**Question****1. Find, which of the following points lie on the line x β 2y + 5 = 0:****(i) (1, 3)****(ii) (0, 5)****(iii) (-5, 0)****(iv) (5, 5)****(v) (2, -1.5)****(vi) (-2, -1.5)****Solution:**

(i) Given that, the line is π₯β 2π¦ + 5 = 0

Here, π₯ = 1 and π¦ = 3

Put the value of x and y in given equation,

(1)β 2(3) + 5 = 0

1β 6 + 5 = 0

β5 + 5 = 0

0 = 0

Here, πΏπ»π = π
π»π

Hence, the point (1,3) lies on the line π₯β 2π¦ + 5 = 0.

(ii) Given that, the line is π₯β 2π¦ + 5 = 0

Here, π₯ = 0 and π¦ = 5

Put the value of x and y in given equation,

(0)β 2(5) + 5 = 0

0β 10 + 5 = 0

β10 + 5 = 0

β5 = 0

πΏπ»π β π
π»π

Hence, the point (0,5) does not lies on the line π₯β 2π¦ + 5 = 0.

(iii) Given that, the line is π₯β 2π¦ + 5 = 0

Here, π₯ = β5 and π¦ = 0

Put the value of x and y in given equation,

(β5)β 2(0) + 5 = 0

β5β 0 + 5 = 0

β5 + 5 = 0

0 = 0

πΏπ»π = π
π»π

Hence, the point (-5,0) lies on the line π₯β 2π¦ + 5 = 0.

(iv) Given that, the line is π₯β 2π¦ + 5 = 0

Here, π₯ = 5 and π¦ = 5

Put the value of x and y in given equation,

(5)β 2(5) + 5 = 0

5β 10 + 5 = 0

10 β 10 = 0

0 = 0

πΏπ»π = π
π»π

Hence, the point (5,5) lies on the line π₯β 2π¦ + 5 = 0.

(v) Given that, the line is π₯β 2π¦ + 5 = 0

Here, π₯ = 2 and π¦ = β1.5

Put the value of x and y in given equation,

(2)β 2(β1.5) + 5 = 0

2 + 3 + 5 = 0

10 = 0

πΏπ»π β π
π»π

Hence, the point (2,-1.5) does not lies on the line π₯β 2π¦ + 5 = 0.

(vi) Given that, the line is π₯β 2π¦ + 5 = 0

Here, π₯ = β2 and π¦ = β1.5

Put the value of x and y in given equation,

(β2)β 2(β1.5) + 5 = 0

β2 + 3 + 5 = 0

6 = 0

πΏπ»π β π
π»π

Hence, the point (-2,-1.5) does not lies on the line π₯β 2π¦ + 5 = 0.

**Question****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) 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

Here, x = 2 and y = 3

Put the value of x and y in given equation,

2 + (β2) = 0

0 = 0

Hence, the given statement is true.

(iii) The given line is π¦ β 7 = 0

Here, π₯ = 8 and π¦ = 7

Put the value of x and y in given equation,

7 β 7 = 0

0 = 0

Hence, the given statement is true.

(iv) The given line is π₯ + 3 = 0

Here, π₯ = β3 and π¦ = 0

Put the value of x and y in given equation,

β3 + 3 = 0

0 = 0

Hence, the given statement is true.

(v) The given line is 2π₯ β π¦ = 3

Here, π₯ = 2 and π¦ = π

Put the value of x and y in given equation,

2(2) β π = 3

4 β π = 3

βπ = 3 β 4

βπ = β1

π = 1

Hence, the given statement is false.

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

Here, x = k and y = 6

Put the value of x and y in given equation,

2(k) – 6/3 = 7

2(k) – 2 = 7

2k-2 = 7

2k = 7 + 2

2k = 9

k = 9/2

k = 4.5

Hence, the value of k is 4.5.

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

It is given that, the equation of the line is 9x + 4y = 3.

Here, x =3 and y =-k

Put the value of x and y in given equation,

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

27 β 4k = 3

4k = 27 β 3

4k = 24

k = 24/4

k = 6

Hence, the value of k is 6.

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

It is given that, the equation is 3x/5 – 2y/3+1 = 0.

Here, x=m and y = 2m – 1

Put the value of x and y in given equation,

β11π + 10 = β1 Γ 15

β11π + 10 = β15

β11π = β15 β 10

β11π = β25

11π = 25

π = 25/11

π = 2(3/11)

Hence, the value of m is 2(3/11)

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

It is given that, the line 3x β 5y = 6 bisect the join of (5, -2) and (-1, 2). The co-ordinates of the mid-point of AB are

Put the value of x and y in given equation,

3x β 5y = 6

3(2) β 5(0) = 6

6 β 0 = 6

6 = 6

Hence, the line 3π₯β 5π¦ = 6 bisect the join of (5,-2) and (-1,2).

