# NCERT Solutions for Class 11 Maths

Find 100% accurate solutions for NCERT Class XI Maths. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!

1.    Let A(-4, 5), B(0, 7), C(5, -5) and D(-4, -2) be the vertices of the quadrilateral

Join the vertices A and C to obtain the diagonal AC

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD………………..(i)

But we have area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is

2.    Given: Length of side of equilateral triangle = 2a. The base of triangle lies along y-axis and the mid-point of base is at origin so that the coordinates of vertices are (0, a) and (0, -a).

We have for an equilateral triangle line from the vertex to the mid point of the base will be perpendicular to the base.So we can say the third vertex lies on X axis (positive or negative)

Now let the third vertex be (±x, 0).

It is known that area of an equilateral triangle = , where a is the common length of the sides

∴ Area of the equilateral triangle with side

Therefore the third vertex can be (√3a, 0) or (-√3a, 0)

∴ The vertices of triangle are (0, a), (0, -a) and (√3a, 0) or (0, a), (0, -a) and (-√3a, 0).

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

(i) We have equation of a line parallel to y-axis is X = K, a constant

PQ is parallel to y-axis, then x2 = x1 ⇒ x2 – x1 = 0

(ii) We have equation of a line parallel to x-axis is Y = K, a constant

PQ is parallel to x-axis, then y2 – y1 = 0,

4.    Let P be any point on the x-axis which is equidistant from Q (7, 6) and R (3, 4).

We have distance between two points (x1, y1) and (x2, y2) is

According to question, PQ = PR

Squaring both sides, x2 – 14x + 85 = x2 – 6x + 25

Therefore, the required point is

5.    Here, mid-point of the line segment joining P (0, -4) and Q (8, 0) is

using mid point formula

We have slope of a line passing through two points (x1, y1) and (x2, y2) is given by

Therefore, slope of the line passing through the points (0, 0) and (4, -2) =

Hence, slope of the required line is -1/2.

6.    Let A (4, 4), B (3, 5) and (-1, -1) C be the three vertices of ΔABC.

We have slope of a line passing through two points (x1, y1) and (x2, y2) is given by

Now, slope of AB × Slope of AC = -1 × 1 = -1

This shows that AB ⊥ AC. Thus ΔABC is right angled at point A.

7.

If the line makes an angle of 30° with the positive direction of y-axis then the line will make an angle of (90° + 30°) = 120° with the positive direction of x-axis.

∴ Slope of the line = tan 120° = tan(90° + 30°) = -cot 30° = -√3.

8.    Let A (x, -1), B (2, 1) and C (4, 5) be three collinear points.

∴ Slope of

Slope of BC

According to question, Slope of AB = Slope of BC

9.    Let A (-2, -1), B (4, 0), C (3, 3) and D (-3, 2) be vertices of a quadrilateral ABCD.

To prove a quadrilateral is a parallelogram it is enough to show that both pairs of opposite sides are parallel

We have slope of a line passing through two points (x1, y1) and (x2, y2) is given by

Here Slope of AB = Slope of DC

⇒ AB || DC

And Slope of BC = Slope of AD

Therefore, ABCD is a parallelogram.

10.   Let A (3, -1) and B(4 -2) be two points. Let θ be the angle which AB makes with positive direction of x-axis.

∴ Slope of AB = tan θ……………..(i)

We have slope of a line passing through two points (x1, y1) and (x2, y2) is given by

∴ Slope of AB

From (i) and (ii) we get tan θ = -1

⇒ tan θ = – tan 45°

⇒ tan θ = tan (180° – 45°)

⇒ tan θ = tan 135°

⇒ θ = 135°

11.   Given:   Let the slopes of two lines be m and 2m.

We know that if θ is the acute angle between two lines with slopes m1 and m2 respectively then

When m = -1 the slopes of the lines are -1 and -2

when m = the slopes of the lines are   and -1

when m = 1 the slopes of the lines are 1 and 2

when m =  the slopes of the lines are  and 1

Therefore, the slopes of lines are -1 and -2 or 1 and -1 and  or 1 and

12.   Let A(x1, y1) and B, (h, k) be two points. It is given that Slope of AB = m

We have slope of a line passing through two points (x1, y1) and (x2, y2) is given by

13.  Let A (h, 0), B(a, b) and C(0, k) be three points lie on the line.

We have slope of a line passing through two points (x1, y1) and (x2, y2) is given by

Slope of AB = Slope of BC (give)

14.   Given: The points on the line are A (1985, 92) and B (1995, 97).

We have slope of a line passing through two points (x1, y1) and (x2, y2) is given by

Slope of AB

Let the population in year 2010 be y crores. Then C(2010, y) lies on the line AB.

