# Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.10 | Set 1

• Last Updated : 30 Apr, 2021

### 2x – y + 3 = 0 and x + y – 5 = 0

Solution:

Given equation of lines are:

2x – y + 3 = 0  …..(1)

x + y – 5 = 0  ……(2)

2x – y + 3 = 0

⇒ y = 2x + 3

Now put this value in equation(2), we get

x + y – 5 = 0

x + (2x + 3) – 5 = 0

x + 2x + 3 – 5 = 0

3x – 2 = 0

x = 2/3

Now put the value of x in equation(1), we get

y = 2x + 3 = (2 × 2)/3 + 3 = 4/3 + 3 = 13/2

Hence, the points of intersection is (2/3, 13/3)

### bx + ay = ab and ax + by = ab

Solution:

Given equation of lines are:

bx + ay = ab  …..(1)

ax + by = ab   …..(2)

bx + ay = ab ⇒ Now put this value in equation(2), we get

ax + by = ab

a((ab – ay)/(b)) + by = ab

a2b – a2y + b2y = ab2

y(b2 – a2) = ab(b – a) Now put the value of x in equation(1), we get Hence, the points of intersection is ( )

### y = m1x + a/m1 and y = m2x + a/m2

Solution:

Given equation of lines are:

y = m1x + a/m  …..(1)

y = m2x + a/m2   …..(2)

From equation (1) and (2)

m1x Hence, the point of intersection is ### Question 2 (i). Find the coordinates of the vertices of a triangle, the equations of whose sides are x + y – 4 = 0, 2x – y + 3 = 0 and x – 3y + 2 = 0.

Solution:

Given equations:

x + y – 4 = 0  …..(1)

2x – y + 3 = 0   …..(2)

x – 3y + 2 = 0  …..(3)

On solving eq(1) and (2)

2x – (4 – x) + 3 = 0

2x – 4 + x + 3 = 0

3x – 1 = 0

x = 1/3

Now Put the value of x in eq(1), we get

1/3 + y – 4 = 0

y = 4 – 1/3 = 11/3

So, the first vertex is (1/3, 11/3)

On solving eq(2) and (3), and

Put y = 2x + 3 in eq(3)

x – 3(2x + 3) + 2 = 0

x – 6y – 9 + 2 = 0

-5x = 7

x = -7/5

y = 2x + 3 = 2(-7/5) + 3 = -14/5 + 3 = 1/5

So, the second vertex is (-7/5, 1/5)

Now we find the next vertex

x + y – 4 = 0

x – 3y + 2 = 0

x = 4 – y

4 – y – 3y + 2 = 0

4 – 4y + 2 = 0

-4y = -6

y = 3/2

x = 4 – y

4 – 3/2

8 – 3/2 = 5/2

Hence, the third vertex is (5/2, 3/2)

### Question 2 (ii). Find the coordinates of the vertices if a triangle, the equation of whose sides are y(t1 + t2) = 2x + 2at1t2, y(t2 + t3) = 3x + 2at2t3 and y(t3 + t1) = 2x + 2a + t1t3

Solution:

Given equations:

y(t1 + t2) = 2x + 2at1t2  …..(1)

y(t2 + t3) = 2x + 2at2t3  …..(2)

y(t3 + t1) = 2x + 2at1t3  …..(3)

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

(x1, y1) = (at22, 2at2)

On solving eq (2) and (3), we get

(x2, y2) = (at32, 2at3)

On solving eq(1) and (3), we get

(x3, y3) = (at12, 2at1

Hence, the vertices of the triangle are (at22, 2at2), (at32, 2at3), and (at12, 2at1).

