Question 1. Find the values of k for which the line (k – 3)x – (4 – k2)y + k2 – 7k + 6 = 0 is(a) Parallel to the x-axis
(b) Parallel to the y-axis
(c) Passing through the origin.
Solution:
We are given the line, (k – 3)x – (4 – k2)y + k2 – 7k + 6 = 0
=> (4 – k2)y = (k – 3)x + k2 – 7k + 6
=> y =[Tex]\frac{(k–3)x}{4–k^2} [/Tex]+[Tex]\frac{k^2–7k+6}{4-x^2}[/Tex]
Now the equation of the line is of the form y = mx + c where m is the slope of the line and c is its y-intercept.
So, we get m =[Tex]\frac{k–3}{4–k^2} [/Tex]and c =[Tex]\frac{k^2–7k+6}{4-x^2}[/Tex]
(a) Parallel to the x-axis
If the line is parallel to the x-axis, then we have,
Slope of the line = Slope of the x-axis
=> m = 0
=>[Tex]\frac{k–3}{4–k^2} [/Tex]= 0
=> k – 3 = 0
=> k = 3
Therefore, if the given line is parallel to the x-axis, then the value of k is 3.
(b) Parallel to the y-axis
If the line is parallel to the y-axis, then we have,
Slope of the line = Slope of the y-axis
=> m = ∞ (undefined)
=>[Tex]\frac{k–3}{4–k^2} [/Tex]= ∞
=> k2 − 4 = 0
=> k2 = 4
=> k = ±2
Therefore, if the given line is parallel to the y-axis, then the value of k is ± 2.
(c) Passing through the origin.
If the line is passing through the origin,
Y-intercept = 0
=> c = 0
=> k2 – 7k + 6 = 0
=> (k – 6) (k – 1) = 0
=> k = 1 or k = 6
Therefore, if the given line is passing through the origin, then the value of k is either 1 or 6.
Question 2. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and –6, respectively.Solution:
Let’s suppose the intercepts cut by the given lines on the axes are a and b. According to the question, we have,
=> a + b = 1 . . . . (1)
=> ab = – 6 . . . . (2)
By solving both the equations we get
a = 3 and b = –2 or a = –2 and b = 3
We know that the equation of the line whose intercepts are a and b axes is,
bx + ay – ab = 0
When a = 3 and b = –2
So the equation of the line is – 2x + 3y + 6 = 0, i.e. 2x – 3y = 6.
When a = –2 and b = 3
So the equation of the line is 3x – 2y + 6 = 0, i.e. –3x + 2y = 6.
Therefore, the required equation of the lines are 2x – 3y = 6 and –3x + 2y = 6.
Question 3. What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units?Solution:
Suppose (0, b) is the point on the y-axis whose distance from line x/3 + y/4 = 1 is 4 units.
The line can be written as 4x + 3y – 12 = 0
By comparing our equation to the general equation of line Ax + By + C = 0, we get
A = 4, B = 3 and C = –12
Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from (x1, y1) is given by,
d =[Tex]\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}[/Tex]
We are given d = 4. For the point (0, b), the value of d becomes,
=>[Tex]\frac{|4(0)+3b-12|}{\sqrt{3^2+4^2}} [/Tex]= 4
=>[Tex]\frac{|3b-12|}{5} [/Tex]= 4
=> |3b – 12| = 20
=> 3b – 12 = 20 or 3b – 12 = –20
=> b = 32/3 or b = –8/3
Therefore, (0, 32/3) and (0, –8/3) are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units.
Question 4. Find the perpendicular distance from the origin to the line joining the points (cos θ, sin θ) and (cos Ø, sin Ø).Solution:
The equation of the line joining the points (cos θ, sin θ) and (cos Ø, sin Ø) is given by,
=> y – sin θ =[Tex]\frac{sin Ø-sin θ}{cos Ø-cos θ} [/Tex](x – cosθ)
=> y(cos Ø – cos θ) – sin θ(cos Ø – cos θ) = x(sin Ø – sin θ) – cos θ(sin Ø – sin θ)
=> x(sin Ø – sin θ) + y(cos Ø – cos θ) + cos θ sin Ø – sin θ cos θ – sin θ cos Ø + sin θ cos θ = 0
=> x(sin Ø – sin θ) + y(cos Ø – cos θ) + sin (Ø – θ) = 0
So, we get, A = sin Ø – sin θ, B = cos Ø – cos θ and C = sin (Ø – θ).
Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from the origin (0, 0) is given by,
d =[Tex]\frac{|C|}{\sqrt{A^2+B^2}}[/Tex]
=[Tex]\frac{|sin(Ø – θ)|}{\sqrt{(sinØ–sinθ)^2+(cosØ–cosθ)^2}}[/Tex]
=[Tex]\frac{|sin(Ø – θ)|}{\sqrt{sin^2Ø+sin^2θ-2sinØsinθ+cos^2Ø+cos^2θ-2cosØcosθ}}[/Tex]
=[Tex]\frac{|sin(Ø – θ)|}{\sqrt{2-2(sinØsinθ+cosØcosθ)}}[/Tex]
=[Tex]\frac{|sin(Ø – θ)|}{\sqrt{2(1-cos(Ø-θ))}}[/Tex]
=[Tex]\frac{|sin(Ø – θ)|}{\sqrt{2(2sin^2(\frac{Ø-θ}{2}))}}[/Tex]
=[Tex]\frac{|sin(Ø – θ)|}{|2sin(\frac{Ø-θ}{2})|}[/Tex]
Therefore,[Tex]\frac{|sin(Ø – θ)|}{|2sin(\frac{Ø-θ}{2})|} [/Tex]is the perpendicular distance from the origin to the given line.
Question 5. Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.Solution:
Two given lines are
x – 7y + 5 = 0 . . . . (1)
3x + y = 0 . . . . (2)
By solving equations (1) and (2) we get
x = −5/22 and y = 15/22
(−5/ 22, 15/22) is the point of intersection of lines (2) and (3)
Now the equation of any line parallel to the y-axis is of the form
x = a . . . . (1)
If the line x = a passes through point (−5/22, 15/22) we get a = −5/22.
Therefore, the required equation of the line is x = −5/22.
Question 6. 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:
Given line is, x/4 + y/6 = 1.
=> 3x + 2y – 12 = 0
=> y = −3/2 x + 6, which is of the form y = mx + c
Here the slope of the given line = −3/2
So the slope of line perpendicular to the given line = −1/(−3/2) = 2/3
Suppose the given line intersects the y-axis at (0, y). So, the equation of the given line becomes,
=> y/6 = 1
=> y = 6
Hence, the given line intersects the y-axis at (0, 6). We know that the equation of the line that has a slope of 2/3 and passes through point (0, 6) is given by,
=> (y – 6) = 2/3 (x – 0)
=> 3y – 18 = 2x
=> 2x – 3y + 18 = 0
Therefore, the required equation of the line is 2x – 3y + 18 = 0.
Question 7. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.Solution:
It is given that
y – x = 0 . . . . (1)
x + y = 0 . . . . (2)
x – k = 0 . . . . (3)
Here the point of intersection of lines (1) and (2) is x = 0 and y = 0.
And the point of intersection of lines (2) and (3) is x = k and y = – k, lines (3) and (1) is x = k and y = k.
So the vertices of the triangle formed by the three given lines are (0, 0), (k, –k) and (k, k).
Here the area of triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is
A =[Tex]\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|[/Tex]
=[Tex]\frac{1}{2}|0 (-k – k) + k (k – 0) + k (0 + k)|[/Tex]
=[Tex]\frac{1}{2}|k^2+k^2|[/Tex]
=[Tex]\frac{1}{2}|2k^2|[/Tex]
= k2 square units
Therefore, k2 sq. units is the area of the triangle formed by the given lines.
Question 8. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.Solution:
It is given that
3x + y – 2 = 0 . . . . (1)
px + 2y – 3 = 0 . . . . (2)
2x – y – 3 = 0 . . . . (3)
By solving equations (1) and (3) we get
x = 1 and y = –1
It is given that the three lines intersect at one point and the point of intersection of lines (1) and (3) will also satisfy line (2).
=> p (1) + 2 (–1) – 3 = 0
=> p – 2 – 3 = 0
=> p = 5
Therefore, the value of p is 5.