**Question****7. (i) The line y = 3x β 2 bisects the join of (a, 3) and (2, -5), find the value of a.** **(ii) The line x β 6y + 11 = 0 bisects the join of (8, -1) and (0, k). Find the value of k.Solution:**

(i) It is given that, the line π¦ = 3π₯β 2 bisect the join of (a, 3) and (3, -5). The co-ordinates of the mid-point of AB are

**Question****8. (i) The point (-3, 2) lies on the line ax + 3y + 6 = 0, calculate the value of a.(ii) The line y = mx + 8 contains the point (-4, 4), calculate the value of m.Solution:**

(i) It is given that, the point (-3, 2) lies on the line ax + 3y + 6 = 0.

Here, π₯ = β3 and π¦ = 2

Put the value of x and y in given equation,

π(β3) + 3(2) + 6 = 0

β3π + 12 = 0

3π = 12

π = 4

Hence, the value of a is 4.

(ii) It is given that, the line y = mx + 8 contains the point (-4, 4).

Here, π₯ = β4 and π¦ = 4

Put the value of x and y in given equation,

4 = β4π + 8

4π = 4

π = 1

Hence, the value of m is 1.

**Question****9. The point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3. Does P lie on the line x β 5y + 15 = 0?Solution:**

It is given that, point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3. By section formula,

(π₯, π¦) = 0,3

Here, π₯ = 0 and π¦ = 3

Put the value of x and y in given equation,

π₯β 5π¦ + 15 = 0

0β 5(3) + 15 = 0

0β 15 + 15 = 0

β 15 + 15 = 0

0 = 0

Hence, the point P lies on the line π₯β 5π¦ + 15 = 0.

**Question****10. The line segment joining the points (5,-4) and (2,2) is divided by the point Q in the ratio 1: 2. Does the line x β 2y = 0 contain Q?** **Solution:**It is given that, point P divides the join of (5, -4) and (2, 2) in the ratio 1: 2. By section formula,

(x,y)=4,-2

Here, x=4 and y=-2

Put the value of x and y in given equation,

x β 2y = 0

4 β 2(-2) = 0

4 + 4 = 0

8 β 0

Hence, the point Q does not lie on the line x – 2y = 0.

**Question****11. Find the point of intersection of the lines:4x + 3y = 1 and 3x β y + 9 = 0. If this point lies on the line (2k β 1)x β 2y = 4; find the value of k.**

**Solution:**

It is given that,

4x + 3y = 1β¦..β¦.(1)

3x β y + 9 = 0 β¦β¦β¦(2)

From equation (1) we get, the value of x,

4x + 3y = 1

π₯ = β2

Here, π₯ = β2 and π¦ = 3

Put the value of x and y in given equation,

(2πβ 1)π₯β 2π¦ = 4.

(2πβ 1)(β2)β 2(3) = 4

β4π + 2β 6 = 4

β4π = 8

π = β2

Hence, the value of k is -2.

**Question****12. Show that the lines 2x + 5y = 1,x β 3y = 6 and x + 5y + 2 = 0 are concurrent.** **Solution:**When two or more lines meet at a single location, they are said to be concurrent.

The point of intersection of the first two lines,

2π₯ + 5π¦ = 1 β¦ . β¦ . (1)

π₯β 3π¦ = 6 β¦ β¦ . . (2)

Form equation (2) we get the value of (2),

π₯β 3π¦ = 6

π₯ = 6 + 3π¦ β¦ β¦ β¦ . (3)

Put the value of (3) in equation (1)

2π₯ + 5π¦ = 1

2(6 + 3π¦) + 5π¦ = 1

12 + 6π¦ + 5π¦ = 1

11π¦ = 1 β 12

11π¦ = β11

π¦ = β1

Put the value of y in equation (3)

π₯ = 6 + 3π¦

π₯ = 6 + 3(β1)

π₯ = 6 β 3

π₯ = 3

Hence, the point of intersection is (3, -1).

x + 5y + 2 = 0, then the given lines will be concurrent.

Here, x = 3 and y = -1

Put the value of x and y in equation π₯ + 5π¦ + 2 = 0

π₯ + 5π¦ + 2 = 0

3 + 5(β1) + 2 = 0

5β 5 = 0

0 = 0

(3, -1) also lie on the third line.