### y = m1 + c1, y = m2 + c2, x = 0

Solution:

Given equations

y = m1x + c  …..(1)

y = m2x + c2  …..(2)

x = 0   …..(3)

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

(x1, y1) = ( )

On solving eq (2) and (3), we get

(x2, y2) = (0, c2)

On solving eq (1) and (3), we get

(x3, y3) = (0, c1)

Area of triangle formed by above vertices is sq. units

### y = 0, x = 2, and x + 2y = 3

Solution:

Given equations

y = 0   …..(1)

x = 2   …..(2)

x + 2y = 3   …..(3)

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

Point (x1, y1) = (2, 0)

On solving eq (2) and (3), we get

2 + 2y = 3

y = 1/2

y = 2

Point (x2, y2) = (2, 1/2)

On solving eq (1) and (3), we get

Point (x3, y3) = (3, 0)

As we know that the area of triangle is (A) = 1/2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

A = 1/2[2(1/2 – 0) + 2(0 – 0) + 3( 0 – 1/2)]

A = 1/2[|1 – 3/2|]

A = 1/4 sq. units

### x + y – 6 = 0, x – 3y – 2 = 0 and 5x – 3y + 2 = 0

Solution:

Given equations

x + y – 6 = 0   …..(1)

x – 3y – 2 = 0   …..(2)

5x – 3y + 2 = 0   …..(3)

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

(x1, y1) = (5, 1)

On solving eq (2) and (3), we get

(x2, y2) = (-1, -1)

On solving eq (3) and (1), we get

(x3, y3) = (2, 4)

As we know that the area of triangle is (A) = 1/2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

A = 1/2[|-25 – 3 + 4|]

A = 12 sq. units

### 3x + 2y + 6 = 0, 2x – 5y + 4 = 0, x – 3y – 6 = 0

Solution:

Let us considered ABC is a triangle, in which A, B, and C are the vertices of the triangle

So, according to the question the equations are:

3x + 2y + 6 = 0   …..(1)

2x – 5y + 4 = 0   …..(2)

x – 3y – 6 = 0   …..(3)

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

x = -2 and y = 0

So the coordinates of A is (-2, 0)

On solving eq (3) and (2), we get

x = -42 and y = -16

So the coordinates of B is (-42, -16)

On solving eq (1) and (3), we get

x = -6/11 and y = -24/11

So the coordinates of C is (-6/11, -24/11)

Now let us assume that P, Q, and R be the mid-point of side AB, BC and AC

So, the coordinates of the mid-points are:

P = ((-2 -42)/2, (0 – 16)/2) = (-22, -8)

Q = ((-42 – 6/11)/2, (-16 – 24/11)/2) = (-234/11, -100/11)

R = ((-2 -6/11)/2, (0 – 24/11)/2) = (-14/11, -12/11)

Now we find the median of AP is

= y + 8/x + 22 = = y + 8/x + 22 = -88 + 24/-242 + 6 = 16/59

= 16x – 59y + 352 – 572 = 0

= 16x – 59y – 120 = 0

Similarly, the median of BQ is 25x – 53y + 50 = 0

And the median of CR is 41x – 112y – 70 = 0

### Question 5. Prove that lines y = √3x + 1, y = 4, and y = -√3x + 2 form an equilateral triangle.

Solution:

Let us considered ABC is a triangle in which A, B, and C are the vertices of the triangle

The equation of lines are:

y = √3x + 1       …..(1)

y = 4      …..(2)

y = -√3x + 2     …..(3)

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

4 = √3x + 1

x = 4 – 1/√3 = 3/√3 = √3

So, the coordinates of point A is (√3, 4)

On solving eq (2) and (3), we get

4 = -√3x + 2

√3x = -2

x = -2/√3

x = -2√3/3

So, the coordinates of point B is (-2√3/3, 4)

On solving eq (1) and (3), we get

√3x + 1 = -√3x + 2

2√3x = 1

x = 1/2√3 = √3/6

y = √3(√3/6 + 1) = 3/2

So, the coordinates of point C is(√3/6, 3/2)

Now we have,

AB =  = 5√3/3 units

BC AC Here, AB = BC = AC

Hence, ABC triangle is equilateral

### Question 6 (i). Classify 2x + y – 1 = 0 and 3x + 2y + 5 = 0 as coincident, parallel or intersecting.

Solution:

The equations of lines are:

2x + y – 1 = 0

3x + 2y + 5 = 0

Now on writing above equations in the form y = mx + c

y = -2x + 1,

y = -3x/2 – 5/2

So, the slope of both the equations are

m1 = -2 and m2 = -3/2

Now

m1 ≠ m2 and m1m2 ≠ -1

Hence, the lines are intersecting

### Question 6(ii). Classify x – y = 0 and 3x – 3y + 5 = 0 as coincident, parallel or intersecting.

Solution:

The equations of lines are:

x – y = 0

3x – 3y + 5 = 0

Now on writing above equations in the form y = mx + c

y = x

y = x + 5/3

So, the slope of both the equations are

m1 = 1 and m2 = 1

Slopes of both lines are equal

Hence, the lines are parallel

### Question 6 (iii). Classify 3x + 2y – 4 = 0 and 6x + 4y – 8 = 0 as coincident, parallel or intersecting.

Solution:

The equations of lines are:

3x + 2y – 4 = 0

6x + 4y – 8 = 0

Now on writing above equations in the form y = mx + c

y = -3x/2 + 2

y = -3x/2 + 2

So, the slope of both the equations are

m1 = -3/2 and m2 = -3/2

So, m1 = m2 = -3/2

Hence, the lines are coincident

### Question 7. Find the equation of the joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0.

Solution:

Given equation of lines are

4x + y – 1 = 0        …..(1)

7x – 3y – 35 = 0       …..(2)

From eq(1), we get

y = 1 – 4x

Now put the value of y in eq(2), we get

7x – 3(1 – 4x) – 35 = 0

7x – 3 + 12x – 35 = 0

19x = 38

x = 2

y = 1 – 4x = 1 – 8 = -7

Now we have to find the equation of line the joining the point P(2, -7) and Q(3, 5)

The equation of line PQ is

y – y1 = m(x – x1)

y – y1 (x – x1)

y – (-7) = y + 7 = 12(x – 2)

y – 12x = -31

12x – y – 31 = 0

Hence, the equation of line is 12x – y – 31 = 0

### Question 8. Find the equation of the line passing through the point of intersection of the lines 4x – 7y – 3 = 0 and 3x – 3y + 1 = 0 that has equal intercepts on the axes.

Solution:

Given equation of lines are

4x – 7y = 3

2x – 3y = -1

On solving the above equations, we get the point of intersection,

x = -B, y = -5

Now, the point of intersection of given lines is (-8, -5) equation

of line making equal intercepts (a) on the coordinates axes is, x + y = a

-8 – 5 = a

a = -13

Hence, the equation of line is x + y = -13

### Question 9. Show that area of the triangle formed by the lines y = m1x, y = m2x, and y = c is equal to c2/4(√33 + √11), where m1, m2 are the roots of the equation x2 + (√3 + 2)x + √3 – 1 = 0.

Solution:

Given equation of lines are

y = m1x        …..(1)

y = m2x         …..(2)

y = c        …..(3)

On solving eq (1) and (2), we get A(0, 0)

On solving eq (1) and (3), we get A(c/m1, c)

On solving eq (2) and (3), we get A(c/m2, c)

Now we find the area of triangle when three vertices are given is

= 1/2(x2(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)) Given m1 and m2 are roots of x2 + (√3 + 2)x + √3 – 1 = 0

Product of roots = m1m2 = √3 – 1

|m2-m1|= |m2-m1|= Area = On rationalizing denominator, we get Hence proved

### Question 10.If the straight line passes through the point of intersection of the lines x + y = 3 and 2x – 3y = 1 and is parallel to x – y – 6 = 0, find a and b.

Solution:

Given equation of lines are

x + y = 3

2x – 3y = 1

On solving both the equations we get the intersection point (2, 1)

It is given that the line x/a + y/b = 1 is parallel to line x – y – 6 = 0

so, the slope = 1

Now we find the equation of line passing through (2, 1) and having slope = 1 is

y – y1 = m(x – x1)

y – 1 = 1(x – 2)

y – 1 = x – 2

y – x = -2 + 1

y – x = -1

x – y = 1

On comparing with  x/a + y/b = 1, we get

a = 1, b = -1

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