Question 9. If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0.Solution:
It is given that
y = m1x + c1 . . . . (1)
y = m2x + c2 . . . . (2)
y = m3x + c3 . . . . (3)
By subtracting equation (1) from (2), we get,
=> 0 = (m2 – m1) x + (c2 – c1)
=> (m1 – m2) x = c2 – c1
=> x =[Tex]\frac{c_2-c_1}{m_1–m_2}[/Tex]
And y =[Tex]\frac{m_1(c_2-c_1)}{m_1–m_2} [/Tex]+ c1
=> y =[Tex]\frac{m_1c_2-m_1c_1+m_1c_1-m_2c_1}{m_1–m_2}[/Tex]
=> y =[Tex]\frac{m_1c_2-m_2c_1}{m_1–m_2}[/Tex]
Hence, ([Tex]\frac{c_2-c_1}{m_1–m_2} [/Tex],[Tex]\frac{m_1c_2-m_2c_1}{m_1–m_2} [/Tex]) is the point of intersection of lines (1) and (2).
As the given three lines are concurrent, this point must satisfy the equation (3).
=>[Tex]\frac{m_1c_2-m_2c_1}{m_1–m_2}=\frac{m_3(c_2-c_1)}{m_1–m_2}+c_3[/Tex]
=>[Tex]\frac{m_1c_2-m_2c_1}{m_1–m_2}=\frac{m_3c_2-m_3c_1+m_1c_3-m_2c_3}{m_1–m_2}[/Tex]
=>[Tex]m_1c_2-m_2c_1-m_3c_2+m_3c_1-m_1c_3+m_2c_3=0[/Tex]
=> m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0
Hence proved.
Question 10. Find the equation of the lines through the point (3, 2) which makes an angle of 45° with the line x – 2y = 3.Solution:
Suppose a is the slope of the line which passes through the point (3, 2).
Given line is x – 2y = 3.
y = 1/2 x – 3/2 which is of the form y = mx + c.
So, the slope of the given line b = 1/2
We know that the angle between the required line and line x – 2y = 3 is 45o. The angle is given by,
tan θ =[Tex]|\frac{\frac{1}{2}-a}{1+\frac{a}{2}}|[/Tex]
=> tan 450 =[Tex]|\frac{\frac{1}{2}-a}{1+\frac{a}{2}}|[/Tex]
=>[Tex]|\frac{\frac{1}{2}-a}{1+\frac{a}{2}}| [/Tex]= 1
=>[Tex]\frac{1-2a}{2+a}=\pm1[/Tex]
=> 2 + a = 1 – 2a or 2 + a = – 1 + 2a
=> a = –1/3 or a = 3
When a = 3, the equation of the line passing through (3, 2) and having a slope 3 is,
=> y – 2 = 3 (x – 3)
=> y – 2 = 3x – 9
=> 3x – y = 7
When a = –1/3, the equation of the line passing through (3, 2) and having a slope –1/3 is
=> y – 2 = –1/3 (x – 3)
=> 3y – 6 = – x + 3
=> x + 3y = 9
Therefore, the equations of the lines are 3x – y = 7 and x + 3y = 9.
Question 11. Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.Solution:
Suppose the equation of the line having equal intercepts on the axes as
=> x/a + y/a = 1
=> x + y = a . . . . (1)
By solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we get,
x = 1/13 and y = 5/13
(1/13, 5/13) is the point of intersection of two given lines.
Putting in (1), we get,
=> a = 1/13 + 5/13
=> a = 6/13
Here the equation (1) becomes
=> x + y = 6/13
=> 13x + 13y = 6
Therefore, the required equation of the line is 13x + 13y = 6.
Question 12. Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is[Tex]\frac{y}{x}=\frac{m\pm tanθ}{1\mp tanθ}[/Tex].Solution:
Suppose y = m1x is the equation of the line passing through the origin. Hence, slope of this line is,
m1 =[Tex]\frac{y}{x}[/Tex]
Given line is y = mx + c whose slope is m. It makes an angle θ with the line y = m1x.
