# Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.12 | Set 2

### Question 11. Find the equation of the right bisector of the line segment joining the points (a, b) and (a_{1}, b_{1}).

**Solution:**

Let us considered P(a, b) and Q(a

_{1}, b_{1}) are the given points and O be the mid-point of line PQ.So the coordinate O = ((a + a

_{1})/, (b + b_{1})/2)Slope of line PQ =

And the slope of right bisector of AB(m’) =

The equation of required line is

y – y

_{1 }= m'(x – x_{1})2x(a

_{1 }– a) + 2y(b_{1 }– b) + a^{2 }+ b^{2 }= a_{1}^{2}+ b_{1}^{2}2x(a

_{1 }– a) + 2y(b_{1 }– b) + (a^{2 }+ b^{2})^{ }– (a_{1}^{2}+ b_{1}^{2}) = 0

Hence, the equation of line is2x(a_{1 }– a) + 2y(b_{1 }– b) + (a^{2 }+ b^{2})^{ }– (a_{1}^{2}+ b_{1}^{2}) = 0

### Question 12. Find the image of the point (2, 1) with respect to the line mirror x + y – 5 = 0.

**Solution:**

Let us considered P(2, 1) be the image of Q(a, b) are the given points and O be the mid-point of line PQ.

So the coordinate O = ((2 + a)/2, (1 + b)/2)

And O point lie on the line x + y – 5 = 0

(2 + a)/2 + (1 + b)/2 – 5 = 0

a + b = 7 ………(1)

Here, x + y – 5 = 0 is perpendicular to PQ

So, (Slope of PQ) x (slope of AB) = -1

b – 1 = a – 2

b – a = -1 ………(2)

On solving eq(1) and (2), we get

a = 5 and b = 2

Hence, the image of (1, 2) in x + y – 5 = 0 is (4, 3)

### Question 13. If the image of the point (2, 1) with respect to the line mirror be (5, 2), find the equation of the mirror.

**Solution:**

Let us considered A(5, 2) be image of B(2, -1) and O be the mid point of AB

So the coordinate O = (7/2, 3/2)

Let us considered PQ be the mirror and line AB perpendicular to PQ

So, (Slope of PQ) x (slope of AB) = -1

(2 – 1/5 – 2) x (slope of AB) = -1

(slope of AB) = -3

So the equation of the mirror is

y – 3/2 = -3(x – 7/2)

2y – 3 = -6x + 21

Hence, the equation of the mirror is3x + y – 12 = 0

### Question 14. Find the equation to the straight line parallel to 3x – 4y + 6 = 0 and passing through the middle point of the join points (2, 3) and (4, -1).

**Solution:**

It is given that A(2, 3) and B(4, -1) and O be the mid point of AB

So the coordinate O = (3, 1)

It is given that the equation to the straight line parallel to 3x – 4y + 6 = 0

So,

y = 3x/4 + 3/2

On comparing y = mx + c, we get

m = 3/4

Now put the value of m and (x_{1}, y_{1}) is eq(1), we get

The required equation of line is

y – y_{1 }= m(x – x_{1})

y – 1 = 3/4(x – 3)4y – 4 = 3x – 9

3x – 4y = 5

Hence, the equation of line is3x – 4y = 5

### Question 15. Prove that the lines 2x – 3y + 1 = 0, x + y = 3, 2x – 3y = 2 and x + y = 4 form a parallelogram.

**Solution:**

AS we know that in a parallelogram opposite sides are parallel and parallel sides have equal slope.

So, the slope of line 2x – 3y + 1 = 0

m

_{1 }= 2/3 …….(1)The slope of line x + y = 3

m

_{2 }= -1 …….(2)The slope of line 2x – 3y – 2 = 0

m

_{3 }= 2/3 …….(3)The slope of line x + y = 4

m

_{4 }= -1 …….(4)From (1), (2), (3) and (4), we get

We conclude that the opposite sides of ABCD have same slope

Hence, the given quadrilateral is parallelogram.

### Question 16. Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point where it meets the y-axis.

**Solution:**

Let us considered the required equation of line is

y – y_{1 }= m(x – x_{1}) ……….(1)The required line is perpendicular to the given line x/4 + y/6 = 1

When x = 0

y/6 = 1

y = 6

So, the point (x

_{1}, y_{1}) is (0, 6)It is given that

the required equation of line isperpendicular to the line x/4 + y/6 = 1So,

(slope of required line) x (slope of given line) = -1

m1 = -1/(-6/4) = 4/6 = 2/3

Now put the value of m’ and (x_{1}, y_{1}) is eq(1), we get

(y – 6) = 2/3(x – 0)2x – 3y = -18

2x – 3y + 18 = 0

Hence, the equation of line is2x – 3y + 18 = 0

### Question 17. The perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2). Find the values of m and c.

**Solution:**

Let us considered point O (0, 0) and P (-1, 2) and OP is perpendicular to the given line y = mx + c

So, (slope of OP) x (slope of line)=-1

And point P lie on the line so,

2 = (1/2)(-1) + c

c = 2 + 1/2 = 5/2

Hence, the value of c = 5/2 and m = 1/2

### Question 18. Find the equation of the right bisector of line segment joining the points (3, 4) and (-1, 2).

**Solution:**

Let us considered P(3, 4) and Q(-1, 2) are the given point and O be the mid point of AB

So, the coordinate O = (1, 3)

And the slope of line PQ is

The right bisector of PQ is -2

The equation of the required line is

y – 3 = (-2)(x – 1)

y – 3 = -2x + 2

2x + y – 5 = 0

Hence, the required equation of line is 2x + y – 5 = 0

### Question 19. The line through (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.

**Solution:**

Let us considered P (h, 3) and Q(4, 1) are the given points

Now the slope the line 7x – 9y – 19 = 0 is 7/9

It is given that line PQ is perpendicular to 7x – 9y – 19 = 0

so

7/9 x (1 – 3)/(4 – h) = -1

9h = 22

h = 22/9

Hence, the value of h is 22/9

### Question 20. Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

**Solution:**

Let us considered A(5, 2) be image of B(a, b) and O be the mid point of AB

So, the coordinate of O = ((3 + a)/2, (8 + b)/2)

O point lies on line x + 3y = 7

So, (3 + a)/2 + 3 x ((8 + b)/2) = 7

a + 13b + 13 = 0 ……(1)

It is given that line AB is perpendicular to mirror PQ

So, slope of AB x slope of PQ = -1

(b – 8/a – 3) x -1/3 = -1

3a – b – 1 = 0 ……(2)

On solving eq(1) and (2), we get

a = -1, b = -4

Hence, the image is (-1, -4)

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