Solved Example: Straight Lines

``` Page 1

JEE Solved Example on Straight Lines
JEE Mains
Q1: The opposite angular points of a square are ( ?? , ?? ) and ( ?? , - ?? ) . Then the co-ordinates of other
two points are
(A) D(
?? ?? ,
?? ?? ) , ?? ( -
?? ?? ,
?? ?? )
(B) ?? (
?? ?? ,
?? ?? ) , ?? (
?? ?? ,
?? ?? )
(C) ?? (
?? ?? ,
?? ?? ) , ?? ( -
?? ?? ,
?? ?? )
(D) None of these
Ans: (C)
Obviously, slope of AC = 5 / 2.
Let ?? be the slope of a line inclined at an angle of 45
°
to ???? , then tan ? 45
°
= ±
?? -
5
2
1 + ?? ·
5
2
? ?? = -
7
3
·
3
7
.
Thus, let the slope of ???? or ???? be 3/7and that of ???? or ???? be -
7
3
. Then equation of ???? is 3x -
7y + 19 = 0.
Also the equation of ???? is 7x + 3y - 4 = 0.
On solving these equations, we get, B ( -
1
2
,
5
2
).
Now let the coordinates of the vertex ?? be ( h , ?? ) . Since the middle points of ???? and ???? are same,
therefore
1
2
( ?? -
1
2
) =
1
2
( 3 + 1 ) ? ?? =
9
2
,
1
2
( ?? +
5
2
) =
1
2
( 4 - 1 )
? ? ?? =
1
2
. Hence, ?? = (
9
2
,
1
2
) .

Q2: If the lines ?? = ?? ?? + ?? and ?? ?? = ?? + ?? are equally inclined to the line ?? = ???? + ?? , then ?? =
(A)
?? + ?? v ?? ??
(B)
?? - ?? v ?? ??
(C)
?? ± ?? v ?? ??
(D)
?? ± ?? v ?? ??
Ans: (D)
?? 1
= 3 , ?? 2
=
1
2
and ?? 3
= ??
Let the angle between first and third line is ?? , and between second and third is ?? 2
, then
Page 2

JEE Solved Example on Straight Lines
JEE Mains
Q1: The opposite angular points of a square are ( ?? , ?? ) and ( ?? , - ?? ) . Then the co-ordinates of other
two points are
(A) D(
?? ?? ,
?? ?? ) , ?? ( -
?? ?? ,
?? ?? )
(B) ?? (
?? ?? ,
?? ?? ) , ?? (
?? ?? ,
?? ?? )
(C) ?? (
?? ?? ,
?? ?? ) , ?? ( -
?? ?? ,
?? ?? )
(D) None of these
Ans: (C)
Obviously, slope of AC = 5 / 2.
Let ?? be the slope of a line inclined at an angle of 45
°
to ???? , then tan ? 45
°
= ±
?? -
5
2
1 + ?? ·
5
2
? ?? = -
7
3
·
3
7
.
Thus, let the slope of ???? or ???? be 3/7and that of ???? or ???? be -
7
3
. Then equation of ???? is 3x -
7y + 19 = 0.
Also the equation of ???? is 7x + 3y - 4 = 0.
On solving these equations, we get, B ( -
1
2
,
5
2
).
Now let the coordinates of the vertex ?? be ( h , ?? ) . Since the middle points of ???? and ???? are same,
therefore
1
2
( ?? -
1
2
) =
1
2
( 3 + 1 ) ? ?? =
9
2
,
1
2
( ?? +
5
2
) =
1
2
( 4 - 1 )
? ? ?? =
1
2
. Hence, ?? = (
9
2
,
1
2
) .

Q2: If the lines ?? = ?? ?? + ?? and ?? ?? = ?? + ?? are equally inclined to the line ?? = ???? + ?? , then ?? =
(A)
?? + ?? v ?? ??
(B)
?? - ?? v ?? ??
(C)
?? ± ?? v ?? ??
(D)
?? ± ?? v ?? ??
Ans: (D)
?? 1
= 3 , ?? 2
=
1
2
and ?? 3
= ??
