# Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.5

Last Updated : 18 Mar, 2021

### (i) (0, 0) and (2, -2)

Solution:

Here, the two points are (x1, y1) = (0, 0) and (x2, y2) = (2, -2).

So the equation of the line in two point form is

y – y1 = (y2 – y1) / (x2 – x1)(x – x1)

y – 0 = [(-2 – 0)/(2 – 0)] (x – 0)

â‡’ y = (-1)x

â‡’ y = -x

### (ii) (a, b) and (a + c sin Î±, b + c cos Î±)

Solution:

Here, the two points are (x1, y1) = (a, b) and (x2, y2) = (a + c sin Î±, b + c cos Î±).

So the equation of the line in two point form is

y – b = (b + c cos Î± – b)/(a + c sin Î± – a)(x- a)

â‡’ y – b = (cos Î±/sin Î±)(x – a)

â‡’ y – b = cot Î±(x – a)

### (iii) (0, -a) and (b, 0)

Solution:

Here, the two points are (x1, y1) = (0, -a) and (x2, y2) = (b, 0).

So the equation of the line in two point form is

y – (-a) = (0 – (-a))/(b – 0)(x – 0)

â‡’ b(y + a) = ax

â‡’ ax – by = ab

### (iv) (a, b) and (a + b, a – b)

Solution:

Here, the two points are (x1, y1) = (a, b) and (x2, y2) = (a + b, a – b).

So the equation of the line in two point form is

y – b = (a – b – b)/(a + b – a)(x – a)

â‡’ y – b = (a – 2b)/b (x – a)

â‡’ y – b = (a – 2b)x + a2 – 2ab

â‡’ (a – 2b)x – by + b2 + 2ab – a2 = 0

### (v) (at1,a/t1) and (at2, a/t2)

Solution:

Here, the two points are (x1, y1) = (at1, a/t1) and (x2, y2) = (at2, a/t2).

So the equation of the line in two point form is

y – a/t1 = (a/t2 – a/t1) / (at2 – at1)(x – at1)

â‡’ y – a/t1 = ((at1 – at2)/t1t2)/(at2 – at1)(x – at1)

â‡’ t1t2y – at2 = (-1)(x – at1)

â‡’ t1t2y + x = at1 + at2

### (vi) (a cos Î±, a sin Î±) and (a cos Î², a sin Î²)

Solution:

Here, the two points are (x1, y1) = (a cos Î±, a sin Î±) and (x2, y2) = (a cos Î², a sin Î²).

So the equation of the line in two point form is

y – a sin Î± =(a sin Î² – a sin Î±)/(a cos Î² – a cos Î±)(x – a cos Î±)

â‡’ y – a sin Î± = [2cos (Î² + Î±)/2 sin (Î² – Î±)/2] / [-2sin (Î² + Î±)/2 sin (Î² – Î±)/2](x – a cos Î±)

â‡’ y – a sin Î± = -[cos (Î² + Î±)/2]/[sin (Î² + Î±)/2](x – a cos Î±)

â‡’ x cos (Î² + Î±)/2 + y sin (Î² + Î±)/2 = a(sin Î± [sin (Î² + Î±)/2] + cos Î± [cos (Î² + Î±)/2])

â‡’ x cos (Î² + Î±)/2 + y sin (Î² + Î±)/2 = a cos (Î± – Î²/2 – Î±/2)

â‡’ x cos (Î² + Î±)/2 + y sin (Î² + Î±)/2 = a cos (Î±/2 – Î²/2)

### (i) (1, 4), (2, -3) and (-1, -2)

Solution:

Given: Points A (1, 4), B (2, -3) and C (-1, -2).

Equation of the line passing through the two points (x1, y1) and (x2, y2)

y â€“ y1 = [(y2 – y1)/(x2 – x1)] (x â€“ x1

Equation of the sides AB,

By using the above formula

y â€“ 4 = [(-3 – 4)/(2 – 1)](x â€“ 1)

â‡’ y â€“ 4 = -7x + 7

â‡’ y + 7x = 11

Similarly

Equation of the sides BC

â‡’ y + 3 = [(-2 – (-3))/(-1 – 2)](x â€“ 2)

y + 3 = (-1/3) (x â€“ 2)

â‡’ 3y + 9 = -x + 2

â‡’ 3y + x = â€“ 7

â‡’ x + 3y + 7 = 0

And equation of the sides CA

â‡’ y + 2 = [(4 – (-2))/(1 – (-1))](x + 1)

â‡’ y + 2 = 3(x + 1)

â‡’ y + 2 = 3x + 3

â‡’ y â€“ 3x = 1

The equation of the required sides of the triangle are y + 7x = 11, x+ 3y + 7 =0 and y â€“ 3x = 1

### (ii) (0, 1), (2, 0) and (-1, -2)

Solution:

Given, Points A (0, 1), B (2, 0) and C (-1, -2).

