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

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 14B.

**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 slope of the line whose inclination is :**

**(i) 0°**

**(ii) 30°**

**(iii) 72° 30′**

**(iv) 46°**

**Solution:**

(i) Slope of line whose inclination is 0° = tanθ = tan 0° = 0

(ii) Slope of line whose inclination is 30° = tan 30° = \(\frac { 1 }{ \surd 3 }\)

(iii) Slope of line whose inclination is 72° 30′ = tan 72°30’ = 3. 1716 (Using tables)

(iv) Slope of line whose inclination is 46° = tan 46° = 1.0355 (Using tables)

**Question 2.**

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

**(i) 0**

**(ii) √3**

**(iii) 0.7646**

**(iv) 1.0875**

**Solution:**

Slope of a line = tanθ. Where θ is the inclination

(i) When slope is θ. then tanθ = 0 ⇒ θ = 0°

(ii) When slope is θ, then tanθ = √3 ⇒ θ = 60°

(iii) When slope is 0.7646, then tanθ = 0.7646 ⇒ θ = 37°24′ (Using tables)

(iv) When slope is 1.0875, then tanθ = 1.0875 ⇒ θ = 47°24′ (Using tables)

**Question 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 of a line which passes

**Question 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:**

**Question 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:**

**Question 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 two points (0, 2) and (-3, -1)

**Question 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 the points

**Question 8.**

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

**Solution:**

AB and CA are perpendicular to each other

Hence, ΔABC is a right-angled triangle.

**Question 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:**

Slopes of AB and DC are equal

AB || DC Similarly slope of BC and slope of DA are equal.

BC || DA

Hence ABCD is a parallelogram.

**Question 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:**

QR || PS.

Hence PQRS is a parallelogram.

**Question 11.**

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

**Solution:**

The given points are P (a, b + c), Q (b, c + a) and R (c, a + b)

We know that, these points P, Q, R are collinear if Slope of PQ = Slope of QR

Slope of PQ = Slope of QR

P, Q and R are collinear.

**Question 12.**

**Find x, if the slope of the line joining (x, 2) and (8, -11) is \(\frac { -3 }{ 4 }\).**

**Solution:**

Slope of line joining (x, 2) and (8, -11) is

**Question 13.**

**The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slopes of all its sides.**

**Solution:**

ΔABC is an equilateral

Each angle is equal to 60°

Side AB is parallel to x-axis

Slope of AB = slope of x-axis = 0.

Slope of AC = tan A = tan 60° = √3

Slope of CB = tan B = tan 120° = tan (180°- 60°) = – tan 60° = -√3

Slopes of AB, BC and CA are 0, -√3, √3

**Question 14.**

**The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides. Also, find :**

**(i) the slope of the diagonal AC**

**(ii) the slope of the diagonal BD.**

**Solution:**

ABCD is a square in which AB || DC || x-axis.

AD || BC || y-axis

Slope of AB and DC = 0

and slope of AD and BC = not defined (tan90° is not defined)

AC and BD are the diagonals of square ABCD.

Now slope of AC = tan 45° = 1

and slope of BD = tan 135° = tan (180° – 45°) = – tan 45° = -1

**Question 15.**

**A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find :**

**(i) the slope of the altitude of AB**

**(ii) the slope of the median AD and**

**(iii) the slope of the line parallel to AC.**

**Solution:**

Vertices of ΔABC are A (5, 4), B (-3, -2), and C (1, -8)

**Question 16.**

**The slope of the side BC of a rectangle ABCD is \(\frac { 2 }{ 3 }\). Find**

**(i) The slope of the side AB,**

**(ii) the slope of the side AD.**

**Solution:**

ABCD is a rectangle in which

**Question 17.**

**Find the slope and the inclination of the line AB if**

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

**(ii) A = (0, -√3) and B = (3, 0)**

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

**Solution:**

**Question 18.**

**The points (-3, 2), (2, -1) and (a, 4) are collinear, Find ‘a’.**

**Solution:**

Points are collinear.

Slope of (-3, 2) and (2, -1) = Slope of (2, -1) and (a, 4)

Now, Slope of (-3, 2) and (2, -1) will be

**Question 19.**

**The points (k, 3), (2, -4) and (-k + 1, -2) are collinear. Find k.**

**Solution:**

Points (k, 3), (2, -4) and (-k + 1, -2) are collinear

Slope of (k, 3) and (2, -4) = slope of (2, -4) and (-k + 1, -2)

Now, slope of (k, 3) and (2, -4)

**Question 20.**

**Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.**

**Which segment appears to have the steeper slope, AB or AC ?**

**Justify your conclusion by calculating the slopes of AB and AC.**

**Solution:**

**Question 21.**

**Find the value(s) of k so that PQ will be parallel to RS. Given :**

**(i) P (2, 4), Q (3, 6), R (8, 1) and S (10, k)**

**(ii) P (3, -1), Q (7, 11), R (-1, -1) and S (1, k)**

**(iii) P (5, -1), Q (6, 11), R (6, -4k) and S (7, k²)**

**Solution:**

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

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