=> tan θ =[Tex]|\frac{m_1-m}{1+m_1m}|[/Tex]
=>[Tex]|\frac{\frac{y}{x}-m}{1+\frac{my}{x}}| [/Tex]= tan θ
=>[Tex]\pm\left(\frac{\frac{y}{x}-m}{1+\frac{my}{x}}\right) [/Tex]= tan θ
=>[Tex]tanθ+\frac{y}{x}mtanθ=\frac{y}{x}-m [/Tex]or[Tex]tanθ+\frac{y}{x}mtanθ=-\frac{y}{x}+m[/Tex]
=>[Tex]\frac{y}{x}(1-mtanθ)=m+tanθ [/Tex]or[Tex]\frac{y}{x}(1+mtanθ)=m-tanθ[/Tex]
=>[Tex]\frac{y}{x}=\frac{m+tanθ}{1-mtanθ} [/Tex]or[Tex]\frac{y}{x}=\frac{m-tanθ}{1+mtanθ}[/Tex]
=>[Tex]\frac{y}{x}=\frac{m\pm tanθ}{1\mp tanθ}[/Tex]
Hence proved.
Question 13. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?Solution:
We know that the equation of the line joining (–1, 1) and (5, 7) is given by,
=> y – 1 =[Tex]\frac{7-1}{5+1} [/Tex](x + 1)
=> y – 1 = x + 1
=> x – y + 2 = 0 . . . . (1)
We are given the line, x + y = 4 . . . . (2)
Solving (1) and (2), we get the point of intersection of these lines,
x = 1 and y = 3
Suppose the point (1, 3) divides the line joining (–1, 1) and (5, 7) in the ratio 1 : k.
Using section formula, we get,
(1, 3) =[Tex]\left(\frac{k(-1)+1(5)}{1+k},\frac{k(1)+1(7)}{1+k}\right)[/Tex]
(1, 3) =[Tex]\left(\frac{5-k}{1+k},\frac{k+7}{1+k}\right)[/Tex]
We get,
=> – k + 5 = 1 + k
=> 2k = 4
=> k = 2
Therefore, the line joining the points (–1, 1) and (5, 7) is divided by the line x + y = 4 in the ratio 1 : 2.
Question 14. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.Solution:
We are given,
2x – y = 0 . . . . (1)
4x + 7y + 5 = 0 . . . . (2)
Solving (1) and (2), we get the point of intersection of lines,
x = –5/18 and y = –5/9.
 So, we get B (–5/18, –5/9) and we have A (1, 2). Now we have to find the distance AB.
Using the distance formula, we get,
AB =[Tex]\sqrt{\left(1+\frac{5}{18}\right)^2+\left(2+\frac{5}{9}\right)^2}{}[/Tex]
=[Tex]\sqrt{\left(\frac{23}{18}\right)^2+\left(\frac{23}{9}\right)^2}[/Tex]
=[Tex]\sqrt{\left(\frac{23}{9}\right)^2\left(\frac{1}{2}\right)^2+\left(\frac{23}{9}\right)^2}[/Tex]
=[Tex]\frac{23}{9}\sqrt{\frac{1}{4}+1}[/Tex]
=[Tex]\frac{23}{9}\sqrt{\frac{5}{4}}[/Tex]
=[Tex]\frac{23\sqrt{5}}{18} [/Tex]units
Therefore, the required distance is[Tex]\frac{23\sqrt{5}}{18} [/Tex]units.
Question 15. Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.Solution:
Suppose y = mx + c is the line passing through the point (–1, 2). So we get
=> 2 = m (–1) + c
=> 2 = –m + c
=> c = m + 2
Putting the value of c in the equation, we get,
y = mx + m + 2 . . . . (1)
Now the given line is
x + y = 4 . . . . (2)
By solving both the equations we get the point of intersection of these lines,
x =[Tex]\frac{2-m}{m+1} [/Tex], y =[Tex]\frac{5m+2}{m+1}[/Tex]
Now the point[Tex]\left(\frac{2-m}{m+1},\frac{5m+2}{m+1}\right) [/Tex]is at a distance of 3 units from the point (–1, 2).
