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Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.18

  • Last Updated : 08 May, 2021

Question 1: Find the equation of the straight lines passing through the origin and making an angle of 45° with the straight line √3x+y = 11.

Solution:

As line 1 passes through origin, there won’t be any intercept, and it will be in the form of y = mx (m as slope)

For line 2: √3x+y = 11, slope is M = -√3

Here, it is given that these two lines make an angle of 45° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = 45°, m=m and M = -√3

tan 45° = |\frac{-\sqrt{3}-m}{1+m(-\sqrt{3})}|

1 = |\frac{-\sqrt{3}-m}{1-\sqrt{3}m}|

We will have two cases,

1 = \frac{-\sqrt{3}-m}{1-\sqrt{3}m}   and 1 = -\frac{-\sqrt{3}-m}{1-\sqrt{3}m}

1-√3m = -√3-m and 1-√3m = √3+m

√3m-m = 1+√3 and √3m+m = 1-√3

m = \frac{1+\sqrt{3}}{\sqrt{3}-1}   and m = \frac{1-\sqrt{3}}{\sqrt{3}+1}

By rationalizing, we get

m = \frac{(1+\sqrt{3})^2}{3-1}   and m = -\frac{(1-\sqrt{3})^2}{3-1}

m = \frac{1+3+2\sqrt{3}}{2}   and m = -\frac{1+3-2\sqrt{3}}{2}

m = 2+√3 and m = -(2-√3)

m = √3+2 and m = √3-2

Hence, the equation of line will be,

y = (√3+2)x and y = (√3-2)x

Question 2: Find the equation to the straight lines which pass through the origin and are inclines at an angle of 75° to the straight line x+y+√3(y-x)=a.

Solution:

As line 1 passes through origin, there won’t be any intercept and it will be in the form of y = mx (m as slope)

For line 2: x+y+√3(y-x)=a, 

x+y+√3y-√3x=a

x(1-√3)+y(1+√3)=a

slope is M = \frac{\sqrt{3}-1}{\sqrt{3}+1}

After rationalizing, we get

M = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{(3+1-2\sqrt{3})^2}{2}   = 2-√3

Here, it is given that these two lines make an angle of 75° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = 75°, m=m and M = 2-√3

tan 75° = |\frac{(2-√3)-m}{1+m(2-√3)}|

tan (45°+30°)= |\frac{(2-√3)-m}{1+m(2-√3)}|

Using the trigonometric identity, 

tan (a+b) = \frac{tan \hspace{0.1cm}a + tan\hspace{0.1cm} b}{1-(tan\hspace{0.1cm} a) (tan\hspace{0.1cm} b)}

\frac{tan 45\degree + tan 30\degree}{1-(tan 45\degree) (tan 30\degree)} = |\frac{(2-\sqrt{3})-m}{1+m(2-\sqrt{3})}|

\frac{1 + \frac{1}{\sqrt{3}}}{1-(1) (\frac{1}{\sqrt{3}})} = |\frac{(2-√3)-m}{1+m(2-√3)}|\\ \frac{\frac{√3+1}{\sqrt{3}}}{\frac{√3-1}{\sqrt{3}}} = |\frac{(2-√3)-m}{1+m(2-√3)}|\\ \frac{√3+1}{√3-1} = |\frac{(2-√3)-m}{1+m(2-√3)}|\\ \frac{(√3+1)^2}{3-1} = |\frac{(2-√3)-m}{1+m(2-√3)}|\\ \frac{(3+1+2√3)^2}{2} = |\frac{(2-√3)-m}{1+m(2-√3)}|\\ 2+√3 = |\frac{(2-√3)-m}{1+m(2-√3)}|

We will have two cases,

2+√3 = \frac{(2-\sqrt{3})-m}{1+m(2-\sqrt{3})}   and 2+√3 = -\frac{(2-\sqrt{3})-m}{1+m(2-\sqrt{3})}

(2+√3)(1+m(2-√3)) = 2-√3-m and (2+√3)(1+m(2-√3)) = -(2-√3-m)

(2+√3+m(22-(√3)2) = 2-√3-m and (2+√3+m(22-(√3)2) = √3-2+m

2+√3+m(4-3) = 2-√3-m and 2+√3+m(4-3) = √3-2+m

2+√3+m+m = 2-√3 and 2+√3+m-m = √3-2

2m = -2-√3+2-√3 and 2+√3 = √3-2

2m = -2√3 and m is not defined

m = -√3 and m is not defined

Hence, the equation of line will be,

y = -√3x and x = 0

Question 3: Find the equation of the straight lines passing through (2,-1) and making an angle of 45° with the line 6x+5y-8=0.