The equation of the line passing through the two points (x1, y1) and (x2, y2

y â€“ y1 = [(y2 – y1)/(x2 – x1)] (x â€“ x1

Equation of the sides AB,

By using the above formula

y â€“ 1 = [(0 – 1)/(2 – 0)](x â€“ 0)

â‡’ y â€“ 1 = -1/2 x

â‡’ 2y â€“ 2 = -x

â‡’ x + 2y = 2

Equation of the sides BC

y â€“ 0 = [(-2 – 0)/(-1 – 2)](x â€“ 2)

â‡’ y â€“ 0 = (-2/3)(x â€“ 2)

â‡’ 3y = -2x + 4

â‡’ 2x â€“ 3y = 4

Equation of the sides CA

â‡’ y – (-2) = [(1 + 2)/(1 + 0)](x – (-1))

â‡’ y + 2 = 3(x + 1)

â‡’ y + 2 = 3x + 3

â‡’ y â€“ 3x = 1

The required equation of sides of the triangle are x + 2y = 2, 2x â€“ 3y = 4, and y â€“ 3x = 1

### Question 3.  Find the equations of the medians of a triangle the coordinate of whose vertices are (-1, 6), (-3, -9), and (5, -8).

Solution:

Let A (âˆ’1, 6), B (âˆ’3, âˆ’9) and C (5, âˆ’8) be the coordinates of the given triangle and

D, E, and F be midpoints of AB, BC and AC, respectively.

Coordinate of midpoint of AB = D[(-1 – 3)/2, (6 – 9)/2]

= D(-2, -3/2)

Median CD passes through C (5, -8) and D(-2, -3/2)

So, by using the formula (y â€“ y1) = [(y2 – y1)/(x2 – x1)] (x â€“ x1

(y – (-8)) = [(-3/2 – (-8))/(-2 – 5)] (x – 5)

â‡’ y + 8 = -13/14(x-5)

â‡’ 14y + 112 = -13x + 65

â‡’ 13x + 14y + 47 = 0

Coordinate of midpoint of BC = E[(-3 + 5)/2, (-9 – 8)/2]

= E(1, -17/2)

Median AE passes through A (-1, 6) and E (1, -17/2)

Using the two point formula

â‡’ y – 6 = -29/4 (x + 1)

â‡’ 4y â€“ 24 = -29x â€“ 29

â‡’ 29x + 4y + 5 = 0

Coordinate of midpoint of AC = F[(-1 + 5)/2, (6 – 8)/2]

= E(2, -1)

Similarly, Median BF passes through B (-3,-9) and F (2,-1)

By using the formula,

y – (-9) = [(-1 – (-9))/(2 – (-3))] (x – (-3))

â‡’ y + 9 = 8/5 (x + 3)

â‡’ 5y + 45 = 8x + 24

â‡’ 8x â€“ 5y â€“ 21 = 0

The equation of the required medians of triangle are: 29x + 4y + 5 = 0, 8x â€“ 5y â€“ 21 = 0, and 13x + 14y + 47 = 0

### Question 4. Find the equations to the diagonals of the rectangle the equations of whose sides are x = a, x = aâ€™, y = b, and y = bâ€™.

Solution:

Given,

Equation of sides of the rectangle are x = a, x = aâ€™, y = b and y = bâ€™

Thus, the vertices of the rectangle are A (a, b), B (aâ€™, b), C (aâ€™, bâ€™) and D (a, bâ€™)

Equation of the diagonal passing through A (a, b) and C (aâ€™, bâ€™)

By using the formula (y â€“ y1) = [(y2 – y1)/(x2 – x1)] (x â€“ x1

â‡’ y – b = [(b’ – b)/(a’ – a)] (x – a)

â‡’ (aâ€™ â€“ a)y â€“ b(aâ€™ â€“ a) = (bâ€™ â€“ b)x â€“ a(bâ€™ â€“ b)

â‡’ (aâ€™ â€“ a) â€“ (bâ€™ â€“ b)x = baâ€™ â€“ abâ€™

Similarly, equation of diagonal passing through B (aâ€™, b) and D (a, bâ€™)

By using the two point formula,

â‡’ y – b = [(b’ – b)/(a – a’)] (x – a’)

â‡’ (a â€“ a’)y â€“ b(a â€“ aâ€™) = (bâ€™ â€“ b)x â€“ aâ€™ (bâ€™ â€“ b)

â‡’ (aâ€™ â€“ a)y + (bâ€™ â€“ b)x = aâ€™bâ€™ â€“ ab

The required equation of diagonals are y(aâ€™ â€“ a) â€“ x(bâ€™ â€“ b) = aâ€™b â€“ abâ€™ and y(aâ€™ â€“ a) + x(bâ€™ â€“ b) = aâ€™bâ€™ â€“ ab

### Question 5. Find the equation of the side BC of the triangle ABC Whose vertices are A(-1, -2), B(0, 1) and C(2, 0) respectively. Also, find the equation of median through (-1, -2).