From distance formula, we get,
=>[Tex]\sqrt{\left(\frac{2-m}{m+1}+1\right)^2+\left(\frac{5m+2}{m+1}-2\right)^2}=3[/Tex]
=>[Tex]\sqrt{\left(\frac{2-m+m+1}{m+1}\right)^2+\left(\frac{5m+2-2m-2}{m+1}\right)^2}=3[/Tex]
=>[Tex]\sqrt{\frac{9}{(m+1)^2}+\frac{9m^2}{(m+1)^2}}=3[/Tex]
=>[Tex]\frac{9}{(m+1)^2}+\frac{9m^2}{(m+1)^2}=9[/Tex]
=>[Tex]\frac{1}{(m+1)^2}+\frac{m^2}{(m+1)^2}=1[/Tex]
=>[Tex]\frac{1+m^2}{(m+1)^2}=1[/Tex]
=> 1 + m2 = m2 + 1 + 2m
=> 2m = 0
=> m = 0
As the slope of the required line is zero, the line must be parallel to the x-axis.
Question 16. The hypotenuse of a right-angled triangle has its ends at the points (1, 3) and (−4, 1). Find the equation of the legs (perpendicular sides) of the triangle.Solution:
Suppose △ABC is right-angled triangle where ∠ C = 90o and m is the slope of AC.
So the slope of BC = –1/m. And the equation of AC is given by,
=> y – 3 = m (x – 1)
=> x – 1 = 1/m (y – 3)
We know the equation of BC is given by,
=> y – 1 = – 1/m (x + 4)
=> x + 4 = – m (y – 1)
If m = 0, we get the equations of perpendicular sides,
y – 3 = 0 and x + 4 = 0 or,
y = 3 and x = –4
If m = ∞, we get the equations,
x – 1 = 0 and y – 1 = 0 or,
x = 1 and y = 1
Question 17. 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:
We are given that
x + 3y = 7 . . . . (1)
Suppose B (a, b) is the image of point A (3, 8). Hence line (1) is the perpendicular bisector of AB.
 Slope of line (1) = −1/3
And slope of line AB =[Tex]\frac{b-8}{a-3}[/Tex]
We know, AB and line (1) are perpendicular, so we have,
=>[Tex](\frac{b-8}{a-3})×\frac{-1}{3}=-1[/Tex]
=>[Tex]\frac{b-8}{3a-9}=1[/Tex]
=> 3a − b = 1 . . . . (2)
Now, mid-point of AB =[Tex]\left(\frac{a+3}{2},\frac{b+8}{2}\right) [/Tex]. As the mid-point is satisfying line (1), we get,
=>[Tex](\frac{a+3}{2})+3(\frac{b+8}{2})=7[/Tex]
=> a + 3 + 3b + 24 = 14
=> a + 3b = −13 . . . . (3)
Solving (2) and (3), we get, a = –1 and b = –4
Therefore, the image of the given point with respect to the given line is (–1, –4).
Question 18. If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.Solution:
We are given,
y = 3x + 1 . . . . (1)
2y = x + 3 . . . . (2)
y = mx + 4 . . . . (3)
So the slope of line (1), a = 3,
line (2), b = 1/2
line (3), c = m
It is given that the lines (1) and (2) are equally inclined to line (3) which implies that the angle between lines (1) and (3) equals the angle between lines (2) and (3).
=>[Tex]|\frac{a-c}{1+ac}|=|\frac{b-c}{1+bc}|[/Tex]
=>[Tex]|\frac{3-m}{1+3m}|=|\frac{\frac{1}{2}-m}{1+\frac{m}{2}}|[/Tex]
=>[Tex]|\frac{3-m}{1+3m}|=|\frac{1-2m}{2+m}|[/Tex]
=>[Tex]\frac{3-m}{1+3m}=\pm\left(\frac{1-2m}{2+m}\right)[/Tex]
=>[Tex]\frac{3-m}{1+3m}=\frac{1-2m}{2+m} [/Tex]or[Tex]\frac{3-m}{1+3m}=\frac{2m-1}{2+m}[/Tex]
=>[Tex]–m^2+m+6=1+m–6m^2 [/Tex]or[Tex]–m^2+m+6=-1-m+6m^2[/Tex]
=> 5m2 + 5 = 0 (ignored) or 7m2 − 2m − 7 = 0
=> 7m2 − 2m − 7 = 0
=> m =[Tex]\frac{2\pm2\sqrt{1+49}}{14}[/Tex]
=> m =[Tex]\frac{1\pm5\sqrt{2}}{7}[/Tex]
Hence the required value of m is[Tex]\frac{1\pm5\sqrt{2}}{7}[/Tex].