Slope of BC

Since points A, B and C lie on the line(collinear points),we have Slope of AB = Slope of BC

Therefore, population in 2010 will be 104.5 crores.

15.   Equation for x-axis is y = 0 since the y-coordinate of every point on the x-axis is 0.

Equation for y-axis is x = 0 since the x-coordinate of every point on the y-axis is 0.

16.  We have the equation of a line passing through (x0, y0) and slope m is y – y0 = m(x – x0)

Given: x0 = -4, y0 = 3 and m =

∴ Equation of required line is

⇒ 2y – 6 = x + 4

⇒ x – 2y + 10 = 0.

17.   We have the equation of a line passing through (x0, y0) and slope m is y – y0 = m(x – x0)

Given: x0 = 0, y0 = 0 and slope = m

∴ Equation of the required line is y – 0 = m(x – 0) ⇒ y = mx.

18.   We have the equation of a line passing through (x0, y0) and slope m is y – y0 = m(x – x0)

Given: x0 = 0, y0 = 2√3 and θ = 75°

Now, m = tan 75°

19.   We have if a line with slope m makes x-intercept d, then equation of the line is y = m(x – d)

Given: m = -2 and d = -3

∴ Required line is y = -2(x + 3)

⇒ 2x + y + 6 = 0.

20.  We have if a line with slope m males y-intercept c, then equation of the line is y = mx + c.

Given:  and c = 2

∴ Equation of the required line is

21.   We have the equation of a line passing through the points (x1, y1) and (x2, y2) is

Given: (x1, y1) = (-1, 1) and (x2, y2) = (2, -4)

∴ Equation of the required line is

22.   If the length of the perpendicular from origin to the line is p and the angle made by the perpendicular with the positive direction of x-axis is , then the equation of the line is x cosω + y sinω = p

Given: p = 5 and ω = 30°

∴ Equation of the required line is x cos 30° + y sin 30° = 5

23.  Given: P (2, 1), Q(-2, 3) and R (4, 5) are the vertices of ΔPQR. Let RS be the median through vertex R. Then S is the mid-point of PQ.

∴ Using midpoint formula coordinates of S are

We have the equation of a line passing through the points (x1, y1) and (x2, y2) is

∴ Equation of required median RS is

24.  Slope of the line passing through te points A(2,5) and B(-3, 6) is

Since, the required line is perpendicular to AB, therefore slope of required line m = 5

Equation of the required line passing through point (-3, 5) having slope 5 is

y – y0 = m(x – x0)

⇒ y – 5 = 5(x + 3)

⇒ y – 5 = 5x + 15

⇒ 5x – y + 20 = 0.

25.  Let point C divides the join of A (1, 0) and B (2, 3) in the ratio 1 : n.

Now using section formula we have coordinates of C are

Since, the required line is perpendicular to AB, slope of required line

We have the equation of a line passing through (x0, y0) and slope m is y – y0 = m(x – x0)

∴ The equation of the line passing through point  having slope  is

26.  Let equal intercepts on the coordinate axes be α

∴ Equation of the line in intercept form is

Since this line passes through point (2, 3), we get α = 2 + 3 = 5

Therefore, the equation of required line is

⇒ x + y = 5.