Hence, the given lines are concurrent

**Exercise 14B**

**Question****1. Find the slope of the line whose inclination is:(i) 0Β°(ii) 30Β°(iii) 72Β°30β(iv) 46Β° Solution:**

(i) tan 0Β°=0

(ii) tan 30Β°=1/β3

(iii) tan 72Β°30β=3.1716

(iv) tan 46Β°=1.0355

**Question****2. Find the inclination of the line whose slope is:(i) 0(ii) β3(iii) 0.7646(iv) 1.0875**

**Solution:**

(i) ta nβ‘0Β° = 0

Ζ = 0Β°

(ii) tan Ζ = β3

Ζ = 60Β°

(iii) tan Ζ = 0.7646

Ζ = 37Β° 24′

(iv) tan Ζ =1.0875

Ζ = 47Β° 24′

**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:**

**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:**

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

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

Slop of the line perpendicular to AB = -1/-1

Slop of the line perpendicular to AB = -1/-1

Slop of the line perpendicular to AB = 1

**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:**

It is given that, the line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a).

According to question, lines are parallel,

1 = πβ5/5

1 Γ 5 = π β 5

5 = π β 5

5 + 5 = π

π = 10

Hence, the value of a is 10.

**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:**

It is given that, the line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1)

By cross-multiplication,

-1(6) = 6(2 – a)

-6 =12 – 6a

6a = 12 + 6

a =18/6

a =3

Hence, the value of a is 3.

**Question****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:**It is given that, the points are A(4,-2), B(-4,4) and C(10,6) are the vertices of 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:**It is given that, the points A(4, 5), B(1, 2), C(4, 3) and D(7, 6) are the vertices of 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:**

It is given that, A(-2, 4), B(4, 8), C(10, 7) and D(11, -5) are the vertices of a quadrilateral.

Let us assumed that, P, Q, R and S be the mid-points of AB, BC, CD and DA .

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

It is given that, the points P (a, b + c), Q (b, c + a) and R (c, a + b).

**Question****12. Find x, if the slop of the line joining (x,2) and (8,-11) is -3/4. Solution:**

It is given that, the slop of the line joining (x,2) and (8,-11) is (-3)/4.

Let us assumed that, A = (x,2) and B= (8,-11)

**Question****13. The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slope of all its sides.**

Slope of AB = 0. As the slope of any line parallel to x-axis is 0.

ΞABC is an equilateral triangle β A = 60Β°

Slope of AC = tan 60Β° = β3. As the slope of any line parallel to x-axis is 0.

Slope of BC = -tan 60Β° = ββ3. As the slope of any line parallel to x-axis is 0.

**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:**It is given that, the side AB of a square ABCD is parallel to the x-axis.

The slop of any line parallel to x-axis is 0.

Slope of AB = 0

According to figure,

AB | | CD

Slope of AB = Slope of CD = 0

BC is perpendicular to AB,

Slope of BC = β1/πππππ ππ π΄π΅

Slope of BC = β1/0

Slope of BC = not defined

AD is perpendicular to AB,

Slope of AD = β1/πππππ ππ π΄π΅

Slope of AD = β1/0

Slope of AD = not defined

(i) Diagonal of a square is 45Β°, Diagonal AC makes an angle of 45Β° with the positive direction of x-axis.

Slope of AC = tan45Β°

Slope of AC = 1

(ii) Diagonal of a square is 45Β°, Diagonal AC makes an angle ofβ45Β° with the positive direction of x-axis.

Slope of AC = tan (β45Β°)

Slope of AC = β1

**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:**

**Question****16. The slope of the side BC of a rectangle ABCD is 2/3. Find:(i) the slope of the side AB.(i) the slope of the side AD.**

**Solution:**

(i) It is given that, the slope of the side BC of a rectangle ABCD is 2/3.

Here, BC is perpendicular to AB.

**Q17. Find the slope and the inclination of the line AB if:(i) A = (-3,-2) and B = (1,2)(ii) A = (0,-β3) and B = (3,0)(ii) A = (-1,2β3) and B = (-2,β3)**

**Solution:**

(i) Co-ordinates of A (-3,-2) and B (1,2)

**Question****18. The points (-3, 2), (2, -1) and (a, 4) are collinear. Find a.** **Solution:**It is given that, points A(-3, 2), B(2, -1) and C(a, 4) are collinear.

**Question****19. The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K.** **Solution:**It is given that, the points A(K, 3), B(2, -4) and C(-K+1, -2) are collinear.

**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:**

The formula of slope of the line = π¦_{2}βπ¦_{1}/π₯_{2}βπ₯_{1}

Co-ordinates of A(1, 1) and B(4, 7)

Slope of AB = 7β1/4β1

Slope of AB = 6/3

Slope of AB = 2

Co-ordinates of A(1, 1) and C(4, 10)

Slope of AC = 10β1/4β1

Slope of AC = 9/3

Slope of AC = 3

We can say that, the line with greater slope is steeper.