Question 19. If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.Solution:
We are given lines,
x + y – 5 = 0 . . . . (1)
3x – 2y + 7 = 0 . . . . (2)
Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from (x, y) is given by,
d =[Tex]\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}[/Tex]
So, perpendicular distance(d1) of a line x + y – 5 = 0 from (x, y) is,
[Tex]d_1=\frac{|x+y-5|}{\sqrt{1^2+1^2}}[/Tex]
=[Tex]\frac{|x+y-5|}{\sqrt{2}}[/Tex]
And the perpendicular distance(d2) of a line 3x – 2y + 7 = 0 from (x, y) is,
[Tex]d_2=\frac{|3x-2y+7|}{\sqrt{3^2+(-2)^2}}[/Tex]
=[Tex]\frac{|3x-2y+7|}{\sqrt{13}}[/Tex]
According to the question,
=> d1 + d2 = 10
=>[Tex]\frac{|x+y-5|}{\sqrt{2}}+\frac{|3x-2y+7|}{\sqrt{13}}=10[/Tex]
=>[Tex]\sqrt{13}|x+y-5|+\sqrt{2}|3x-2y+7|-10\sqrt{26}=0[/Tex]
=>[Tex]\sqrt{13}(x+y-5)+\sqrt{2}(3x-2y+7)-10\sqrt{26}=0[/Tex]
=>[Tex]\sqrt{13}x+\sqrt{13}y-5\sqrt{13}+3\sqrt{2}x-2\sqrt{2}y+7\sqrt{2}-10\sqrt{26}=0[/Tex]
=>[Tex](\sqrt{13}+3\sqrt{2})x+(\sqrt{13}-2\sqrt{2})y+(7\sqrt{2}-5\sqrt{13}-10\sqrt{26})=0[/Tex]
As the equation represents a line, P must move on a line.
Hence, proved.
Question 20. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.Solution:
We are the given the lines,
9x + 6y – 7 = 0 . . . . (1)
3x + 2y + 6 = 0 . . . . (2)
Let (a, b) be the point lying on the line which is equidistant from (1) and (2).
Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from (x, y) is given by,
[Tex]d = \frac{|Ax+By+C|}{\sqrt{A^2+B^2}}[/Tex]
So, perpendicular distance(d1) of a line 9x + 6y – 7 = 0 from (a, b) is,
[Tex]d_1=\frac{|9a+6b-7|}{\sqrt{9^2+6^2}}[/Tex]
=[Tex]\frac{|9a+6b-7|}{3\sqrt{13}}[/Tex]
And the perpendicular distance(d2) of a line 3x + 2y + 6 = 0 from (a, b) is,
[Tex]d_2=\frac{|3a+2b+6|}{\sqrt{3^2+2^2}}[/Tex]
=[Tex]\frac{|3a+2b+6|}{\sqrt{13}}[/Tex]
According to the question,
=> d1 = d2
=>[Tex]\frac{|9a+6b-7|}{3\sqrt{13}} = \frac{|3a+2b+6|}{\sqrt{13}}[/Tex]
=> |9a + 6b − 7| = 3 |3a + 2b + 6|
=> 9h + 6k – 7 = 3 (3h + 2k + 6) or 9h + 6k – 7 = – 3 (3h + 2k + 6)
=> 9h + 6k – 7 = 9h + 6k + 18 or 9h + 6k – 7 = –9h – 6k – 18
=> – 7 = 18 (ignored) or 18h + 12k + 11 = 0
=> 18h + 12k + 11 = 0
Therefore, the required equation of the line is 18x + 12y + 11 = 0.
Question 21. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.Solution:
 The perpendicular line from the axes divides the ∠ BAC = 90o in two equal parts, each having value of θ.
Therefore, θ = 45o. And hence lines AC and AB subtend equal angles (θ) at x-axis.