27.   Let the intercepts be a and b.

Given a + b = 9 ⇒ x = 9 – b

Equation of the line in intercept form is

Since this line passes through (2,2), we get

28.  Given:

We have the equation of a line passing through (x0, y0) and slope m is y – y0 = m(x – x0)

∴ Equation of the line passing through point (0, 2) having slope -√3 is

Now we have to find the equation of the line parallel to √3x + y – 2 = 0 and crossing the y-axis at (0,-2)

Since parallel lines have same slope, slope of required line = -√3

Therefore, the equation of required line is y – y0 = m(x – x0)

⇒ y – (-2) = -√3 (x – 0)

⇒ y + 2 = -√3x

⇒ √3x + y + 2 = 0.

29.  Let O(0,0) be the origin and P be the given point (-2,9)

Now slope of the line OP =

Since the required line is perpendicular to OP, it’s Slope =

We have the equation of a line passing through (x0, y0) and slope m is y – y0 = m(x – x0)

∴ Equation of the required line is

30.  According to Question, when C=20, the value of L is 124.942 and when C=110, the value of L is 125.134

So the point (20, 124.942) and (110, 125.134) ll satisfy the linear function of L and C.

Considering C along x-axis and L along y axis, we have two points stated above in XY-plane. .i.e. (20, 124.942) and (110, 125.134)

Linear equation of L and C ll be the line passing through these two points.

31.  Let x be the demand in litres and y be the selling price per litre

Now the linear relationship between demand and selling price is the equation of the line passing through the points (980, 14) and (1220, 16)

We have the equation of a line passing through the points (x1 – y1) and (x2 – y2) is

Here, (x1, y1) = (980, 14) and (x2, y2) = (1220, 16)

∴ Linear relation relationship between selling price and demand is given by

Hence the owner could sell 1340 litres of milk weekly at rupees 17 per litre.

32.  Let A(x, 0) and B(0, y) be two points where the line intersect x and y-axis respectively and P(a, b) be the mid-point of AB.

Hence the coordinates of A and B are (2a,0) and (0,2b) respectively

We have the equation of a line passing through the points (x1, y1) and (x2, y2) is

On dividing both sides by ab, we get

Hence proved.

33.  Let A (x, 0) and B(0, y) be two points where the line intersect x and y-axes respectively and let R(h, k) be the point that divides AB in the ratio 1 : 2.

Therefore coordinates of A and B are (, 0) and (0, 3k)

Hence we have the intercepts on the coordinate axes are  and 3k respectively

∴ Using intercept form equation of the required line is

34.  We have the equation of a line passing through the points (x1, y1) and (x2, y2) is

Here, (x1, y1) = (3, 0) and (x2, y2) = (-2, -2)

Therefore equation of a line joining the points (3, 0) and (-2, -2) is

Putting the coordinates of third point in in the above equation , we have 2 × 8 – 5 × 2 = 6

⇒ 16 – 10 = 6

⇒ 6 = 6

Therefore,line joining the points (3,0) and (-2, -2) also passes through the point (8,2)

Hence we have the given points are collinear.

35.  (i) Given: x + 7y = 0

Comparing with y = mx + c, we have  and c = 0

(ii) Given: 6x + 3y – 5 = 0

Comparing with y = mx + c, we have m = -2 and

(iii) Given: y = 0

⇒ y = 0x + 0     …(i)

Comparing with y = mx + c, we have m = 0 and c = 0.

36.  (i) Given: 3x + 2y – 12 = 0

Comparing with  we have a = 4 and b = 6

(ii) Given: 4x – 3y = 6

Comparing with  we have  and b = -2

(iii) Given: 3y + 2 = 0

Comparing with  we have  and no intercept with x-axis.

37.  (i) Given: x – √3y + 8 = 0

⇒ x + √3y = 8

Dividing both sides by we have

∴ Equation of line in normal form is

Comparing with x cos α + y sin α = p, we have  and p = 4

(ii) Given: y – 2 = 0 ⇒ y = 2

⇒ 0x + y = 2

Dividing both sides by  we have 0x + y = 2

Putting cos α = 0 and sin α = 1

∴ Equation of line in normal form is

Comparing with x cos α + y sin α = p, we have  and p = 2

(iii) Given: x – y = 4

Dividing both sides by  we have

∴ Equation of line in normal form is

Comparing with x cos α + y sin α = p, we have  and p = 2√2.