Hence, AC has steeper slope.

**Exercise 14C**

**Question****1. Find the equation of a line whose: y-intercept = 2 and slope = 3.Solution:**

It is given that, y-intercept c = 2 and slope m = 3

Substituting the values of c and m in the equation π¦ = ππ₯ + π

Hence, the required equation is π¦ = 3π₯ + 2

**Question****2. Find the equation of a line whose: y-intercept = -1 and inclination ππΒ°****Solution:**

It is given that, y-intercept c = -1 and inclination = 45Β°

The slope of m is tan 45Β° = 1

Substituting the values of c and m in the equation π¦ = ππ₯ + π

Hence, the required equation is π¦ = π₯ β 1.

**Question****3. Find the equation of the line whose slope isΒ β4/3Β and which passes through (-3, 4).****Solution:**

It is given that, the slope isΒ β4/3.

The equation passes through (-3,4) = (π₯_{1}, π¦_{1})

Substitution the values in π¦ β π¦_{1}Β = π(π₯ β π₯_{1})

π¦ β 4 = β4

3 (π₯ β (β3))

π¦ β 4 = β4

3 (π₯ + 3)

3(π¦ β 4) = β4(π₯ + 3)

3π¦ β 12 = β4π₯ β 12

4π₯ + 3π¦ = 0

Hence, the required equation is 4π₯ + 3π¦ = 0

**Question****4. Find the equation of a line which passes through (5,4) and makes an angle of ππΒ° with the positiveΒ direction of the x-axis.****Solution:**

The slope of the line tan60Β° is β3.

The equation passes through (5,4) = (π₯_{1}, π¦_{1})

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 4 = β3(π₯ β 5)

π¦ β 4 = β3π₯ β 5β3

π¦ = β3π₯ β 5β3 + 4

Hence, the required equation is π¦ = β3π₯ β 5β3 + 4.

**Question****5. Find the equation of the line passing through:****(i) (0, 1) and (1, 2)****(ii) (-1, -4) and (3, 0).****Solution:**

(i) The given co-ordinates of are (0, 1) and (1, 2)

The formula of slope of the line =Β π¦_{2}βπ¦_{1}/π₯_{2}βπ₯_{1}

Slope of the line =Β 2β1/1β0

Slope of the line =Β 1/1

Slope of the line = 1

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 1 = 1(π₯ β 0)

π¦ β 1 = π₯

π¦ = π₯ + 1

Hence, the required equation is π¦ = π₯ + 1.

(ii) The given co-ordinates of are (-1, -4) and (3, 0)

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β (β4) = 1(π₯ β (β1))

π¦ + 4 = 1(π₯ + 1)

π¦ + 4 = π₯ + 1

π¦ = π₯ β 3

Hence, the required equation is π¦ = π₯ β 3

**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:**

(i) It is given that, the co-ordinates of two points P and Q are (2, 6) and (-3, 5)

(ii) The equation of PQ,

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 6 = 1

5 (π₯ β 2)

5(π¦ β 6) = π₯ β 2

5π¦ β 30 = π₯ β 2

5π¦ β π₯ = β2 + 30

5π¦ β π₯ = 28

5π¦ = 28 + π₯

Hence, the required equation is 5π¦ = 28 + π₯.

(iii) Let us assumed that, the co-ordinates of the point where PQ intersects the x-axis at point π΄(π₯, 0).

Put the value of π¦ = 0 in the above equation.

5π¦ = 28 + π₯

5(0) = 28 + π₯

0 = 28 + π₯

π₯ = β28

Hence, the value of x is -28.

**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 the y-axis.****Solution:**

(i) It is given that, co-ordinates of two points A and B are (-3,4) and (2,-1).

The formula of slope of the line =Β π¦_{2}βπ¦_{1}/π₯_{2}βπ₯_{1}

Slope of the line =Β β1β4/2+3

Slope of the line =Β β5/5

Slope of the line = β1

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β (β1) = β1(π₯ β 2)

π¦ + 1 = β1(π₯ β 2)

π¦ + 1 = βπ₯ + 2

π¦ + 1 = βπ₯ + 2

π₯ + π¦ = 2 β 1

π₯ + π¦ = 1

(ii) Let us assumed that, the co-ordinates of the point where AB intersects the y-axis at point π΄(0, π¦).

Put the value of π₯ = 0 in the above equation.

π₯ + π¦ = 1

0 + π¦ = 1

π¦ = 1

Hence, the value of y is 1.

**Question****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:**It is given that, the angle is 45Β°

Slope of line AB = π‘ππ 45Β°

Slope of line AB = 1

The line AB passes through P(3,4).