Solution:

Equation of side BC of the triangle ABC is

(y – 1) = [(0 – 1)/(2 – 0)] (x – 0)

â‡’ (y – 1) = (-1/2)x

â‡’ x + 2y – 2 = 0

Mid point of (0, 1) and (2, 0)

= [(0 + 2)/2, (1 + 0)/2]

= (1, 1/2)

Equation of median through (-1, -2) and (1, 1/2)

(y – (-2)) = [(1/2 + 2)/(1 + 1)](x – (-1))

â‡’ y + 2 = 5/4(x + 1)

â‡’ 4y + 8 = 5x + 5

â‡’ 5x – 4y – 3 = 0 is the required equation of median.

### Question 6. By using the concept of equation of line, prove that the three points (3, 0), (-2, -2), and (8, 2) are collinear.

Solution:

To show that the points(-2, -2), (8, 2) and (3, 0) are collinear,

The equation of the line passing through points (3, 0), (-2, -2) is

(y âˆ’ 0) = [(âˆ’2 âˆ’ 0)/(âˆ’2 âˆ’ 3)](x âˆ’ 3)

y = -2/âˆ’5(x âˆ’ 3)

5y = 2x âˆ’ 6

2x âˆ’ 5y = 6

Now, putting the third point in the above equation to check if it satisfies or not.

L.H.S. = 2 Ã— 8 âˆ’ 5 Ã— 2

= 16 âˆ’ 10

= 6 = R.H.S.

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

Hence, points (3,0), (-2, -2) and (8, 2) are collinear.

### Question 7. Prove that the line y- x+ 2 = 0 divides the join of points (3, -1) and (8, 9) in the ratio 2:3.

Solution:

Suppose the line y- x+ 2=0 divides the join of points (3, -1) and (8, 9) in the ratio 2:3.

The coordinate of point of division must lie on the given line.

Coordinate of point of division = [(8Ã—2+ 3Ã—3) / (2+3), (9Ã—2-3) / (2+ 3)]

= (25/5, 15/ 5) = (5, 3)

Putting this coordinate in y – x + 2 = 0

= 3 – 5 + 2

= -2 + 2 = 0

Thus, we see that the coordinate satisfies the r = equation of line which

proves that line divides the join of points in the ratio of 2:3.

### Question 8. Find the equation to the straight line which bisects the distance between the points (a, b) and (aâ€², bâ€²) and also bisects the distance between the points (âˆ’a, b) and (aâ€², âˆ’bâ€²).

Solution:

The points which bisects the given points are (x1, y1) = (a + aâ€²/2, b + bâ€²/2) and (x2, y2) = (-a + aâ€²/2, b âˆ’ bâ€²/2)

Using the two point formula

(y âˆ’ y1) = [(y2 âˆ’ y1)/(x2 âˆ’ x1)](x âˆ’ x1

[y – (b + b’)/2] =

â‡’ (2y – b – b’)/2 = b’/a (2x – a – a’)/2

â‡’ 2ay – ab – ab’ = 2b’x – ab’ – a’b’

â‡’ 2ay – 2b’x + a’b’ – ab = 0 Is the required equation of the straight line

### Question 9. In what ratio is the line joining the points (2, 3) and (4, -5) divided by the line passing through the points (6, 8) and (-3, -2).

Solution:

Equation of the line joining (6, 8) and (-3, -2)

y – 8 = [(-2 – 8)/(-3-  6)](x – 6)

â‡’ Y – 8 = 10/9(x – 6)

â‡’ 9Y – 72 = 10x – 60

â‡’ 10x – 9y + 12 = 0

Let the ratio in which the points (2, 3) and (4, -5) divided by k : 1

P[(4k + 2)/(k + 1), (-5k + 3)/(k + 1)]

We put this point in the above straight line as it satisfies the above equation

10 (4k + 2)/(k + 1) – 9(-5k + 3)/ (k + 1) + 12 = 0

40k + 20 + 45k – 27 + 12k + 12 = 0

97k + 5 = 0

k = -5/97

or k = 5/97 (External)

### Question 10. The vertices of a quadrilateral are A(-2, 6), B(1, 2) C(10, 4), and D(7, 8). Find the equation of its diagonal.