Solution:

As line 1 passes through (2,-1), then it will be in the form of 

y-(-1) = m(x-2)  (m as slope)

y+1 = m(x-2)

For line 2: 6x+5y-8=0

5y = -6x + 8

y = \frac{-6}{5}x + \frac{8}{5}

slope is M = \frac{-6}{5}

Here, it is given that these two lines make an angle of 45° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = 45°, m=m and M = \frac{-6}{5}

tan 45° = |\frac{\frac{-6}{5}-m}{1+m(\frac{-6}{5})}|

1 = |\frac{\frac{-6-5m}{5}}{1-m(\frac{6}{5})}|\\ 1 = |\frac{\frac{-6-5m}{5}}{\frac{5-6m}{5}}|\\ 1 = |\frac{-6-5m}{5-6m}|

We will have two cases,

1 = \frac{-6-5m}{5-6m}   and 1 = -\frac{-6-5m}{5-6m}

5-6m = -6-5m and 5-6m = -(-6-5m)

6m-5m = 5+6 and 5-6m = 5m+6

m = 11 and 5m+6m = 5-6

m = 11 and m = \frac{-1}{11}

Hence, the equation of line will be,

y+1 = 11(x-2) and y+1 = \frac{-1}{11}(x-2)

11x-y-23=0 and 11y+x+9 = 0

Question 4: Find the equation to the straight lines which pass through the point (h,k) and are inclined at angle tan-1 m to the straight line y=mx+c.

Solution:

As line 1 passes through (h,k), then it will be in the form of

y-(k) = M(x-h)  (M as slope)

For line 2: y=mx+c (m as slope)

Here, it is given that these two lines make an angle of tan-1 m between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = tan-1 m,

tan (tan-1 m) = |\frac{M-m}{1+mM}|

m = |\frac{M-m}{1+mM}|

We will have two cases,

m = \frac{M-m}{1+mM}   and m = -\frac{M-m}{1+mM}

m(1+mM) = M-m and m(1+mM) = -(M-m)

m+m2M = M-m and m+m2M = m-M

2m = M-m2M and m2M = -M

M = \frac{2m}{1-m^2}   and M = 0

Hence, the equation of line will be,

y-(k) = 0(x-h) and y-k = \frac{2m}{1-m^2}(x-h)

y-k = 0 and (y-k)(1-m2) = (2m)(x-h)

Question 5: Find the equation to the straight lines passing through the point (2,3) and inclined at 45° to the line 3x+y-5=0.

Solution:

As line 1 passes through (2,3), then it will be in the form of

y-3 = M(x-2)  (M as slope)

For line 2: 3x+y-5=0

y = -3x+5

slope m = -3

Here, it is given that these two lines make an angle of 45° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = 45°,

tan 45° = |\frac{M-(-3)}{1+(-3)M}|

1 = |\frac{M+3}{1-3M}|

We will have two cases,

1 = \frac{M+3}{1-3M}   and 1 = -\frac{M+3}{1-3M}

1-3M = M+3 and 1-3M = -(M+3)

M+3M = 1-3 and 1-3M = -M-3

4M = -2 and 3M-M = 1+3

M = \frac{-2}{4}   and 2M = 4

M = \frac{-1}{2}   and M = 2

Hence, the equation of line will be,

y-3 = \frac{-1}{2}(x-2)   and y-3 = 2(x-2)

2y-6 = -(x-2) and y-3 = 2x-4

x+2y-8=0 and 2x-y-1=0

Question 6: Find the equation to the sides of an isosceles right-angled triangle the equation of whose hypotenuse is 3x+4y=4 and the opposite vertex is the point (2,2).

Solution:

As △ABC is an isosceles right angled triangle at B.

∠A = ∠C = 45°

We can say that, AB and BC makes 45° with AC.

Let the slope of AB as m1 and BC as m2.

AB: (y-2) = m1(x-2)

BC: (y-2) = m2(x-2)

Slope of AC : 3x+4y=4

y = \frac{-3}{4}x + 1

Slope of AC is \frac{-3}{4}.

