x – x_{1})

⇒ y – 4 = -1(x + 1)

⇒ y – 4 = -x – 1

⇒ x + y = 3

**(ii)** For point A (on x-axis), y = 0.

So, putting y = 0 in the equation of PQ, we have

x = 3

Hence, the co-ordinates of point A are (3, 0).

For point B (on y-axis), x = 0.

So, putting x = 0 in the equation of PQ, we have

y = 3

Hence, the co-ordinates of point B are (0, 3).

**(iii)** M is the mid-point of AB.

Thus, the co-ordinates of point M are

(3+0/2, 0+3/2) = (3/2, 3/2)

**6. (1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.**

**Solution**

Given, A = (1, 5) and C = (-3, -1) of rhombus ABCD.

We know that in a rhombus, diagonals bisect each other at right angle.

Let’s take O to be the point of intersection of the diagonals AC and BD.

Then, the co-ordinates of O are

Slope of AC = (-1 – 5)/ (-3 – 1) = -6/-4 = 3/2

Then, the equation of the line AC is

y – y_{1} = m (x – x_{1})

⇒ y – 5 = (3/2) (x – 1)

⇒ 2y – 10 = 3x – 3

⇒ 3x – 2y + 7 = 0

Now, the line BD is perpendicular to AC

Slope of BD = -1/(slope of AC) = -2/3

And, (x_{1}, y_{1}) = (-1, 2)

Hence, equation of the line BD is

y – y_{1} = m (x – x_{1})

⇒ y – 2 = (-2/3) (x + 1)

⇒ 3y – 6 = -2x – 2

⇒ 2x + 3y = 4

**7. Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.**

**(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.**

**(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.**

**Solution**

Given, A (3, 2), B (6, -2) and C (2, -5)

Now, by distance formula

AB = √[(6 – 3)^{2} + (-2 – 2)^{2}] = √(9 + 16) = 5

BC = √[(2 – 6)^{2} + (-5 + 2)^{2}] = √(16 + 9) = 5

Thus, AC = BC

Then,

Slope of AB = (-2 – 2)/ (6 – 3) = -4/3

Slope of BC = (-5 + 2)/ (2 – 6) = -3/-4 = ¾

Slope of AB × Slope of BC = -4/3 × ¾ = -1

Hence, AB ⊥ BC

∴ A, B, C can be the vertices of a square.

**(i)** Slope of AB = (-2 – 2)/ (6 – 3) = -4/3 = slope of CD

So, the equation of CD is

y – y_{1} = m (x – x_{1})

⇒ y + 5 = -4/3(x – 2)

⇒ 3y + 15 = -4x + 8

⇒ 3y = -4x – 7

⇒ 4x + 3y + 7 = 0 **… (1)**

Now, slope of BC = (-5 + 2)/ (2 – 6) = -3/-4 = ¾ = Slope of AD

So, the equation of the line AD is

y – y_{1} = m (x – x_{1})

⇒ y – 2 = (3/4) (x – 3)

⇒ 4y – 8 = 3x – 9

⇒ 3x – 4y = 1 **…. (2)**

Now, D is the point of intersection of CD and AD.

Solving (1) and (2),

4×(1) + 3×(2)

⇒ 16x + 12y + 9x – 12y = -28 + 3

⇒ 25x = -25

⇒ x = -1

Putting value of x in (1), we get

4(-1) + 3y + 7 = 0

⇒ 3y = -3

⇒ y = -1

∴ the co-ordinates of point D are (-1, -1).

**(ii) **From the equation (2)

The equation of the line AD is,

3x – 4y = 1

Slope of BD = (-1 + 2)/(-1 – 6) = (1/-7) = (-1/7)

The equation of the diagonal BD is

y – y_{1} = m (x – x_{1})

⇒ y + 1 = -1 / 7 (x + 1)

⇒ 7y + 7 = – x – 1

⇒ x + 7y + 8 = 0

**8. A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.**

**Solution**

The given line equation is

x = 3y + 2 **… (1)**

⇒ 3y = x – 2

⇒ y = 1/3 x – 2/3

So, slope of this line is 1/3.

And, the required line intersects the given line at right angle.

Thus, slope of the required line = -1/(1/3) = -3

And. the required line passes through (0, 0) = (x_{1}, y_{1})

So, the equation of the required line is:

y – y_{1} = m(x – x_{1})

⇒ y – 0 = -3(x – 0)

⇒ 3x + y = 0 **… (2)**

Next,

Point X is the intersection of the lines (1) and (2).

Using (1) in (2), we get,

3(3y + 2) + y = 0

⇒ 9y + 6 + y = 0

⇒ 10y = -6

⇒ y = -6/10 = -3/5

And, finally

x = 3(-3/5) + 2

= -9/5 + 2

= 1/5

Thus, the co-ordinates of the point X are (1/5, -3/5).

**9. A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.**

**Solution**

Let the line intersect the x-axis at point A (x, 0) and y-axis at point B (0, y).