Substitution the values in π¦ β π¦

_{1}= π(π₯ β π₯

_{1})

π¦ β 4 = 1(π₯ β 3)

π¦ β 4 = π₯ β 3

π¦ = π₯ β 3 + 4

π¦ = π₯ + 1

Again,

The angle is 60Β°

Slope of line CD = π‘ππ 60Β°

Slope of line CD = β3

The line AB passes through P(3,4).

Substitution the values in π¦ β π¦

_{1}= π(π₯ β π₯

_{1})

π¦ β 4 = β3(π₯ β 3)

π¦ β 4 = π₯β3 β 3β3

π¦ = π₯β3 β 3β3 + 4

**Question****9. In ΞABC,Β AΒ = (π, π),Β BΒ = (7,Β 8) and πͺ = (1, β10). Find the equation of the median through A.Solution:**

It is given that, the vertices of ΞABC,Β AΒ = (3,Β 5),Β BΒ = (7,Β 8) andΒ CΒ = (1, βΒ 10).

Slope of the line = β6

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 5 = β6(π₯ β 3)

π¦ β 5 = β6π₯ + 18

6π₯ + π¦ = 18 + 5

6π₯ + π¦ = 18 + 5

6π₯ + π¦ = 23

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

**Solution:**It is given that, ABCD is a parallelogram.

β A + β B = 180Β°

60Β° + β B = 180Β°

β B = 180Β° β 60Β°

β B = 120Β°

Slope of line CD = tan 60Β°

Slope of line CD = tan (90Β° + 30Β°)

Slope of line CD = cot 30Β°

Slope of line CD = β3

Substitution the values in π¦ β π¦

_{1}= π(π₯ β π₯

_{1})

π¦ β 5 = β3(π₯ β 7)

π¦ β 5 = β3π₯ β 7β3

π¦ = β3π₯ β 7β3 + 5

It is given that, AB is parallel to the x-axis.

CD||AB

Slope of line AB = 0

Slope of line AB = Slope of line CD

Substitution the values in π¦ β π¦

_{1}= m(π₯ β π₯

_{1})

π¦ β 5 = 0(π₯ β 7)

π¦ β 5 = 0

π¦ = 5

**Question****11. Find the equation of the straight line passing through origin and the point of intersection of the lines XΒ + 2Y =Β 7 and XβΒ Y = 4.Solution:**

It is given that, the point of intersection of the lines π₯ + 2π¦ = 7 and π₯β π¦ = 4.

π₯ + 2π¦ = 7 β¦ β¦ β¦ . (π)

π₯β π¦ = 4 β¦ β¦ β¦ . . (ππ)

From equation (ii) get the value of x,

π₯ = 4 + π¦ β¦ β¦ β¦ . . (πππ)

Put the value of x in equation (i)

(4 + π¦) + 2π¦ = 7

4 + π¦ + 2π¦ = 7

3π¦ = 3

π¦ = 3/3

π¦ = 1

Put the value of y in equation (iii)

π₯ = 4 + 1

π₯ = 5

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

The formula of slope of the line =Β π¦

_{2}βπ¦

_{1}/π₯

_{2}βπ₯

_{1}

Slope of the line =Β 1β0/5β0

Slope of the line =Β 1/5

Substitution the values in π¦ β π¦

_{1}= π(π₯ β π₯

_{1})

π¦ β 0 = 1/5 (π₯ β 0)

5π¦ = π₯

5π¦ = π₯

0 = π₯ β 5π¦

π₯ β 5π¦ = 0

**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:**

It is given that, the co-ordinated of Ξπ΄π΅πΆ are (4,7), (-2,3) and (0,1).

Let us assumed that, AD is the median through vertex A

Slope of the line AD = 1

Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)

π¦ β 2 = 1(π₯ β (β1))

π¦ β 2 = 1(π₯ + 1)

π¦ β 2 = π₯ + 1

π¦ β π₯ = 1 + 2

π¦ = 3 + π₯

Slope of the line AC = 1β7/0β4

Slope of the line AC = β6/β4

Slope of the line AC = 3/2

Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)

π¦ β 3 = 3/2 (π₯ β (β2))

2(π¦ β 3) = 3(π₯ + 2)

2π¦ β 6 = 3π₯ + 6

2π¦ β 3π₯ = 6 + 6

2π¦ = 3π₯ + 12

**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:**

It is given that, The co-ordinates A(0, 3), B(4, 4) and C(8, 0).