Solution:

Given A(âˆ’2, 6), B(1, 2), C(10, 4) and D(7, 8)

Diagonals are AC and BD

Equation of line AC

y âˆ’ 6 = [(4 – 6)/(10 – (-2))] (x âˆ’ (-2))

â‡’ y âˆ’ 6 = (-2/12)(x + 2)

â‡’ 6y – 36 = -x – 2

â‡’ x + 6y âˆ’ 34 = 0

Equation of line BD

y âˆ’ 2 = [(8 – 2)/(7 – 1)] (x âˆ’ 1)

â‡’ y – 2 =(6/6)(x- 1)

â‡’ y – 2 = x – 1

â‡’ x âˆ’ y + 1 = 0

So the required equation are x + 6y âˆ’ 34 = 0 and x âˆ’ y + 1 = 0

### Question 11. The length L (in centimeters ) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.

Solution:

Let us assume C to be along the x-axis and L along the y-axis

We have two points (20, 124.942) and (110, 125.134)

The linear relation between L and C is given by the equation of

the line passing through points (20, 124.942) and (110, 125.134)

(Lâˆ’ 124.942) = (125.134 âˆ’ 124.942) / (110 âˆ’ 20)(C âˆ’ 20)

â‡’ L âˆ’124.942 = 0.192/90(C âˆ’ 20)

â‡’ L = 0.192/ 90(C âˆ’ 20) + 124.942 is the required linear relation.

### Question 12. The owner of a milk store finds that, he can sell 980 liters of milk each week of Rs 14 per liter and 1220 liters of milk at Rs 16 per liter. Assuming a linear relationship between selling price and demand, how many liters could he sell weekly at Rs 17 per liter?

Solution:

Given,

980 liters of milk sells out at Rs 14 per liter

1220 liters of milk sells out at Rs 16 per liter

Let us assume amount of milk on y-axis and selling price on x-axis.

We can say the line passes through A(14, 980), B(16, 1220) and C(17, x)

Since there is a linear relationship between selling price and demand

Therefore, Slope of AB= Slope BC

(1220 – 980)/(16 – 14) = (x – 1220)/(17 – 16)

â‡’ 240/2 = x – 1220 / 1

x = 1340 liters of milk he could sell weekly at Rs 17 per liter

### Question 13. Find the equation of the bisector of angle A of the triangle whose vertices are A(4, 3), B(0, 0), and C(2, 3).

Solution:

Let AD be the bisector of angle A

Then BD : DC = AB : AC

|AB| = âˆš((4 – 0)2 + (3 – 0)2) = 5

|AC| = âˆš((4 – 2)2 + (3 – 3)2) = 2

â‡’ BD/DC = AB/AC = 5/2

i.e. D divides BC in the ratio of 5:2

Now, coordinate of D are \frac{(5×2 +0)}{5+2}, \frac{(5×3+0)}{5+2} = (10/7, 15/7)

The required equation of the bisector AD=

Using the two point formula

y – 3 = [)](x – 4)

â‡’ y – 3 = [(15 – 21)/(10 – 28)](x – 4)

â‡’ y – 3 = 1/3 (x – 4)

â‡’ 3y – 9 = x – 4

â‡’ x – 3y + 5 = 0

### Question 14. Find the equations to the straight line which go through the origin and trisect the portion of the straight line 3x + y = 12 which is intercepted between the axes of coordinates.

Solution:

To find the interception point we put x = 0 and y = 0.

3x + y = 12

x = 0 â‡’ y = 12

y = 0 â‡’ x = 4

So the line is intercepted between A(4, 0) and B(0, 12)

Let the points P and Q trisect the portion of the line AB

Now P divides AB in 1:2

= P \frac{(1(0) + 2(4))}{(1 + 2)}, \frac{(1(12) + 2(0))}{(1 + 2)}

= P(8/3, 4)

So the equation of line joining origin and P is

y âˆ’ 0 = [(4 – 0)/(8/3 – 0)](x – 0)

y = (12/8)x = (3/2)x

Now Q divides AP in 1:1,

= Q[(8/3 + 0)/2, (4 + 12)/2]

= Q(4/3, 8)

The equation of line joining origin to Q is

y – 0 = [(8 – 0)/(4/3 – 0)](x – 0)

y = (24/4)x = 6x

So, the required equations are y = 6x and 2y = 3x.

### Question 15. Find the equation of the diagonals of the square formed by lines x = 0, y = 0, x = 1, and y = 1.

Solution:

From the graph, we get four points, i.e., A(0, 1), B(1, 1), C(1, 0), and D(0, 0).

Now let us considered D1 is the diagonal formed by joining B and D point,

and D2 is the diagonal formed by A and C points.

So, we find the equation of D1 diagonal:

(y – 1) =

(y – 1) = (1)(x – 1)

y = x

So, we find the equation of D2 diagonal:

(y – 1) =

(y – 1) = (-1)(x)

y + x = 1

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