Let the angle between first and third line is ?? , and between second and third is ?? 2
, then
tan ? ?? 1
=
3 - ?? 1 + 3 ?? and tan ? ?? 2
=
?? -
1
2
1 +
?? 2

But ?? 1
= ?? 2
? ? ?
3 - ?? 1 + 3 ?? =
?? -
1
2
1 +
?? 2

? ? 7 ?? 2
- 2 ?? - 7 = 0 ? ?? =
1 ± 5 v 2
7

Q3: Let ?? ( - ?? , ?? ) , ?? ( ?? , ?? ) and ?? ( ?? , ?? v ?? ) be three points. Then the equation of the bisector of the
angle ???? ?? is
(A)
v ?? ?? ?? + ?? = ??
(B) ?? + v ?? ?? = ??
(C) v ?? ?? + ?? = ??
(D) ?? +
v ?? ?? ?? = ??

Ans: (C)
Slope of QR =
3 v 3 - 0
3 - 0
= v 3 i.e., ?? = 60
°

Clearly, ? ?????? = 120
°

???? is the angle bisector of the angle, so line ???? makes 120
°
with the positive direction of ?? -axis.
Therefore, equation of the bisector of ? ?????? is ?? = tan ? 120
°
?? or ?? = v 3 ?? i.e., v 3 ?? + ?? = 0.

Q4: If for a variable line
?? ?? +
?? ?? = ?? , the condition
?? ?? ?? +
?? ?? ?? =
?? ?? ?? ? ( ?? is a constant) is satisfied, then
locus of foot of perpendicular drawn from origin to the line is
(A) ?? ?? + ?? ?? = ?? ?? / ??
(B) ?? ?? + ?? ?? = ?? ?? ??
(C) ?? ?? + ?? ?? = ?? ??
(D) ?? ?? - ?? ?? = ?? ??
Ans: (C)
Equation of perpendicular drawn from origin to the line
?? ?? +
?? ?? = 1 is ?? - 0 =
?? ?? ( ?? - 0 )
[ ? m of given line =
- ?? ?? , ? m of perpendicular =
?? ?? ]
Page 3

JEE Solved Example on Straight Lines
JEE Mains
Q1: The opposite angular points of a square are ( ?? , ?? ) and ( ?? , - ?? ) . Then the co-ordinates of other
two points are
(A) D(
?? ?? ,
?? ?? ) , ?? ( -
?? ?? ,
?? ?? )
(B) ?? (
?? ?? ,
?? ?? ) , ?? (
?? ?? ,
?? ?? )
(C) ?? (
?? ?? ,
?? ?? ) , ?? ( -
?? ?? ,
?? ?? )
(D) None of these
Ans: (C)
Obviously, slope of AC = 5 / 2.
Let ?? be the slope of a line inclined at an angle of 45
°
to ???? , then tan ? 45
°
= ±
?? -
5
2
1 + ?? ·
5
2
? ?? = -
7
3
·
3
7
.
Thus, let the slope of ???? or ???? be 3/7and that of ???? or ???? be -
7
3
. Then equation of ???? is 3x -
7y + 19 = 0.
Also the equation of ???? is 7x + 3y - 4 = 0.
On solving these equations, we get, B ( -
1
2
,
5
2
).
Now let the coordinates of the vertex ?? be ( h , ?? ) . Since the middle points of ???? and ???? are same,
therefore
1
2
( ?? -
1
2
) =
1
2
( 3 + 1 ) ? ?? =
9
2
,
1
2
( ?? +
5
2
) =
1
2
( 4 - 1 )
? ? ?? =
1
2
. Hence, ?? = (
9
2
,
1
2
) .

Q2: If the lines ?? = ?? ?? + ?? and ?? ?? = ?? + ?? are equally inclined to the line ?? = ???? + ?? , then ?? =
(A)
?? + ?? v ?? ??
(B)
?? - ?? v ?? ??
(C)
?? ± ?? v ?? ??
(D)
?? ± ?? v ?? ??
Ans: (D)
?? 1
= 3 , ?? 2
=
1
2
and ?? 3
= ??
Let the angle between first and third line is ?? , and between second and third is ?? 2
, then
tan ? ?? 1
=
3 - ?? 1 + 3 ?? and tan ? ?? 2
=
?? -
1
2
1 +
?? 2

But ?? 1
= ?? 2
? ? ?
