Straight Line - Model Problems - 6

Problems with complexity

1. The set of lines ax+by+c = 0 where a,b,c satisfy the relation 3a+2b+4c = 0 is concurrent at the point --------------. (JEE, 1982)

Answer (3/4, ½)

3a+2b+4c = 0
Dividing the equation by 4

3a/4+2b/4+4c/4 = 0
=> 3a/4 + b/2+c = 0
x = ¾ and y = ½ satisfied this relation

Hence the set of lines are concurrent at (3/4, ½)

2. The straight lines x+y = 0, 3x+y-4 = 0, and x+3y-4 = 0 form a triangle which is

a. isosceles
b. equilateral
c. right angled
d. none of these

(JEE 1984)
Answer: (a)

Solution:

x+y = 0 … (1) slope = -1
3x+y = 4 … (2) slope = -3/1
x+3y = 4 … (3) slope = -1/3

So no two lines are perpendicular to each other as m1*m2 is not equal to -1 for any two lines.

From (1) and (2) intersection point is

A(2,-2)

From (2) and (3) intersection point is

B(1,1)

From (3) and (1) intersection point is
C(-2,2)

AC² = [(2-(-2)] ² + [-2-2] ² = 32
BC² = [1-(-2)] ²+[1-2] ² = 10
AB²= [2-1] ² + [-2-1] ² = 10
As BC² = AB² which means BC = AB and AC is different, the triangle is a isosceles triangle.

3. Three lines px+qy+r = 0, qx+ry+p =0, and rx+py+q +0 are concurrent if:

a. p+q+r + 0
b. p²+q²+r² = pq+rq+rp
c. p³+q³+r³ = 3pqr
d. none of these

Answer: (a), (b),©

Solution

The condition for concurrency of three lines
a1x +b1y+c1 = 0,
a2x+b2y+c2 = 0 and
a3x+b3y+c3=0 is
The determinant

|a1 b1 c1|
|a2 b2 c2|
|a3 b3 c3|

= 0

Hence the condition for given lines to be concurrent is
|p q r|
|q r p|
|r p q|

= 0
=> p(rq-p²) –q(q²-rp) +r(pq-r²) = 0
=> prq - p³ - q³+qrp +rpq - r³ = 0
= > -p³-q³-r³ = -3pqr
=> p³+q³+r³ = 3pqr
=> p³+q³+r³ - 3pqr = 0
=> (p+q+r)(p²+q²+r²-pq-qr-rp) = 0
=> p+q+r = 0 or
=> p²+q²+r² = pq+qr+rp

4. The points (0,8/3), (1,3) and (82,30) are vertices of:

a. an obtuse angled triangle
b. an acute angled triangle
c. a right angled triangle
d. an isosceles triangle
e. none of these

(JEE 1986)

Answer (e)

Solution

Concepts involed:

The slope of a line passing through (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1), provided x1≠x2.

The equation of a line having slope m and passing through (x1,y1)and (x2,y2) is

(y-y1)/(x-x1) = (y1-y2)/(x1-x2)

The equation of the line passing through (0,8/3) and (1,3) is

(y - 8/3)/(x) = (8/3 – 3)/(0-1)

y- 8/3 = x/3
=> x/3 –y +8/3 = 0
=> x – 3y +8 = 0

Checking whether (82,30) is on this line

82 – 3(30)+8 = 90-90 turns out to be zero.

Hence given points are collinear.

5. State whether the following statement is true or false
The lines 2x +3y +19 = 0 and 9x+6y-17 = 0 cut the coordinate axes in concyclic points.
(JEE 1988)

Answer: True

Concyclic points are points on a circle.

If ac = bd = k, then the equation of the circle through A(a,0), B(0,b), C(c,0)) and D(0,d) is

x² + y² -(a+c)x – (b+d) y +k = 0
In this case the points are A(-19/2, 0), B(0,-19/3), C(17/9,0), D (0,17/6)

ac = -19/2*17/9 = -323/18

bd = -19/3*17/6 = -323/18

Hence the given lines cut the coordinate axes in concyclic points.

(Note: This problem belongs to “circle” topic. But as it is worded it seems to be a problem under straight line topic)