38.  Given equation of the line is 12(x + 6) = 5 (y – 2)

⇒ 12x + 72 = 5y – 10

⇒ 12x – 5y + 82 0

We have the distance of a given point (x1, y1) from a given line Ax + BY + C = 0 is

∴ Perpendicular distance of the point (-1, 1) from the line 12x – 5y + 82 = 0 is

39.  Let the coordinates of the point on x-axis be (α, 0).

Given line is

We have the distance of a given point (x1, y1) from a given line Ax + BY + C = 0 is

Therefore perpendicular distance of the point (α, 0) from the line 4x + 3y – 12 = 0 is

Therefore, the points on x-axis are (8, 0) and (-2, 0).

40.  (i) Given: Two equations 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

Here, a = 15, b = 8, c1 = -34 and c2 = 31

Distance between two parallel lines

units

(ii) Given: Two equations lx + ly + p = 0 and lx + ly – r = 0

Here, a = l, b = l, c1 = p and c2 = -r

Distance between two parallel lines

41.  We have equation of any line parallel to Ax+BY+C=0 is of the form Ax+By+K=0, K is a constant

∴  Equation of a line which is parallel to the line 3x – 4y + 2 = 0 is 3x – 4y + k = 0.

Since the line passes through the point (-2, 3).

We have 3 × (-2) – 4 × 3 + k = 0

⇒ -6 -12 + k = 0

⇒ k = 18

Therefore, the equation of required line is 3x – 4y + 18 = 0.

42.  We have equation of any line parallel to Ax+BY+C=0 is of the form Bx-Ay+K=0, K is a constant

∴  Equation of a line which is parallel to the line x – 7y + 5 = 0 is 7x + y + k = 0.

Since the line passes through the point (3, 0).

We have 7 × 3 + 0 + k = 0

⇒ 21 + k = 0

⇒ k = -21

Therefore, the equation of required line is 7x + y – 21 = 0.

43.  Given: √3x + y = 1

An acute angle θ between the two lines is given by

44.  Slope of the line passing through the points (h, 3) and (4, 1) =

Also slope of the line is 7x – 9y – 19 = 0 is

We have if two lines are perpendicular to each other then products of their slopes = -1

45.  Equation of any line parallel to the line Ax + By + C = 0 is  Ax + By + K = 0   …(i)

Since line (i) passes through (x1, y1), we get Ax1 + By1 + K = 0  …(ii)

Subtracting eq. (ii) from eq. (i), we have A (x – x1) + B (y – y1) = 0

46.  Given: m1 = 2 and θ = 60°

An acute angle θ between the two lines is given by

47.  Let A(3, 4) and B(-1, 2) be the given points

Midpoint of the line segment joining the points A and B =  = (1, 3)

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

Since the required line is perpendicular to the line AB,slope of the required line is -2

We have the equation of a line passing through (x0, y0) and slope m is y – y0 = m(x – x0)

Since the required line passes through point (1, 3) and having slope -2.

We have ,equation of the required line is y – 3 = -2 (x – 1)

⇒ y – 3 = -2x + 2

⇒ 2x + y – 5 = 0.

48.  Let Q be the foot of perpendicular drawn from P(-1, 3) on the line 3x – 4y – 16 = 0

∴ Equation of a line perpendicular to 3x – 4y – 16 = 0 is 4x + 3y + k = 0

Since the line passes through (-1, 3)

∴ 4 × (-1) + 3 × 3 + k = 0

⇒ -4 + 9 + k = 0

⇒ k = -5.

Therefore, Q is a point of intersection of the lines 3x – 4y – 16 = 0 and 4x + 3y – 5 = 0

Solving both the equations, we have

Therefore, coordinates of foot of perpendicular are

49.  Given equation of line is y = mx + c     …(i)

Since the perpendicular from the origin meets the above line (i) at (-1, 2),we have the line joining (0, 0) and (-1, 2) is perpendicular to the line (i)

Now slope of line joining (0, 0) and

and slope of given line = m

But we have if two lines are perpendicular the product of their slopes is equal to -1.