Since, P is the mid-point of AB, we have:

⇒ (x/2, y/2) = (3, 2)

⇒ x = 6, y = 4

Thus, A = (6, 0) and B = (0, 4)

Slope of line AB = (4 – 0)/ (0 – 6) = 4/-6 = -2/3

And, let (x_{1}, y_{1}) = (6, 0)

So, the required equation of the line AB is given by

y – y_{1} = m(x – x_{1})

⇒ y – 0 = (-2/3) (x – 6)

⇒ 3y = -2x + 12

⇒ 2x + 3y = 12

**10. Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x – 2y = 1.**

**Solution**

Given line equations are,

7x + 6y = 71 ⇒ 28x + 24 = 284 **… (1)**

5x – 8y = -23 ⇒ 15x – 24y = -69 **… (2)**

On adding (1) and (2), we have

43x = 215

⇒ x = 5

From (2), we get

8y = 5x + 23 = 25 + 23 = 48

⇒ y = 6

Hence, the required line passes through the point (5, 6).

Given, 4x – 2y = 1

⇒ 2y = 4x – 1

⇒ y = 2x – (1/2)

So, the slope of this line = 2

And, the slope of the required line = -1/2 [As the lines are perpendicular to each other]

Thus, the required equation of the line is

y – y_{1} = m(x – x_{1})

⇒ y – 6 = (-1/2) (x – 5)

⇒ 2y – 12 = -x + 5

⇒ x + 2y = 17

**11. Find the equation of the line which is perpendicular to the line x/a – y/b = 1 at the point where this line meets y-axis.**

**Solution**

The given line equation is,

x/a – y/b = 1

y/b = x/a – 1

y = (b/a)x – b

The slope of this line = b/a

So, the slope of the required line = -1/(b/a) = – a/b

Let the line intersect at point P (0, y) on the y-axis.

So, putting x = 0 in the equation x/a – y/b = 1, we get

0 – y/b = 1

y = -b

Hence, P = (0, -b) = (x_{1}, y_{1})

∴ The equation of the required line is

y – y_{1} = m (x – x_{1})

⇒ y + b = (-a/b) (x – 0)

⇒ by + b^{2} = -ax

⇒ ax + by + b^{2} = 0

**12. O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:**

**(i) the equation of median of triangle OAB through vertex O.**

**(ii) the equation of altitude of triangle OAB through vertex B.**

**Solution**

**(i)** Let’s consider the median through O meets AB at D. So, D will be the mid-point of AB.

Co-ordinates of point D are:

The slope of OD = (1 – 0)/ (-1 – 0) = -1

And, (x_{1}, y_{1}) = (0, 0)

Hence, the equation of the median OD is

y – y_{1} = m(x – x_{1})

⇒ y – 0 = -1(x – 0)

⇒ x + y = 0

**(ii) **The altitude through vertex B is perpendicular to OA.

We have, slope of OA = (5 – 0)/ (3 – 0) = 5/3

Then, the slope of the required altitude = -1/(5/3) = -3/5

Hence, the equation of the required altitude through B is

y – y_{1} = m(x – x_{1})

⇒ y + 3 = (-3/5) (x + 5)

⇒ 5y + 15 = -3x – 15

⇒ 3x + 5y + 30 = 0

**13. Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.**

**Does the line 3x = y + 1 bisect the line segment joining the two given points?**

**Solution**

Let A = (-2, 3) and B = (4, 1)

Slope of AB = m_{1} = (1 – 3)/(4 + 2) = -2/6 = -1/3

So, the equation of line AB is

y – y_{1} = m_{1}(x – x_{1})

⇒ y – 3 = (-1/3) (x + 2)

⇒ 3y – 9 = -x – 2

⇒ x + 3y = 7 **… (1)**

Slope of the given line 3x = y + 1 is 3 = m_{2}.

It’s seen that, m_{1} x m_{2} = -1

Thus, the line through points A and B is perpendicular to the given line.

Given line is 3x = y +1 **… (2)**

The co-ordinates of the mid-point of AB are

Now, Let’s check if point P satisfies the line equation (2)

3(1) = 2 + 1

⇒ 3 = 3

Hence, the line 3x = y + 1 bisects the line segment joining the points A and B.

**14. Given a straight line x cos 30 ^{o} + y sin 30^{o} = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).**

**Solution**

Given line equation, x cos 30^{o} + y sin 30^{o} = 2

So, the slope of this line = -√3

Slope of a line which is parallel to this given line = -√3

Let (4, 3) = (x_{1}, y_{1})

∴ the equation of the required line is given by

y – y_{1} = m_{1 }(x – x_{1})

⇒ y – 3 = -√3 (x – 4)

⇒ √3x + y = 4√3 + 3

**15. Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:**

**(i) perpendicular to the line 2x – y + 7 = 0**

**(ii) parallel to it.**

**Solution**

Given line equation,

(k – 2)x + (k + 3)y – 5 = 0 **…. (1)**

(k + 3)y = -(k – 2)x + 5

Slope of this line = m_{1} = (2 – k)/ (k + 3)