Slope of line perpendicular to BC = 1

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 3 = 1(π₯ β 0)

π¦ = π₯ + 3

Hence, the required equation is π¦ = π₯ + 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:**

Let us assumed that, the co-ordinates are A(1, 4), B(2,3) and C(-1,2)

Slope of line perpendicular to BC = 1

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 2 = 1(π₯ β (β1))

π¦ β 2 = (π₯ + 1)

π¦ β 2 = π₯ + 1

π¦ = π₯ + 1 + 2

π¦ = π₯ + 3

Hence, the required equation is π¦ = π₯ + 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 = -4Solution:**

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

x-intercept = 5 means corresponding point on x-axis is (5,0)

y-intercept = 3 means corresponding point on y-axis is (0,3)

5π¦ = β3(π₯ β 5)

5π¦ = β3π₯ + 15

5π¦ + 3π₯ = 15

Hence, the required equation is 5π¦ + 3π₯ = 15

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

x-intercept = -4 means corresponding point on x-axis is (-4,0)

y-intercept = 6 means corresponding point on y-axis is (0,6)

Slope of the line = 6/4

Slope of the line = 3/2

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 0 = 3

2 (π₯ β (β4))

π¦ = 3

2 (π₯ + 4)

2π¦ = 3π₯ + 12

Hence, the required equation is 2π¦ = 3π₯ + 12

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

x-intercept = -8 means corresponding point on x-axis is (-8,0)

y-intercept = -4 means corresponding point on y-axis is (0,-4)

**Question****16. Find the equation of the line whose slope is β5/6 and x-intercept is 6.Solution:**

It is given that, x-intercept = 6 means corresponding point on x-axis is (6, 0).

Slope of the line = β5/6

Substitution the values in π¦ β π¦

_{1}= π(π₯ β π₯

_{1})

π¦ β 0 = β5

6 (π₯ β 6)

6π¦ = β5(π₯ β 6)

6π¦ = β5π₯ + 30

6π¦ + 5π₯ = 30

**Question****17. Find the equation of the line with x-intercept 5 and a point on it (-3, 2).Solution:**

It is given that, x-intercept = 5 means corresponding point on x-axis is (-3, 2).

Slope of the line = 2/β8

Slope of the line = β1/4

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 0 = β1

4 (π₯ β 5)

4π¦ = β1(π₯ β 5)

4π¦ = βπ₯ + 5

4π¦ + π₯ = 5

Hence, the required equation is 4π¦ + π₯ = 5

**Question****18. Find the equation of the line through (1, 3) and making an intercept of 5 on the y-axis.Solution:**

It is given that, y-intercept = 5 means corresponding point on y-axis is (0, 5).

Slope of the line = β2/1

Slope of the line = β2

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 5 = β2(π₯ β 0)

π¦ β 5 = β2(π₯)

π¦ β 5 = β2π₯

π¦ + 2π₯ = 5

Hence, the required equation is π¦ + 2π₯ = 5

**Question****19. Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axis.****Solution:**

Let us assumed that, AB and CD be two equally inclined lines.

Slope of line AB,

π‘ππ 45Β° = 1

(π₯_{1}, π¦_{1}) = (β2,0)

Substitution the values in π¦ β π¦_{1}Β = π(π₯ β π₯1)

π¦ β 0 = 1(π₯ β (β2))

π¦ = 1(π₯ + 2)

π¦ = π₯ + 2

Let us assumed that, AB and CD be two equally inclined lines.

Slope of line AB,

π‘ππ β 45Β° = β1

(π₯_{1}, π¦_{1}) = (β2,0)

Substitution the values in π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 0 = β1(π₯ + 2)

π¦ = β1(π₯ + 2)

π¦ = βπ₯ β 2

π¦ + π₯ + 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) The equation of the y-axis is x = 0

It is given that the required line through P(5,3) intersects the y-axis at Q and the angle of inclination is 45Λ.

Hence, the slope of line PQ = π‘ππ 45Β° = 1

(ii) The equation of a line passing through the point A(π₯1, π¦1) with slope βmβ is

The equation of the line passing through the point P(5,3) with slope 1 is

(π₯1, π¦1) = (5,3)

Substitution the values in π¦ β π¦1 = π(π₯ β π₯1)

π¦ β 3 = 1(π₯ β 5)

π¦ β 3 = π₯ β 5

π¦ β π₯ = β5 + 3

π¦ β π₯ = β2

π₯ β π¦ = 2

(iii) The equation of the line PQ is π₯ β π¦ = 2

It is given that, the line intersects with y-axis π₯ = 0.

Substituting π₯ = 0 in the equation π₯ β π¦ = 2

Put the value of π₯

0 β π¦ = 2

π¦ = β2

Hence, the co-ordinates points of intersection Q are (0.-2).

