Slide 1: Believe on your Strength Example 3 : Find the joint equation of the pair of lines passing through origin and having slopes 2 and -3. Solution : The equations of lines passing through origin and slopes 2 and -3 are
y = 2x and y = -3x
i.e. 2x – y = 0 and 3x + y = 0
Therefore , their joint equation is
(2x – y)(3x + y) = 0
i.e. 6x2 – xy – y2 = 0 The equation of line passing through origin and having slope m is y = mx
Slide 2: Believe on your Strength Example 4 : Find the joint equation of the pair of lines passing through origin, one of which is parallel to the line 2x – y + 4 = 0 and the other is perpendicular to the line x + 3y – 5 = 0. Solution : The given lines are
2x – y + 4 = 0 ………(i)
and x + 3y - 5 = 0……….(ii)
The slopes of these lines are 2 and -1/3
The slope of the line parallel to line (i) is 2 and
Slope of the line perpendicular to line (ii) is 3.
?The equations of the lines through origin and having
these slopes are
y = 2x and y = 3x
i.e. 2x – y = 0 and 3x – y = 0
? The required joint equation is
6x2 – 5xy + y2 = 0
Slide 3: Believe on your Strength Example 5 : Find the separate equations of the lines represented by x2 + 2xy coseseca + y2 = 0 Solution : The given equation is
x2 + 2xy coseseca + y2 = 0
This can be written as
? By quadratic formula ?
Slide 4: Believe on your Strength ? The separate equations of the lines are and
Slide 5: Believe on your Strength Example 6 : Find the acute angle between lines represented by 3x2- 4?3xy + 3y2 = 0 The acute angle between the lines is given by
Slide 6: Believe on your Strength ? ?
Slide 7: Believe on your Strength Example 7 : Find the joint equation of pair of lines, through the point (3,4)and having slopes 1 and -2. Solution: The equations of lines passing through the point (3, 4) and having slopes 1 and -2 are
y - 4 = 1 (x - 3) and
y - 4 = -2 (x - 3).
i.e. (x- 3) - (y - 4) = 0 and 2 (x - 3) + (y- 4) = 0
Therefore their joint equation is
[(x - 3) - (y - 4)][2(x - 3) + (y - 4)] = 0
i.e. 2x2 – xy – y2 - 8x + lly - 10 = 0
Slide 8: Believe on your Strength Example 8 : Find the joint equation of pair of lines passing through the point (-2,3) and parallel to pair of lines given by 3x2 + 5xy + 7y2 =O. Solution :Let m1 and m2 be the slopes of the lines given by 3X2 + 5xy + 7y2 = 0
m1 + m2 = and m1m2 =
The equations of the lines passing through the point (-2, 3) and having slopes m1 and m2 are
y- 3 = m1(x + 2) and (y-3) = m2(x + 2)
i.e. m1(x+ 2) - (y- 3) = 0 and m2(x+ 2) - (y- 3) = 0’ ?
Slide 9: Believe on your Strength Therefore their joint equation is
[m1(x+ 2) - (y - 3)][m2(x + 2)- (y - 3)] = 0
m1m2 (x + 2)2- (m1 + m2)(x + 2) (y - 3) + (y - 3)2 = 0
(x+2)2- (x+2)(Y-3) + (Y-3)2 = 0
i.e. 3x2 + 5Xy + 7Y2- 3x - 32y + 45 = 0
is the required equation.
Slide 10: Believe on your Strength
Slide 11: Believe on your Strength Example 4 : Show that the lines represented by
ax2 + 2hxy + by2 = 0 are equally inclined with the
(i) x-axis, if h = 0, (ii) co-ordinate axes, if a = +b Solution : Let m1 and m2 be the slopes of the lines represented by
ax2 + 2hxy + by2 = 0 ……..(1)
Then, m1 + m2 = and m1.m2 =
Let q be the inclination of a line whose slope is m1. then m1 = tanq
(i) Since the lines represented by (1) are equally inclined with the X-axis.
p - q is the inclination of other straight line.
Slide 12: Believe on your Strength m2= tan (p - q) = -tan q
m1 + m2= tan q - tan q = 0.
= 0
i.e. h = O.
Slide 13: Believe on your Strength (ii) since the lines represented by (I) are equally co-ordinate axes.
? the inclination of the other line is I - eor I + e.
1t . 1t
:. m2= tan ("2- e) = cot e orm2= tan ("2+ e) r -cot e.
:. m,m2= tan e cot e = lor mlm2= tan e (-cot e) =-1.
a i
i.e. m,m2=:t1. :'b=:tl.Hencea= :tb. . ~
Slide 14: Believe on your Strength