3 - ?? 1 + 3 ?? =
?? -
1
2
1 +
?? 2

? ? 7 ?? 2
- 2 ?? - 7 = 0 ? ?? =
1 ± 5 v 2
7

Q3: Let ?? ( - ?? , ?? ) , ?? ( ?? , ?? ) and ?? ( ?? , ?? v ?? ) be three points. Then the equation of the bisector of the
angle ???? ?? is
(A)
v ?? ?? ?? + ?? = ??
(B) ?? + v ?? ?? = ??
(C) v ?? ?? + ?? = ??
(D) ?? +
v ?? ?? ?? = ??

Ans: (C)
Slope of QR =
3 v 3 - 0
3 - 0
= v 3 i.e., ?? = 60
°

Clearly, ? ?????? = 120
°

???? is the angle bisector of the angle, so line ???? makes 120
°
with the positive direction of ?? -axis.
Therefore, equation of the bisector of ? ?????? is ?? = tan ? 120
°
?? or ?? = v 3 ?? i.e., v 3 ?? + ?? = 0.

Q4: If for a variable line
?? ?? +
?? ?? = ?? , the condition
?? ?? ?? +
?? ?? ?? =
?? ?? ?? ? ( ?? is a constant) is satisfied, then
locus of foot of perpendicular drawn from origin to the line is
(A) ?? ?? + ?? ?? = ?? ?? / ??
(B) ?? ?? + ?? ?? = ?? ?? ??
(C) ?? ?? + ?? ?? = ?? ??
(D) ?? ?? - ?? ?? = ?? ??
Ans: (C)
Equation of perpendicular drawn from origin to the line
?? ?? +
?? ?? = 1 is ?? - 0 =
?? ?? ( ?? - 0 )
[ ? m of given line =
- ?? ?? , ? m of perpendicular =
?? ?? ]
? by - ax = 0
?
?? ?? -
?? ?? = 0
Now, the locus of foot of perpendicular is the intersection point of line
?? ?? +
?? ?? = 1.
and
?? ?? -
?? ?? = 0.
To find locus, squaring and adding (i) and (ii)
(
?? ?? +
?? ?? )
2
+ (
?? ?? -
?? ?? )
2
= 1
? ?? 2
(
1
?? 2
+
1
?? 2
) + ?? 2
(
1
?? 2
+
1
?? 2
) = 1
? ?? 2
(
1
?? 2
) + ?? 2
(
1
?? 2
) = 1 , [ ?
1
?? 2
+
1
?? 2
=
1
?? 2
]
? ?? 2
+ ?? 2
= ?? 2
.

Q5: If the circle ?? ?? + ?? ?? - ?? ?? - ?? ?? + ?? = ?? does not touch or intersect the axes and the point
( ?? , ?? ) lies inside the circle, then
(A) ?? < ?? < ??
(B) ?? < ?? < ????
(C) ?? < ?? < ????
(D) none of these
Ans: (C)
The centre of the given circle is ( 2 , 3 ) and the radius = v 4 + 9 - ?? i.e. v 13 - k.
Since the given circle does not touch or intersect the coordinate axes and the point
( 2 , 2 ) lies inside the circle
? ?? -cooridnate of centre > radius i.e. 2 > v 13 - k,
?? -coordinate of centre > radius i.e. 3 > v 13 - k and
4 + 4 - 8 - 12 + ?? < 0
? 4 > 13 - ?? , 9 > 13 - ?? and - 12 + ?? < 0
? ?? > 9 , ?? > 4 and ?? < 12 ? 9 < ?? < 12

Q6: If ?? ?? , ?? ?? be the inclination of tangents drawn from the point ?? to the circle ?? ?? + ?? ?? = ?? ?? with
?? -axis, then the locus of ?? , it is given that ?? ???? ? ?? ?? + ?? ???? ? ?? ?? = ?? , is
(A) ?? ( ?? ?? - ?? ?? ) = ?? ????
(B) ?? ( ?? ?? - ?? ?? ) = ?? ?? - ?? ??
(C) ?? ( ?? ?? - ?? ?? ) = ?? ????
