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Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.10 | Set 1
  • Last Updated : 30 Apr, 2021

Question 1. (i) Find the point of intersection of the lines:

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)

Question 1(ii). Find the point of intersection of the pairs of lines:

bx + ay = ab and ax + by = ab 

Solution:

Given equation of lines are:

bx + ay = ab  …..(1)

ax + by = ab   …..(2)

 bx + ay = ab ⇒ x=\frac{ab-ay}{b}

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)

y=\frac{ab(b-a)}{b^2-a^2}=\frac{ab}{b+a}

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

x=\frac{ab-\frac{a(ab)}{a+b}}{b}=\frac{ab}{a+b}

Hence, the points of intersection is (\frac{ab}{a+b},\frac{ab}{a+b} )

(iii) Find the point of intersection of the pairs of lines:

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+\frac{a}{m_1}=m_2+\frac{a}{m_2}\\ x(m_1-m_2)=\frac{a}{m_2}-\frac{a}{m_1}=a(\frac{m_1-m_2}{m_1m_2})\\ x=\frac{a}{m_1m_2}\\ y=m_1x+\frac{a}{m_1}\\ =m_1(\frac{a}{m_1m_2})+\frac{a}{m_1}\\ =\frac{a}{m_2}+\frac{a}{m_1}\\ =a(\frac{m_1+m_2}{m_1m_2})

Hence, the point of intersection is (\frac{a}{m_1m_2},a(\frac{1}{m_1}+\frac{1}{m_2}))

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). 

Question 3 (i). Find the area of the triangle formed by the lines

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) = (\frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2}   )

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 

=\frac{1}{2}[(\frac{c_2-c_1}{m_1-m_2} \times c_1)-(\frac{c_2-c_1}{m_1-m_2}\times c_2)]\\ =\frac{(c_2-c_1)^2}{2(m_1-m_2)} sq. units

Question 3 (ii). Find the area of the triangle formed by the lines 

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

Question 3 (iii). Find the area of the triangle formed by the lines

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

Question 4. Find the equation of the medians of a triangle, the equation of whose sides are:

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 = \frac{-8+\frac{24}{11}}{-22+\frac{6}{11}}\\

= 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 = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

=\sqrt{(\frac{-2√3}{3}-√3)^2+(4-4)^2}\\ =\sqrt{(\frac{5√3}{3})}

= 5√3/3 units

BC =\sqrt{(\frac{√3}{6}+\frac{2√3}{3})^2+(\frac{3}{2}-4)}\\ =\sqrt{(5√3/6)^2+(-5/2)^2}\\ =\sqrt{\frac{75}{36}+\frac{25}{4}}\\ =\sqrt{\frac{75}{36}+\frac{225}{36}}\\ =\sqrt{\frac{300}{36}}\\ =\frac{10}{6}√3\\ =\frac{5}{3}√3

AC =\sqrt{(\frac{\sqrt{3}}{6}-\sqrt{3})^2+(\frac{3}{2}-4)^2}\\ =\sqrt{(\frac{5\sqrt{3}}{6})^2+(\frac{-5}{2})^2}\\ =\sqrt{\frac{75}{36}+\frac{25}{4}}\\ =\sqrt{\frac{75}{36}+\frac{225}{36}}\\ =\sqrt{\frac{300}{36}}\\ =\sqrt{\frac{10}{6}\sqrt{3}}\\ =\sqrt{\frac{5}{3}\sqrt{3}}

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 \frac{y_2-y_1}{x_2-x_1}   (x – x1)

y – (-7) = \frac{5-(-7)}{3-2}(x-2)

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,

\frac{x}{a}+\frac{y}{a}=1

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))

=\frac{1}{2}[|\frac{c^2}{m_1}-\frac{c^2}{m_2}|]=\frac{c^2}{2}[|\frac{m_2-m_1}{m_1m_2}|]

Given m1 and m2 are roots of x2 + (√3 + 2)x + √3 – 1 = 0

Product of roots = m1m2 = √3 – 1

|m2-m1|=\sqrt{(m_2+m_1)^2-4m_1m_2}=\sqrt{(√3+2)^2-4√3+4}

|m2-m1|=\sqrt{3+4+4√3-4√3+4}=\sqrt{11}

Area = \frac{c^2}{2}[\frac{√11}{√3-1}]

On rationalizing denominator, we get 

\frac{c^2}{4}[√33+√11]

Hence proved

Question 10.If the straight line \frac{x}{a}+\frac{y}{b}=1   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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