Slope of AC = tan θ =[Tex]\frac{3}{5-a} [/Tex]. . . . (1)
Slope of AB = tan (180−θ) =[Tex]\frac{2}{1-a}[/Tex]
=> −tan θ =[Tex]\frac{2}{1-a}[/Tex]
=> tan θ =[Tex]\frac{2}{a-1} [/Tex]. . . . (2)
From (1) and (2), we get,
=>[Tex]\frac{3}{5-a} = \frac{2}{a-1}[/Tex]
=> 3a – 3 = 10 – 2a
=> a = 13/5
Therefore, the coordinates of point A are (13/5, 0).
Question 22. Prove that the product of the lengths of the perpendiculars drawn from the points[Tex](\sqrt{a^2-b^2},0) [/Tex]and[Tex](-\sqrt{a^2-b^2},0) [/Tex]to the line[Tex]\frac{x}{a}cosθ+\frac{y}{b}sinθ=1 [/Tex]is b2.Solution:
We are given the line,
[Tex]\frac{x}{a}cosθ+\frac{y}{b}sinθ=1[/Tex]
=> bx cos θ + ay sin θ – ab = 0 . . . . (1)
Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from (x, y) is given by,
[Tex]d = \frac{|Ax+By+C|}{\sqrt{A^2+B^2}}[/Tex]
So, perpendicular distance(d1) of (1) from point[Tex](\sqrt{a^2-b^2},0) [/Tex]is,
[Tex]d_1= \frac{|bcosθ(\sqrt{a^2-b^2})+asinθ(0)-ab|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}[/Tex]
=[Tex]\frac{|bcosθ(\sqrt{a^2-b^2})-ab|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}[/Tex]
And the perpendicular distance(d2) of (1) from point[Tex](-\sqrt{a^2-b^2},0) [/Tex]is,
[Tex]d_2= \frac{|bcosθ(-\sqrt{a^2-b^2})+asinθ(0)-ab|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}[/Tex]
=[Tex]\frac{|bcosθ(\sqrt{a^2-b^2})+ab|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}[/Tex]
Now, L.H.S. = d1d2
=[Tex]\frac{|bcosθ(\sqrt{a^2-b^2})-ab|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}×\frac{|bcosθ(\sqrt{a^2-b^2})+ab|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}[/Tex]
=[Tex]\frac{|b^2cos^2θ(a^2-b^2)-a^2b^2|}{b^2cos^2θ+a^2sin^2θ}[/Tex]
=[Tex]\frac{b^2|cos^2θ(a^2-b^2)-a^2|}{b^2cos^2θ+a^2sin^2θ}[/Tex]
=[Tex]\frac{b^2|a^2cos^2θ-b^2cos^2θ-a^2|}{b^2cos^2θ+a^2sin^2θ}[/Tex]
=[Tex]\frac{b^2|a^2(cos^2θ-1)-b^2cos^2θ|}{b^2cos^2θ+a^2sin^2θ}[/Tex]
=[Tex]\frac{b^2|-a^2sin^2θ-b^2cos^2θ|}{b^2cos^2θ+a^2sin^2θ}[/Tex]
=[Tex]\frac{b^2|-(b^2cos^2θ+a^2sin^2θ)|}{b^2cos^2θ+a^2sin^2θ}[/Tex]
=[Tex]\frac{b^2(b^2cos^2θ+a^2sin^2θ)}{b^2cos^2θ+a^2sin^2θ}[/Tex]
= b2
= R.H.S.
Hence, proved.
Question 23. A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.Solution:
We are given that,
2x – 3y + 4 = 0 . . . . (1)
3x + 4y – 5 = 0 . . . . (2)
6x – 7y + 8 = 0 . . . . (3)
It is given that the person is standing at the intersection of the paths represented by lines (1) and (2).
By solving equations (1) and (2) we get
x = –1/17 and y = 22/17
As perpendicular distance is the shortest distance, the person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point (–1/17, 22/17).
Now, the slope of the line (3) = 6/7
Hence, the slope of the line perpendicular to line (3) = –1/(6/7) = –7/6
So the equation of line passing through (–1/17, 22/17) and having a slope of –7/6 is given by,
=>[Tex]y-\frac{22}{17}=\frac{-7}{6}(x+\frac{1}{17})[/Tex]
=> 6 (17y – 22) = –7 (17x + 1)
=> 102y – 132 = –119x – 7
=> 1119x + 102y = 125
Therefore, the required path is 119x + 102y = 125.
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