Here, it is given that these two lines make an angle of 45° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = 45°, and M = \frac{-3}{4}

tan 45° = |\frac{\frac{-3}{4}-m}{1+m(\frac{-3}{4})}|

1 = |\frac{\frac{-3-4m}{4}}{1-m(\frac{3}{4})}|\\ 1 = |\frac{\frac{-3-4m}{4}}{\frac{4-3m}{4}}|\\ 1 = |\frac{-3-4m}{4-3m}|

We will have two cases,

1 = \frac{-3-4m}{4-3m}   and 1 = -\frac{-3-4m}{4-3m}

4-3 = -3-4 and 4-3 = -(-3-4)

4m-3 = -3-4 and 4-3 = 3+4

m = -7 and 4+3 = 4-3

m = -7 and m = \frac{1}{7}

Hence, the equation of lines will be,

AB: (y-2) = m1(x-2)

y-2 = \frac{1}{7}(x-2)

7y-x-12=0

BC: (y-2) = m2(x-2)

y-2 = -7(x-2)

7x+y-16=0

Question 7: The equation of one side of an equilateral triangle is x-y=0 and one vertex is (2+√3,5). Prove that a second side is y+(2-√3)x=6 and find the equation of the third side.

Solution:

As equation of one side of equilateral triangle is x-y=0 and one vertex is (2+√3,5),

As, the point (2+√3,5) does not satisfy x-y=0. Then this is the vertex opposite of the line x-y=0

In equilateral triangle,

∠A = ∠B = ∠C = 60°

We can say that, AC and BC makes 60° with AB.

Let the slope of AC as m1 and BC as m2.

AB: (y-5) = m1(x-(2+√3))

BC: (y-5) = m2(x-(2+√3))

Slope of AC : x-y=0

y = x

Slope of AC is 1.

Here, it is given that these two lines make an angle of 60° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = 60°, m = m1 and m2 and M = 1

tan 60° = |\frac{1-m}{1+m(1)}|

√3= |\frac{1-m}{1+m}|

We will have two cases,

√3 = \frac{1-m}{1+m}   and √3 = -\frac{1-m}{1+m}

m = \frac{1-\sqrt{3}}{1+\sqrt{3}}   and √3(1+m) = -(1-m)

m = \frac{1-\sqrt{3}}{1+\sqrt{3}}   and √3+√3m = m-1

m = \frac{1-\sqrt{3}}{1+\sqrt{3}}   and √3m-m = -1-√3

m = \frac{1-\sqrt{3}}{1+\sqrt{3}}   and m = \frac{1+\sqrt{3}}{1-\sqrt{3}}

After rationalizing, we get

m = \frac{(1-\sqrt{3})^2}{1-3}   and m = \frac{(1+\sqrt{3})^2}{1-3}

m = \frac{1+3-2\sqrt{3}}{-2}   and m = \frac{1+3+2\sqrt{3}}{-2}

m = -(2-√3) and m = -(2+√3)

Hence, the equation of lines will be,

AB: (y-5) = m1(x-(2+√3))

y-5 = -(2-√3)(x-(2+√3))

y-5 = -(2-√3)x+ (22-(√3)2)

(2-√3)x+y-6 = 0

BC: (y-5) = m2(x-(2+√3))

(y-5) = -(2+√3)(x-(2+√3))

y-5 = -(2+√3)x+(2+√3)^2

(2+√3)x+y-5 = 4+3+4√3

(2+√3)x+y = 12+4√3

Question 8: Find the equation of the two straight lines passing through (1,2) forming two sides of a square of which 4x+7y=12 is one diagonal.

Solution:

Let the point opposite of diagonal 4x+7y=12 be C(1,2)

Here, △BCD form an isosceles right angled triangle at C.

∠B = ∠D = 45°

We can say that, CD and BC makes 45° with BD.

Let the slope of CD as m1 and BC as m2.

CD: (y-2) = m1(x-1)

BC: (y-2) = m2(x-1)

Slope of BD : 4x+7y=12

y = \frac{-4}{7}x + 3

Slope of BD is \frac{-4}{7}.