(D) none of these
Page 4

JEE Solved Example on Straight Lines
JEE Mains
Q1: The opposite angular points of a square are ( ?? , ?? ) and ( ?? , - ?? ) . Then the co-ordinates of other
two points are
(A) D(
?? ?? ,
?? ?? ) , ?? ( -
?? ?? ,
?? ?? )
(B) ?? (
?? ?? ,
?? ?? ) , ?? (
?? ?? ,
?? ?? )
(C) ?? (
?? ?? ,
?? ?? ) , ?? ( -
?? ?? ,
?? ?? )
(D) None of these
Ans: (C)
Obviously, slope of AC = 5 / 2.
Let ?? be the slope of a line inclined at an angle of 45
°
to ???? , then tan ? 45
°
= ±
?? -
5
2
1 + ?? ·
5
2
? ?? = -
7
3
·
3
7
.
Thus, let the slope of ???? or ???? be 3/7and that of ???? or ???? be -
7
3
. Then equation of ???? is 3x -
7y + 19 = 0.
Also the equation of ???? is 7x + 3y - 4 = 0.
On solving these equations, we get, B ( -
1
2
,
5
2
).
Now let the coordinates of the vertex ?? be ( h , ?? ) . Since the middle points of ???? and ???? are same,
therefore
1
2
( ?? -
1
2
) =
1
2
( 3 + 1 ) ? ?? =
9
2
,
1
2
( ?? +
5
2
) =
1
2
( 4 - 1 )
? ? ?? =
1
2
. Hence, ?? = (
9
2
,
1
2
) .

Q2: If the lines ?? = ?? ?? + ?? and ?? ?? = ?? + ?? are equally inclined to the line ?? = ???? + ?? , then ?? =
(A)
?? + ?? v ?? ??
(B)
?? - ?? v ?? ??
(C)
?? ± ?? v ?? ??
(D)
?? ± ?? v ?? ??
Ans: (D)
?? 1
= 3 , ?? 2
=
1
2
and ?? 3
= ??
Let the angle between first and third line is ?? , and between second and third is ?? 2
, then
tan ? ?? 1
=
3 - ?? 1 + 3 ?? and tan ? ?? 2
=
?? -
1
2
1 +
?? 2

But ?? 1
= ?? 2
? ? ?
3 - ?? 1 + 3 ?? =
?? -
1
2
1 +
?? 2

? ? 7 ?? 2
- 2 ?? - 7 = 0 ? ?? =
1 ± 5 v 2
7

Q3: Let ?? ( - ?? , ?? ) , ?? ( ?? , ?? ) and ?? ( ?? , ?? v ?? ) be three points. Then the equation of the bisector of the
angle ???? ?? is
(A)
v ?? ?? ?? + ?? = ??
(B) ?? + v ?? ?? = ??
(C) v ?? ?? + ?? = ??
(D) ?? +
v ?? ?? ?? = ??

Ans: (C)
Slope of QR =
3 v 3 - 0
3 - 0
= v 3 i.e., ?? = 60
°

Clearly, ? ?????? = 120
°

???? is the angle bisector of the angle, so line ???? makes 120
°
with the positive direction of ?? -axis.
Therefore, equation of the bisector of ? ?????? is ?? = tan ? 120
°
?? or ?? = v 3 ?? i.e., v 3 ?? + ?? = 0.

Q4: If for a variable line
?? ?? +
?? ?? = ?? , the condition
?? ?? ?? +
?? ?? ?? =
?? ?? ?? ? ( ?? is a constant) is satisfied, then
locus of foot of perpendicular drawn from origin to the line is
(A) ?? ?? + ?? ?? = ?? ?? / ??
(B) ?? ?? + ?? ?? = ?? ?? ??
(C) ?? ?? + ?? ?? = ?? ??
(D) ?? ?? - ?? ?? = ?? ??
Ans: (C)
Equation of perpendicular drawn from origin to the line
?? ?? +
?? ?? = 1 is ?? - 0 =
?? ?? ( ?? - 0 )
[ ? m of given line =
- ?? ?? , ? m of perpendicular =
?? ?? ]
? by - ax = 0
?
?? ?? -
?? ?? = 0
Now, the locus of foot of perpendicular is the intersection point of line
?? ?? +
?? ?? = 1.
and
?? ?? -
?? ?? = 0.