**(i)** Given, 2x – y + 7 = 0

⇒ y = 2x + 7 = 0

Slope of this line = m_{2} = 2

Given that, line (1) is perpendicular to 2x – y + 7 = 0

m_{1} x m_{2 }= -1

⇒ (2 – k)/ (k + 3) x 2 = -1

⇒ 4 – 2k = -k – 3

⇒ k = 7

**(ii)** Line (1) is parallel to 2x – y + 7 = 0

So, m_{1} = m_{2}

⇒ (2 – k)/(k + 3) = 2

⇒ 2 – k = 2k + 6

⇒ 3k = -4

⇒ k = -4/3

**16. The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find:**

**(i) the equation of line through A and perpendicular to BC.**

**(ii) the co-ordinates of the point, where the perpendicular through A, as obtained in (i), meets BC.**

**Solution**

Slope of BC = (7 + 2)/(11 +1) = 9/12 = 3/4

Then the equation of the line BC is

y – y_{1 }= m(x – x_{1})

⇒ y + 2 = ¾ (x + 1) where x_{1 }= -1 and y_{1} = -2

⇒ 4y + 8 = 3x + 3

⇒ 3x – 4y = 5 **…. (1)**

**(i)** Slope of line perpendicular to BC will be = -1/(3/4) = -4/3

So, the required equation of the line through A (0, 5) and perpendicular to BC is given by

y – y_{1} = m_{1} (x – x_{1})

⇒ y – 5 = (-4/3) (x – 0)

⇒ 3y – 15 = -4x

⇒ 4x + 3y = 15 **…. (2)**

(ii) Hence, the required point will be the point of intersection of lines (1) and (2).

Solving (1) & (2),

(1) ⇒ 9x – 12y = 15

(2) ⇒ 16x + 12y = 60

Now, adding the above two equations, we get

25x = 75

⇒ x = 3

Substituting the value of x in equation (1) we get,

4y = 3x – 5 = 9 – 5 = 4

⇒ y = 1

Thus, the co-ordinates of the required point is (3, 1).

**17. From the given figure, find:**

**(i) the co-ordinates of A, B and C.**

**(ii) the equation of the line through A and parallel to BC.**

**Solution**

(i) A = (2, 3), B = (-1, 2), C = (3, 0)

(ii) Slope of BC = (0 – 2)/ (3 + 1) = -2/4 = -1/2

Slope of the line which is parallel to BC = Slope of BC = -1/2

(x_{1}, y_{1}) = (2, 3)

Hence, the required equation of the line through A and parallel to BC is given by

y – y_{1} = m_{1 }(x – x_{1})

⇒ y – 3 = (-1/2) (x – 2)

⇒ 2y – 6 = -x + 2

⇒ x + 2y = 8

**18. P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.**

**Solution**

We know that, the median, RX through R will bisect the line PQ.

The co-ordinates of point X are

Slope of RX = (1 + 1)/(5 + 2) = 2/7 = m_{1}

(x_{1}, y_{1}) = (-2, -1)

Then, the required equation of the median RX is given by

y – y_{1} = m_{1}(x – x_{1})

⇒ y + 1 = (2/7) (x + 2)

⇒ 7y + 7 = 2x + 4

⇒ 7y = 2x – 3

**19. A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ.**

**Solution**

P is the mid-point of AB. Hence, the co-ordinates of point P are

Q is the mid-point of AC. So, the co-ordinate of point Q are

Slope of PQ = (-8 + 2)/ (4 – 2) = -6/2 = -3

Slope of BC = (-10 – 2)/ (0 + 4) = -12/4 = -3

As, the slope of PQ = Slope of BC,

∴ PQ || BC

Also,

Slope of PB = (-2 – 2)/ (2 + 4) = -2/3

Slope of QC = (-8 + 10)/ (4 – 0) = 1/2

So, PB is not parallel to QC as their slopes are not equal

Thus, PBCQ is a trapezium.

**20. A line AB meets the x-axis at point A and y-axis at point B. The point P (-4, -2) divides the line segment AB internally such that AP: PB = 1: 2. Find:**

**(i) the co-ordinates of A and B.**

**(ii) the equation of line through P and perpendicular to AB.**

**Solution**

(i) Let’s assume the co-ordinates of point A, lying on x-axis be (x, 0) and the co-ordinates of point B (lying on y-axis) be (0, y).

Given,

P = (-4, -2) and AP: PB = 1:2

By section formula, we get

⇒ -4 = 2x/3 and -2 = y/3

⇒ x = -6 and y = -6

Hence, the co-ordinates of A and B are (-6, 0) and (0, -6).

(ii) Slope of AB = (-6 – 0)/ (0 + 6) = -6/6 = -1

Slope of the required line perpendicular to AB = -1/-1 = 1

Here, (x_{1}, y_{1}) = (-4, -2)

∴ the required equation of the line passing through P and perpendicular to AB is given by

y – y_{1} = m(x – x_{1})

⇒ y + 2 = 1(x + 4)

⇒ y + 2 = x + 4

⇒ y = x + 2