Here, it is given that these two lines make an angle of 45° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = 45° and M = \frac{-4}{7}

tan 45° = |\frac{\frac{-4}{7}-m}{1+m(\frac{-4}{7})}|

1 = |\frac{\frac{-4-7m}{7}}{1-m(\frac{4}{7})}|\\ 1 = |\frac{\frac{-4-7m}{7}}{\frac{7-4m}{7}}|\\ 1 = |\frac{-4-7m}{7-4m}|

We will have two cases,

1 = \frac{-4-7m}{7-4m}   and 1 = -\frac{-4-7m}{7-4m}

7-4m = -4-7m and 7-4m = -(-4-7m)

7m-4m = -4-7 and 7-4m = 4+7m

3m = -11 and 7m+4m = 7-4

m = \frac{-11}{3}   and 11m = 3

m = \frac{-11}{3}   and m = \frac{3}{11}

Hence, the equation of lines will be,

CD: (y-2) = m1(x-1)

y-2 = \frac{-11}{3}(x-1)

3y-6 = -11x+11

11x+3y-17=0

BC: (y-2) = m2(x-1)

y-2 = \frac{3}{11}(x-1)

11y-22 = 3x-3

3x-11y+19=0

Question 9: Find the equation of the two straight lines passing through (1,2) and making an angle of 60° with the line x+y=0. Find also the area of the triangle formed by the three lines.

Solution:

As equation of one side of equilateral triangle is x+y=0 and one vertex is (1,2),

As, the point (1,2) does not satisfy x+y=0. Then this is the vertex opposite of the line x+y=0

Hence, the lines make an equilateral triangle,

∠A = ∠B = ∠C = 60°

We can say that, AC and BC makes 60° with AB.

Let the slope of AC as m1 and BC as m2.

AB: (y-2) = m1(x-1)

BC: (y-2) = m2(x-1)

Slope of AC : x+y=0

y = -x

Slope of AC is -1.

Here, it is given that these two lines make an angle of 60° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = 60° and M = -1

tan 60° = |\frac{-1-m}{1+m(-1)}|

√3= |\frac{-1-m}{1-m}|

We will have two cases,

√3 = \frac{-1-m}{1-m}   and √3 = -\frac{-1-m}{1-m}

√3(1-m) = -1-m and √3(1-m) = 1+m

√3-√3m = -1-m and √3-√3m = m+1

√3m-m = √3+1 and +√3m+m = √3-1

m = \frac{\sqrt{3}+1}{\sqrt{3}-1}   and m = \frac{\sqrt{3}-1}{\sqrt{3}+1}

After rationalizing, we get

m = \frac{(\sqrt{3}+1)^2}{3-1}   and m = \frac{(\sqrt{3}-1)^2}{3-1}

m = \frac{1+3+2\sqrt{3}}{2}   and m = \frac{1+3-2\sqrt{3}}{2}

m = 2+√3 and m = 2-√3

Hence, the equation of lines will be,

AB: (y-2) = (2+√3)(x-1)

y-2=(2+√3)x-2-√3

(2+√3)x-y-√3=0 ……………….(i)

BC: (y-2) = (2-√3)(x-1)

y-2=(2-√3)x-2+√3

(2-√3)x-y+√3=0 ……………….(ii)

Using (i) and x+y=0, we get

A = (\frac{-1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2})

C = (1,2)

Using distance formula, AC

AC = \sqrt{(1-\frac{-1-\sqrt{3}}{2})^2+(2-\frac{1+\sqrt{3}}{2})^2}\\ AC = \sqrt{(\frac{2+1+\sqrt{3}}{2})^2+(\frac{4-1-\sqrt{3}}{2})^2}\\ AC = \sqrt{(\frac{3+\sqrt{3}}{2})^2+(\frac{3-\sqrt{3}}{2})^2}\\ AC = \sqrt{\frac{9+3+6\sqrt{3}+9+3-6\sqrt{3}}{4}}\\ AC = \sqrt{\frac{24}{4}}\\ AC = \sqrt{6}

Area of equilateral triangle ABC,

= \frac{\sqrt{3}}{4}(side)^2\\ = \frac{\sqrt{3}}{4}(\sqrt{6})^2\\ = \frac{\sqrt{3}}{4}(6)\\ = \frac{3\sqrt{3}}{2}   sq. unit

Question 10: Two sides of an isosceles triangle are given by the equations 7x-y+3=0 and x+y-3=0 and its third side passes through the point (1,-10). Determine the equation of the third side.

Solution:

Let the equation of the line AB and AC 7x-y+3=0 and x+y-3=0 respectively.

