## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C.

**Other Exercises**

- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14A
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D
- Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E

**Question 1.**

**Find the equation of line whose :**

**(i) y-intercept = 2 and slope = 3,**

**(ii) y-intercept = -1 and slope = \(\frac { -3 }{ 4 }\)**

**Solution:**

(i) The point whose y-intercept = 2 will be (0, 2) and slope (m) = 3.

Equation of line will be

**Question 2.**

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

**(i) y-intercept = -1 and inclination = 45°**

**(ii) y-intercept = 3 and inclination = 30°**

**Solution:**

(i) The point whose y-intercept is -1, will be (0, -1) and inclination = 45°

Slope (m) = tan 45° = 1

Equation will be

**Question 3.**

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

**Solution:**

Slope of the line (m) = \(\frac { -4 }{ 3 }\)

The point from which the line passes (-3, 4)

Equation of line will be y – y_{1} = m (x – x_{1})

**Question 4.**

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

**Solution:**

The line passes through the point (5, 4) and angle of inclination = 60°

slope (m) = tan 60° = √3

Equation of line

y – y_{1} = m (x – x_{1})

⇒ y – 4 = √3 (x – 5)

⇒ y – 4 = √3 x – 5√3

⇒ y = √3 x + 4 – 5√3

**Question 5.**

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

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

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

**(iii) (4, -2) and (5, 2)**

**Solution:**

Two given points are (0, 1) and (1, 2)

Slope of the line passing through these two

**Question 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:**

Two points P (2,-6) and Q (-3, 5) are given.

⇒ 5y – 30 = x – 2

⇒ 5y = x – 2 + 30

⇒ 5y = x + 28 ….(i)

(iii) Co-ordinates of the point where PQ intersects x-axis will be = 0

substituting, the value of y in (i)

5 x 0 = x + 28 ⇒ x + 28 = 0 ⇒ x = -28

Co-ordinates of point are (-28, 0)

**Question 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 they- axis.**

**Solution:**

Slope of the line passing through two points A (-3, 4) and B (2, -1) will be :

Its abscissa = 0

substituting, the value of x = 0 in (i)

0 + y = 1

y = 1

Co-ordinates of point = (0, 1)

**Question 8.**

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

**Solution:**

Two lines AB and CD intersect each other at P (3, 4)

AB inclined at angle of 45° and CD at angle of 60° with the x-axis.

Slope of AB = tan 45° = 1

and slope of CD = tan 60° = √3

Now, equation of line AB will be

y – y_{1} = m (x – x_{1})

⇒ y – 4 = 1 (x – 3)

⇒ y – 4 = x – 3

⇒ y = x – 3 + 4

⇒ y = x + 1

(ii) Equation of CD will be :

y – y_{1} = m (x – x_{1})

⇒ y – 4 = √3 (x – 3)

⇒ y – 4 = √3 x – 3√3

⇒ y = √3 x – 3√3 + 4

⇒ y = √3 x + 4 – 3√3

**Question 9.**

**In ΔABC, A (3, 5), B (7, 8) and C (1, -10). Find the equation of the median through A.**

**Solution:**

AD is median

D is mid point of BC

Equation of AD

y – y_{1} = m (x – x_{1})

⇒ y + 1 = -6 (x – 4)

⇒ y + 1 = -6x + 24

⇒ y + 6x = -1 + 24

⇒ 6x + y = 23

**Question 10.**

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

**Solution:**

ABCD is a ||gm in which AB = CD || x-axis

∠A = 60° and C (7, 5)

(i) CD || AB || x-axis ,

Equation of CD will be

y – y_{1} = m (x – x_{1})

⇒ y – 5 = 0 (x – 7)

⇒ y – 5 = 0

⇒ y = 5

BC || AD

Slope of BC = tan 60° = √3

Equation of BC will be

y – y_{1} = m (x – x_{1})

⇒ y – 6 = √3 (x – 7)

⇒ y – 6 = √3 x – 7√3

⇒ y = √3 x + 6 – 7√3

**Question 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:**

Point of intersection of two lines

x + 2y = 7 ….(i)

x – y = 4 ….(ii)

Subtracting, we get

3y = 3

y =1

Substituting, the value of y in (ii)

x – 1 = 4

⇒ x = 4 + 1 = 5

Point of intersection is (5, 1)

Slope of the line passing through origin (0, 0) and (5, 1)

Equation of line will be

y – y_{1} = m (x – x_{1})