**Exercise 14D**

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

(i) y = 4

(ii) axΒ βby = 0

(iii) 3xΒ –Β 4y = 5

Solution:

(i) π¦ = 4

Substituting this equation with π¦ = ππ₯ + π

Slope of line = 0

Put the value of y in the above equation,

4 = 0π₯ + π

4 = π

π = 4

y-intercept = c = 4

(ii) ππ₯β ππ¦ = 0

ππ₯ = ππ¦

π/π π₯ = π¦

π¦ = (π/π)π₯

Substituting this equation with π¦ = ππ₯ + π

Slope of line =Β π/π

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

It is given that, the equation of a line is π₯ β π¦ = 4

βπ¦ = 4 β π₯

π¦ = π₯ β 4

Substituting this equation with π¦ = ππ₯ + π

Slope of line = 1

Slope of line = 1

y-intercept = c = -4

Let us assumed that, the inclination be Ζ

Slope of line = 1

π‘ππΖ = π‘ππ45Λ

Ζ = 45Λ

Hence, the inclination us 45Λ.

**Question****3. (i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x β 21y + 50 = 0?(ii) Is the line x β 3y = 4 perpendicular to the line 3x β 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) It is given that, 3π₯ + 4π¦ + 7 = 0

3π₯ + 4π¦ + 7 = 0

4π¦ = β7 β 3π₯

Slope of the line is 4/3

Hence, the product of the slope of the two lines is -1, the line are perpendicular to each other.

(ii) It is given that, π₯ β 3π¦ = 4

β3π¦ = β4 β π₯

β3π¦ = β(4 + π₯)

3π¦ = π₯ + 4

π¦ = 1/3 π₯ + 4/3

Slope of the line is 1/3

3π₯β π¦ = 7

β π¦ = β3π₯ β 7

π¦ = 3π₯ + 7

Slope of the line = 3

Product of slope of the two lines 11 β β1

Hence, the lines are not perpendicular to each other.

Slope of the line = β 1/2

Product of slope of the two lines 3 β β1

Hence, the lines are not perpendicular to each other.

(iv) It is given that, the slope of the line by (1,4) and (x,2) is 2.

By cross-multiplication,

2(π₯ β 1) = β2

π₯ β 1 = β1

π₯ = β1 + 1

π₯ = 0

**Question****4. Find the slope of the line which is parallel to:(i) x + 2y + 3 = 0**

**Solution.**

(i) x + 2y + 3 = 0

2y = -x – 3

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

**Solution:**

By cross-multiplication,

y = 2(x + 3)

y = 2x + 6

Slope of the line = 2

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

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

**Question****6. (i) Lines 2x β 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 β ny β 3 = 0 are perpendicular to each other. Find the relation connecting m and n.** **Solution:**

(i) The given equation is 2x-by+3=0

2x + 3 = by

by = 2x + 3

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

The given equation is 2π₯ β π¦ + 5 = 0

βπ¦ = β7 β 2π₯

π¦ = 2π₯ + 7

Slope of the line = 2

ππ₯ + 3π¦ = 4

3π¦ = 4 β ππ₯

3π¦ = βππ₯ + 4

**Question****8. The equation of a line AB is 2x β 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) The given equation is 2π₯β 2π¦ + 3 = 0

β 2π¦ = β2π₯ β 3

2π¦ = 2π₯ + 3

π¦ = π₯ + 3/2

Slope of the line = 1

(ii) Let the required angle be Ζ

Slope of tan Ζ = 1

tan Ζ = tan 45Λ

Ζ = 45Λ

Hence, the value of Ζ is 45Λ

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

The given equation is 4π₯ + 3π¦ = 9

3π¦ = β4π₯ + 9

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

The given equation is π¦ = 3π₯ + 7

Slope of the line = 3

again,

2π¦ + ππ₯ = 3

2π¦ = βππ₯ + 3

**Question****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:**

It is given that, the line through A(-2,3) and B(4,b) is perpendicular to the line 2x – 4y = 5

**Question****12. Find the equation of the line through (-5, 7) and parallel to:****(i) x-axis****(ii) y-axis****Solution:**

(i) The slope of the line parallel to x-axis is 0.

(π₯_{1}, π¦_{1}) = (β5, 7)

Substituting this equation with π¦ β π¦_{1}Β = π(π₯ β π₯_{1})

π¦ β 7 = π(π₯ β (β5))

π¦ β 7 = π(π₯ + 5)

π¦ β 7 = 0(π₯ + 5)

π¦ β 7 = 0

π¦ = 7

Hence, the value of y is 7.

(ii) Here, the slope of the line parallel to y-axis in not defined. The slope of the line is tan90Λ and hence the given

line is parallel to y-axis.