Since the point (-1, 2) lies on the line(i) we get

50.  We have the length of the perpendicular from of a given point (x1, y1) to a given line

Length of perpendicular from origin to line x cos θ – y sin θ – k cos 2θ = 0 is

And Length of perpendicular from origin to line x sec θ + y cos ec θ – k = 0 is

51.  Slope of BC

Since AD ⊥ BC, therefore slope of AD = 1

We have the equation of a line passing through (x0, y0) and slope m is y – y0 = m(x – x0)

∴ Equation of altitude AD is

y – 3 = 1(x – 2)

⇒ x – y + 1 = 0

And Equation of BC is

y + 1 = -1(x – 4)

⇒ x + y – 3 = 0

We have the length of the perpendicular from of a given point (x1, y1) to a given line

∴ Length of AD = perpendicular distance from (2, 3) to the line

∴ Equation of the altitude from A is x – y + 1 = 0 and it’s length = √2 units.

52.  Given: Line

⇒ bx + ay = ab

⇒ bx + ay – ab = 0

Now, is the length of perpendicular from origin to bx + ay – ab = 0.

We have the length of the perpendicular from of a given point (x1, y1) to a given line

53.  Given: Equation of line (k – 3)x – (4 – k2)y + k2– 7k + 6 = 0 can be written as

(4 – k2)y = (k – 3)x + k2 – 7k + 6

(a) If the line parallel to x-axis, then m = 0

(b) If the line parallel to y-axis, then

(c) If the line passes through origin then

⇒ (k – 3) × 0 – (4 – k2) × 0 + k2 – 7k + 6 = 0

⇒ k2 – 7k + 6 = 0

⇒ (k – 1) (k – 6) = 0

⇒ k = 1 or k = 6.

54.  Given: √3x + y + 2 = 0

Comparing equation (i) with x cos θ + y sin θ = p, we get

55.  Let the intercepts be a and b.

Then we have equation of line be  and

it is given that a + b = 1 and ab = -6

Now (a – b)2 = (a + b)2 – 4ab

⇒ (a – b)2 = (1)2 – 4(-6)

⇒ (a – b)2 = 1 + 24 = 25

⇒ a – b = ±5

Solving a + b = 1 and a – b = 5,

we get a = 3, b = -2

∴ Equation of the line is

⇒ -2x + 3y = -6

⇒ 2x – 3y = 6

Solving a + b = 1 and a – b = -5, we get a = -2, b = 3

∴ Equation of the line is

⇒ 3x – 2y = -6

⇒ -3x + 2y = 6

Hence required equations of the lines are 2x – 3y = 6 and -3x + 2y = 6

56.  Let point on y-axis be (0, y)

Given equation of line is

⇒ 4x + 3y = 12

⇒ 4x + 3y – 12 = 0

We have perpendicular distance from (x1, y1) to the line ax + by + c = 0 is

∴ Perpendicular distance from point (0,y) to 4x + 3y – 12 = 0

57.  Equation of the line joining points (cos θ, sin θ) and (cos Φ, sin Φ) is

Now, perpendicular distance from (0, 0) to this line,

58.  The equation of any line parallel to y-axis is of the form x= k,a constant

Given lines are x – 7y + 5 = 0  ….(i)

3x + y = 0        …(ii)

From (ii) we get  y = -3x

Subtracting y in equation (i),

∴ Point of intersection of lines (i) and (ii) is

Since x = k, passes through the point  we get

Therefore the equation of required line is

59.  Given: Equation of line

⇒ 6x + 4y – 24 = 0           …(i)

Slope of given line

Therefore slope of line perpendicular to given line

Let the given line meet the y-axis at (0, y)

Now by substituting x=0 in equation (i),

The given line meets the axis at (0, 6).

Now the equation of a line with slope  and passes through (0,6) is

∴ Equation of required line is 2x – 3y + 18 = 0.