∠B = ∠C

Slope of line AB: 7x-y+3=0

y = 7x+3

Slope m1 = 7

Slope of line AC: x+y-3=0

y = -x+3

Slope m2 = -1

Let the slope of line BC be m, which passes through the point (1,-10)

y-(-10) = m(x-1)

y+10 = m(x-1)

The angle between the lines AB and BC is equal to the angle between the lines AC and BC, say θ

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

By taking positive sign

tan θ = \frac{m-7}{1+7m} = \frac{m-(-1)}{1+m(-1)}

tan θ = \frac{m-7}{1+7m} = \frac{m+1}{1-m}

Solving it, we get

m = -3 or \frac{1}{3}

By taking negative sign

tan θ = \frac{m-7}{1+7m} = -\frac{m-(-1)}{1+m(-1)}

tan θ = \frac{m-7}{1+7m} = -\frac{m+1}{1-m}

Solving it, we get

m2 = -1 (which is not possible)

Line BC: when m = -3

y+10 = -3(x-1)

y+10 = -3x+3

3x+y-13=0

Line BC: when m = \frac{1}{3}

y+10 = \frac{1}{3}(x-1)

3y+30 = x-1

x-3y-31=0

Question 11: Show that the point (3,-5) lies between the parallel lines 2x+3y-7=0 and 2x+3y+12=0 and find the equation of lines through (3,-5) cutting the above lines at an angle of 45°.

Solution:

Let the line 1: 2x+3y-7=0

Line 2: 2x+3y+12=0

As the slope of line 1 and line 2 is same, these lines are parallel.

Slope of lines M = \frac{-2}{3}

To check whether (3,-5) lies between these lines

Taking x=3 and y=-5

(2x+3y-7)(2x+3y+12)<0 (If its true then the point lies between these lines)

(2(3)+3(-5)-7)(2(3)+3(-5)+12)

(6-15-7)(6-15+12)

(-16)(3) <0, which is a negative value.

Hence, the point (3,-5) lies between the lines.

Let the slope of the line be m, which is passing through the point (3,-5)

y-(-5) = m(x-3)

y+5 = m(x-3)

As it is given that the line is cutting above lines at an angle of 45°.

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Taking θ = 45° and M = \frac{-2}{3}

By taking positive sign

tan 45° = |\frac{\frac{-2}{3}-m}{1+m(\frac{-2}{3})}|

1 = |\frac{\frac{-2-3m}{3}}{1-m(\frac{2}{3})}|

1 = |\frac{\frac{-2-3m}{3}}{\frac{3-2m}{3}}|

1 = |\frac{-2-3m}{3-2m}|

We will have two cases,

1 = \frac{-2-3m}{3-2m}   and 1 = -\frac{-2-3m}{3-2m}

3-2m = -2-3m and 3-2m = -(-2-3m)

3m-2m = -2-3 and 3-2m = 2+3m

m = -5 and 3m+2m = 3-2

m = -5 and m = \frac{1}{5}

Hence, the equation of line will be,

when m = -5

y+5 = (-5)(x-3)

y+5 = -5x+15

5x+y-10=0

when m = \frac{1}{5}

y+5 = \frac{1}{5}  (x-3)

5y+25 = x-3

x-5y-28=0

Question 12: The equation of the base of an equilateral triangle is x+y=2 and its vertex is (2,-1). Find the length and equations of its sides.

Solution:

∠A = ∠B = ∠C = 60°

We can say that, AC and AB makes 60° with BC.

Let the slope of AC as m1 and AB as m2 which passes through the point (2,-1).

AB: (y-(-1)) = m1(x-2)

y+1 = m1(x-2)

AC: (y-(-1)) = m2(x-2)

y+1 = m2(x-2)

Slope of BC : x+y=2

y = -x+2

Slope of BC is -1.