(π₯_{1}, π¦_{1}) = (β5, 7)

Required equation of the line is

π₯ β π₯_{1}Β = 0

π₯ β (β5) = 0

π₯ + 5 = 0

**Question****13. (i) Find the equation of the line passing through (5, -3) and parallel toΒ xΒ βΒ 3yΒ =Β 4.(ii) Find the equation of the line parallel to the lineΒ 3x + 2y = 8 and passing through the point (0, 1).Solution:**

(i) π₯ β 3π¦ = 4

β3π¦ = 4 β π₯

β3π¦ = β(β4 + π₯)

3π¦ = π₯ β 4

Slope of the line = 1/3

Required equation of the line passing through (5,-3)

Substituting this equation with π¦ β π¦_{1} = π(π₯ β π₯_{1})

3(π¦ + 3) = π₯ β 5

3π¦ + 9 = π₯ β 5

3π¦ β π₯ = β5 β 9

3π¦ β π₯ = β14

β(π₯ β 3π¦) = β14

π₯ β 3π¦ = 14

π₯ β 3π¦ β 14 = 0

(ii) 3π₯ + 2π¦ = 8

2π¦ = 8 β 3π₯

2π¦ = β3π₯ + 8

**Question****14. Find the equation of the line passing through (-2, 1) and perpendicular to ππ + ππ = π.Solution:**

The given equation is 4π₯ + 5π¦ = 6

4π₯ + 5π¦ = 6

5π¦ = β4π₯ + 6

4(π¦ β 1) = 5(π₯ + 2)

4π¦ β 4 = 5π₯ + 10

4π¦ β 5π₯ = 14

5π₯ β 4π¦ = β14

5π₯ β 4π¦ + 14 = 0

**Question****15. Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).Solution:**

Let the A(6,3) and B(0,3)

The perpendicular bisector of a line is perpendicular to the line and it bisects the line, that it, it passes through the mid-point of the line.

**Question****16. In the following diagram, write down:(i) the co-ordinates of the points A, B and C.(ii) the equation of the line through A and parallel to BC.**

**Solution:**(i) The co-ordinates of point A, B and C are (2,3), (-1,2) and (3,0) respectively.

The equation of the line is π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 3 = β1/2 (π₯ β 2)

2(π¦ β 3) = β1(π₯ β 2)

2π¦ β 6 = βπ₯ + 2

π₯ + 2π¦ = 2 + 6

π₯ + 2π¦ = 8

Hence, the required line is π₯ + 2π¦ = 8.

**Question****17. B(-5, 6) and D(1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.Solution:**

We know that, in a rhombus diagonal bisect each other at right angle. Let O be the point of intersection of the diagonals AC and BD.

The equation of the line is π¦ β π¦_{1} = π(π₯ β π₯_{1})

π¦ β 5 = 3(π₯ β (β2))

π¦ β 5 = 3(π₯ + 2)

π¦ β 5 = 3π₯ + 6

π¦ = 3π₯ + 6 + 5

π¦ = 3π₯ + 11

**Question****18. A = (7, -2) and C = (-1, -6) are the vertices of square ABCD. Find the equations of diagonal BD and of diagonal AC.Solution:**

In a square, diagonals bisect each other at right angle. Let O be the point of intersection of the diagonals AC and BD.

Slope of BD = β2

The equation of the line is π¦ β π¦1 = π(π₯ β π₯1)

π¦ β (β4) = β2(π₯ β 3)

π¦ + 4 = β2(π₯ β 3)

π¦ + 4 = β2π₯ + 6

π¦ + 2π₯ = 6 β 4

2π₯ + π¦ = 2

**Question****19. A (1, -5), B (2, 2) and C (-2, 4) are the vertices of triangle ABC, find the equation of:(i) the median of the triangle through A.(ii) the altitude of the triangle through B.(iii) the line through C and parallel to AB.Solution:**

(i) The median through A will pass through the mid-point of BC. Let AD be the median through A.

Slope of AB = β7

Slope of the line parallel to AB = Slope of AB = 7

The equation of the line is π¦ β π¦_{1} = π(π₯ β x_{1})

π¦ β 4 = 7(π₯ β (β2))

π¦ β 4 = 7(π₯ + 2)

π¦ β 4 = 7π₯ + 14

π¦ β 7π₯ = 14 + 4

π¦ β 7π₯ = 18

Hence, the required equation is π¦ β 7π₯ = 18.

**Question****20. (i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin.Solution:**

(i) The given equation is 2π¦ = 3π₯ + 5

**Question****21. Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5. Find its coordinates of point P. Also, find the equation of the line through P and parallel to 3x + 5y = 7.Solution:**

It is given that, Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5.