60.   Given: Equations of lines are y – x = 0      …(i)

x + y = 0      …(ii)

And x – k = 0    …..(iii)

On solving eq. (i) and (ii), we get the point of intersection C =(0, 0)

On solving eq. (ii) and (iii), we get the point of intersection A = (k, -k)

On solving eq. (i) and (iii), we get the point of intersection B = (k, k)

We have the area of the triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is

Hence area of triangle formed by the three given lines is k2 sq. units

61.   We know three lines a1x + b1y + c1z = 0, a2x + b2y + c2z = 0, and a3x + b3y + c3z =  intersect at one point if a3(b1c2 – b2c1) + b3 (c1a2 – c2a1) + c3 (a1b2 – a2b1) = 0

Given lines are 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0

∴ 2[1×(-3) – 2× (-2)] + (-1)[-2 × p-(-3)×3] + (-3)[3 × 2 – p × 1] = 0

⇒ 2[-3 + 4] -1[-2p + 9] -3[6 – p] = 0

⇒ 2 + 2p – 9 – 18 + 3p = 0

⇒ 5p – 25 = 0

⇒ p = 5

Hence the require value of p is 5.

62.  We know three lines a1x + b1y + c1z = 0, a2x + b2y + c2z = 0, and a3x + b3y + c3z =  intersect at one point if a3(b1c2 – b2c1) + b3 (c1a2 – c2a1) + c3 (a1b2 – a2b1) = 0

Given lines are y = m1x + c1, y = m2x + c2 and y = m3x + c3

⇒ m1x – y + c1 = 0, m2x – y + c2 = 0 and m3x – y + c3 = 0

Since the lines are concurrent, we have

m3(-1.c2 + 1.c1) -1(c1.m2 – c2.m1) + c3(m1. – 1 + m2.1) = 0

⇒ m3 (-c2 + c1) – c1. m2 + c2m1 – c3m1 + c3m2 = 0

⇒ m3(c1 – c2) + m2(c3 – c1) + m1(c2 – c3) = 0

⇒ m1(c1 – c2) + m2(c3 – c1) + m3(c2 – c3) = 0

63.  Let m be the slope of required line which passes through point (3, 2),

Then the equation of required line is    y – 2 = m(x – 3) ……….(i)

The equation of given line x – 2y = 3

∴ Slope of given line

We know that if θ is the acute angle between two lines with slopes m1 and m2 respectively

Then the equation of required line is y – 2 = 3(x – 3)

⇒ y – 2 = 3x – 9

⇒ 3x – y – 7 = 0

Then the equation of required line is

⇒ 3y – 6 = -x + 3

⇒ x + 3y – 9 = 0

Thus the equation of the lines are 3x – y – 7 = 0 and x + 3y – 9 = 0.

64.  Let the equal intercepts be a. Equation of a line in intercept form is

Given: The required equation passes through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0

Consider 4x + 7y = 3  …(i) and 2x – 3y = -1   …(ii)

Multiplying equation(ii) by 2, we get 4x – 6y = -2         …(iii)

Subtracting equation (iii) from equation(i),

Substituting y in equation(ii) we get

Hence the point of intersection of (i) and (ii) is

Now since the required line passes through the point

we have,

∴ From equation(A) we get the required equation of the line is

65.  Let m1 be the slope of required line which passes through (0, 0).

∴ Equation of line is y – 0 = m1(x – 0)    ………(i)

⇒ y = m1x         …(i)

Now, let θ be the angle between y = mx + c and y = m1x

We know that if θ is the acute angle between two lines with slopes m1 and m2 respectively then

Therefore from (ii) and (iii) we get

66.  Given: Equation of line x + y – 4 = 0

Let the given line divide the line joining A(-1, 1)  and B (5, 7) in the ratio k : 1 at a point C.

∴ Using section formula we have coordinates of C are

Since the point C lies on the given line.

Therefore, the required ratio is 1 : 2.

67.  First we have to find the point of intersection of given lines 4x + 7y + 5 = 0 …….(i) and 2x – y = 0  …..(ii)

From(ii) we get y = 2x

Substituting y = 2x in equation(i),we get

Hence point of intersection of (i) and (ii) is A

∴ Distance between the points A  and B(1, 2)

68.  Let the required equation of the line be y – y0 = m(x – x0)   ……………(i)

Since equation (i) passes through the point (-1, 2), we get y – 2 = m(x + 1)

⇒ mx – y + (m + 2) = 0     ……………..(ii)

Equation of the given line is x + y = 4    ……….(iii)

From(iii) we get y = 4 -x, substituting this in(ii),we get

Since the slope is zero,we have the required line is parallel to x-axis.