Here, it is given that these two lines make an angle of 60° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = 60°, m = m1 and m2 and M = -1

tan 60° = |\frac{-1-m}{1+m(-1)}|

√3= |\frac{-1-m}{1-m}|

We will have two cases,

√3 = \frac{-1-m}{1-m}   and √3 = -\frac{-1-m}{1-m}

√3(1-m) = -1-m and √3(1-m) = 1+m

√3-√3m = -1-m and √3-√3m = m+1

√3m-m = √3+1 and +√3m+m = √3-1

m = \frac{\sqrt{3}+1}{\sqrt{3}-1}   and m = \frac{\sqrt{3}-1}{\sqrt{3}+1}

After rationalizing, we get

m = \frac{(\sqrt{3}+1)^2}{3-1}   and m = \frac{(\sqrt{3}-1)^2}{3-1}

m = \frac{1+3+2\sqrt{3}}{2}   and m = \frac{1+3-2\sqrt{3}}{2}

m = 2+√3 and m = 2-√3

Hence, the equation of lines will be,

AB: y+1 = m1(x-2)

y+1 = (2+√3)(x-2)

y+1 = (2+√3)x-4-2√3

(2+√3)x-y-5-2√3=0 ……………….(i)

AC: y+1 = m2(x-2)

y+1 = (2-√3)(x-2)

y+1 = (2-√3)x-4+2√3

(2-√3)x-y-5+2√3=0 ……………….(ii)

Using (ii) and x+y=2, we get

C = (\frac{15+\sqrt{3}}{6},-\frac{3+\sqrt{3}}{6})

A = (2,-1)

Using distance formula, AC

AC = \sqrt{(2-\frac{15+\sqrt{3}}{6})^2+(-1-(-\frac{3+\sqrt{3}}{6}))^2}\\ AC = \sqrt{(\frac{12-15-\sqrt{3}}{6})^2+(\frac{-6+3+\sqrt{3}}{6})^2}\\ AC = \sqrt{(\frac{-3-\sqrt{3}}{6})^2+(\frac{-3+\sqrt{3}}{6})^2}\\ AC = \sqrt{\frac{9+3+6\sqrt{3}+9+3-6\sqrt{3}}{36}}\\ AC = \sqrt{\frac{24}{36}}\\ AC = \sqrt{\frac{2}{3}}

AC = AB = BC = \sqrt{\frac{2}{3}}

Question 13: If two opposite vertices of a square are (1,2) and (5,8), find the coordinates of its other two vertices and the equations of its sides.

Solution:

Let’s consider a square ABCD.

According to the property of square, the diagonal bisects the angle.

Hence, ∠AOB = ∠AOD = 45°

Slope of the diagonal AC:

\frac{8-2}{5-1} = \frac{6}{4} = \frac{3}{2}

Let the slope of the line AB and AD be m1 and m2, which passes through the point (1,2)

AB: y-2 = m1(x-1)

AD: y-2 = m2(x-1)

Here, it is given that these two lines make an angle of 45° with AC. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

\mathbf{tan θ = |\frac{M-m}{1+mM}|}

Here, we have θ = 45°, m = m1 and m2 and M = \frac{3}{2}

tan 45° = |\frac{\frac{3}{2}-m}{1+m(\frac{3}{2})}|

1 = |\frac{\frac{3-2m}{2}}{\frac{2+3m}{2}}|

1= |\frac{3-2m}{2+3m}|

We will have two cases,

1 = \frac{3-2m}{2+3m}   and 1 = -\frac{3-2m}{2+3m}

2+3m = 3-2m and 2+3m = -(3-2m)

3m+2m = 3-2 and 2+3m = 2m-3

5m = 1 and 3m-2m = -3-2

m = \frac{1}{5}   and m = -5

m = \frac{1}{5}   or -5

Hence, the equation of lines will be,

AB: y-2 = m1(x-1)

y-2 = \frac{1}{5}(x-1)

5y-10 = x-1

x-5y+9=0 ……………….(i)

AD: y-2 = m2(x-1)

y-2 = -5(x-1)

y-2 = -5x+5

5x+y-7=0 ……………….(ii)

As BC is parallel to AD, So 

The equation BC will be 5x+y+λ=0 and as BC passes through C(5,8), we get

5(5)+(8)+λ=0

33+λ=0

λ = -33

Hence, Equation of BC is 5x+y-33=0 ……………….(iii)

Now as CD is parallel to AB, So

The equation CD will be x-5y+λ=0 and as CD passes through C(5,8), we get

5-5(8)+λ=0

-35+λ=0

λ = 35

Hence, Equation of CD is x-5y+35=0 ………………………(iv)

Solving (i) and (iii), we get

B(6,3)

Solving (ii) and (iv), we get

D(0,7)

Hence, the equation of lines are

AB = x-5y+9=0

BC = 5x+y-33=0

CD = x-5y+35=0

AD = 5x+y-7=0

And the vertices of square are

A(1,2), B(6,3), C(5,8) and D(6,3)


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