⇒ y – 5 = \(\frac { 1 }{ 2 }\) (x – 1)

⇒ 5y – 25 = x – 1

⇒ 5y = x – 1 + 25 = x + 24

⇒ 5y = x + 24

**Question 12.**

**In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A. Also, find the equation of the line through vertex B and parallel to AC.**

**Solution:**

**Question 13.**

**A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.**

**Solution:**

Slope of line through A, perpendicular to BC = -(-1) = 1

Now, the equation of line through A (0, 3) is

y – y_{1} = m (x – x_{1})

y – 3 = 1 (x – 0)

⇒ y – 3 = x

⇒ y = x + 3

**Question 14.**

**Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).**

**Solution:**

Slope of the line joining the points (1, 4) and (2, 3)

Slope of line perpendicular to the above line = 1

Equation of line passing through (-1, 2)

y – y_{1} = m (x – x_{1})

⇒ y – 2 = 1 [x -(-1)]

⇒ y – 2 = x + 1

⇒ y = x + 1 + 2

⇒ y = x + 3

**Question 15.**

**Find the equation of the line, whose :**

**(i) x-intercept = 5 and y-intercept = 3**

**(ii) x-intercept = -4 and y-intercept = 6**

**(iii) x-intercept = -8 and y-intercept = -4**

**(iv) x-intercept = 3 and y-intercept = -6**

**Solution:**

(i) When x-intercept = 5, then point will be (5, 0)

and when y-intercept = 3, then point will be (0, 3)

Slope of the line passing through these points

**Question 16.**

**Find the equation of the line whose slope is \(\frac { -5 }{ 6 }\) and x-intercept is 6.**

**Solution:**

**Question 17.**

**Find the equation of the line with x-intercept 5 and a point on it (-3, 2).**

**Solution:**

x-intercept of the line = 5

Point = (5, 0)

Slope of the line passing through the point (-3, 2)

**Question 18.**

**Find the equation of the line through (1, 3) and making an intercept of 5 on the y- axis.**

**Solution:**

The line makes y-intercept = 5

Point = (0, 5)

Slope of the line passing through the point (1, 3) and (0, 5)

Equation of the line

y – y_{1} = m (x – x_{1})

⇒ y – 3 = -2 (x – 1)

⇒ y – 3 = -2x + 2

⇒ 2x + y = 2 + 3

⇒ 2x + y = 5

**Question 19.**

**Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axes.**

**Solution:**

(i) Slope of line AB = tan 45° = 1

Equation passing through the point (-2, 0)

y – y_{1} = m (x – x_{1})

⇒ y – 0 = 1 (x + 2)

⇒ y = x + 2

⇒ x – y + 2 = 0

(ii) Slope of line CD = tan (-45°) = -1

Equation passing through the point (-2, 0)

y – y_{1} = m (x – x_{1})

⇒ y – 0 = -1 (x + 2)

⇒ y = -x – 2

⇒ y + x + 2 = 0

⇒ x + y + 2 = 0

**Question 20.**

**The line through P (5, 3) intersects y axis at Q.**

**(i) Write the slope of the line.**

**(ii) Write the equation of the line.**

**(iii) Find the co-ordinates of Q.**

**Solution:**

(i) Here θ = 45°

So, slope of the line = tanθ = tan 45° = 1

(ii) Equation of the line through P and Q is

y – 3 = 1 (x – 5)

⇒ y – x + 2 = 0

(iii) Let the coordinates of Q be (0, y)

**Question 21.**

**Write down the equation of the line whose gradient is \(\frac { -2 }{ 5 }\) and which passes through point P, where P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3 : 1.**

**Solution:**

Slope of the line m = \(\frac { -2 }{ 5 }\)

P divides the line AB, whose co-ordinates are (4, -8) and (12, 0) in the ratio of 3 : 1

Co-ordinates of P be

**Question 22.**

**A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC. Find :**

**(i) the co-ordinates of the centroid of ΔABC.**

**(ii) the equation of a line, through the centroid and parallel to AB. [2002]**

**Solution:**

(i) Co-ordinates of vertices of ΔABC are A (1, 4), B (3, 2), C (7, 5)

and let G be the centroid of ΔABC.

Co-ordinates of G are

**Question 23.**

**A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point F in AC; such that AP : CP = 2 : 3.**

**Solution:**

P divides AC in the ratio of 2 : 3

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C are helpful to complete your math homework.

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