69.  Let ABC be a right angled triangle with diagonal AC.

Then we have the legs of the triangle (perpendicular sides) are BA and BC

Let the slope of AB is m ,then since AB and BC are perpendicular to each other

we have, Slope of BC =

We have the equation of a line passing through (x0, y0) and slope m is y – y0 = m(x – x0)

Now since AB passes through A(1, 3) and have slope m equation of AB is y – 3 = m(x – 1)    …(i)

Also BC is a line through C(-4, 1) with slope  hence it’s equation is

y – 1 = (x + 4)   ….(ii)

By putting different values for m in equations (i) and (ii) we get different equations for BA and BC

In particular for m=0,equation of AB is y = 3 [ which is a line parallel to X-axis] and equation of BC is x = -4 [ which is a line parallel to Y-axis].

70.  Let the image of the point A (3, 8) in the line mirror DE be C(α, β). Then AC is perpendicular bisector of DE.

∴ The coordinates of point B are

Since point B lies on the line x + 3y = 7.

Since AC is perpendicular on DE.

∴ Slope of AC × Slope of DE = -1

Solving eq.(i) and (ii), we get α = -1 and β = -4

Therefore, the image of point (3, 8) is (-1, -4).

71.  Given lines are y = 3x + 1   ……..(i)

2y = x + 3         …..(ii) and y = mx + 4    ….(iii)

Now we have slopes of lines (i) ,(ii) and (iii) are 3,1/2,m respectively

We know that if θ is the acute angle between two lines with slopes m1 and m2 respectively then

Let θ be the angle which the line y = mx + 4 makes with the line y = 3x + 1 and 2y = x + 3.

Given that lines y = 3x + 1 and 2y = x + 3 are equally inclined to the lne y = mx + 4

72.  Given: Equations of lines are x + y – 5 = 0   ..(i)

and 3x – 2y + 7 = 0    ..(ii)

Perpendicular distance of point P(x, y) from line (i)

Perpendicular distance of point P(x, y) from line (ii)

Similarly, we can get the equation of a line for any signs (positive or negative) of x + y – 5 and 3x – 2y + 7.

Therefore, we can say that P(x, y) must move on a line.

73.  The equations of parallel lines are 9x + 6 – 7 = 0 and 3x + 2y + 6 = 0

Let A(x1, y1) be any point which is equidistant from the parallel lines.

⇒ 9x1 + 6y1 – 7 = 3(3x1 + 2y1 + 6) ⇒ -7 = 18 which leads to some wrong conclusion

so this case is not possible

Therefore, the required line is 18x + 12y + 11 = 0.

74.  Let the coordinates of the point A be (x,0)

Also let BA be the incident ray and AC be the reflected ray and AD be the normal

We have angle of incidence = angle of reflection ⇒ ∠BAD = ∠CAD

Also let ∠CAX = Φ

Then we have Φ + θ = 90° ⇒ θ = 90° – Φ

Also ∠OAB = 180° – (Φ + 2θ) = 180° – [Φ + 2(90° – Φ)] = 180° – (180° – Φ) = Φ

Therefore, coordinates of point A are

75.  Let p and p be the  length of perpendiculars from

76.  Consider 2x – 3y + 4 = 0    …(i) and 3x + 4y – 5 = 0    …(ii)

The point of intersection of lines (i) and (ii) is given by

It is given that the person is standing at A and wants to reach the path whose equation is 6x – 7y + 8 = 0   …(iii) in the least time

We have the shortest path from point A to line(iii) is the line AB which is the perpendicular distance from  to equation (iii)

∴ Slope of line 6x – 7y + 8 = 0 is

∴ Slope of required line AB is

Therefore, the equation of AB is y – y0 = m(x – x0)

Hence the equation of the path that the person should follow is 119x + 102y = 125.

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