To find locus, squaring and adding (i) and (ii)
(
?? ?? +
?? ?? )
2
+ (
?? ?? -
?? ?? )
2
= 1
? ?? 2
(
1
?? 2
+
1
?? 2
) + ?? 2
(
1
?? 2
+
1
?? 2
) = 1
? ?? 2
(
1
?? 2
) + ?? 2
(
1
?? 2
) = 1 , [ ?
1
?? 2
+
1
?? 2
=
1
?? 2
]
? ?? 2
+ ?? 2
= ?? 2
.

Q5: If the circle ?? ?? + ?? ?? - ?? ?? - ?? ?? + ?? = ?? does not touch or intersect the axes and the point
( ?? , ?? ) lies inside the circle, then
(A) ?? < ?? < ??
(B) ?? < ?? < ????
(C) ?? < ?? < ????
(D) none of these
Ans: (C)
The centre of the given circle is ( 2 , 3 ) and the radius = v 4 + 9 - ?? i.e. v 13 - k.
Since the given circle does not touch or intersect the coordinate axes and the point
( 2 , 2 ) lies inside the circle
? ?? -cooridnate of centre > radius i.e. 2 > v 13 - k,
?? -coordinate of centre > radius i.e. 3 > v 13 - k and
4 + 4 - 8 - 12 + ?? < 0
? 4 > 13 - ?? , 9 > 13 - ?? and - 12 + ?? < 0
? ?? > 9 , ?? > 4 and ?? < 12 ? 9 < ?? < 12

Q6: If ?? ?? , ?? ?? be the inclination of tangents drawn from the point ?? to the circle ?? ?? + ?? ?? = ?? ?? with
?? -axis, then the locus of ?? , it is given that ?? ???? ? ?? ?? + ?? ???? ? ?? ?? = ?? , is
(A) ?? ( ?? ?? - ?? ?? ) = ?? ????
(B) ?? ( ?? ?? - ?? ?? ) = ?? ?? - ?? ??
(C) ?? ( ?? ?? - ?? ?? ) = ?? ????
(D) none of these
Ans: (C)
Any tangent is ?? = ???? ± ?? v 1 + ?? 2

If it passes through ?? ( h , ?? ) , then ( ?? - ?? h )
2
= ?? 2
( 1 + ?? 2
) or ?? 2
( h
2
- ?? 2
) - 2 ?? h ?? +
( ?? 2
- ?? 2
) = 0, where ?? = tan ? ??
c ot ? ?? 1
+ c ot ? ?? 2
= ?? ?
1
?? 1
+
1
?? 2
= ?? ?
?? 1
+ ?? 2
?? 1
?? 2
= ?? ? ?
2 h ?? ( ?? 2
- ?? 2
)
= ?? ? ?? ( ?? 2
- ?? 2
) = 2 ????

Q7: The chord of contact of tangents from a point ?? to a circle passes through ?? . If ?? ?? and ?? ?? are
the lengths of the tangents from ?? and ?? to the circle, then ???? is equal to
(A)
?? ?? + ?? ?? ??
(B)
?? ?? - ?? ?? ??
(C) v ?? ?? ?? + ?? ?? ??
(D) v ?? ?? ?? - ?? ?? ??
Ans: (C)
Let ?? ( ?? 1
, ?? 1
) and ?? ( ?? 2
, ?? 2
) be two points and ?? 2
+ ?? 2
= ?? 2
be the given circle
then the chord of contact of tangents drawn from ?? to the given circle is
xx
1
+ yy
1
= a
2

It will pass through ?? ( ?? 2
, ?? 2
) , if ? ?? 2
?? 1
+ ?? 2
?? 1
= ?? 2

Now, ?? 1
= v ?? 1
2
+ ?? 1
2
- ?? 2
, ? ?? 2
= v ?? 2
2
+ ?? 2
2
- ?? 2

? ? ???? = v ( ?? 2
- ?? 1
)
2
+ ( ?? 2
- ?? 1
)
2
? ? ???? = v( ?? 2
2
+ ?? 2
2
) + ( ?? 1
2
+ ?? 1
2
) - 2 ( ?? 1
?? 2
+ ?? 1
?? 2
)
? ? ???? = v( ?? 2
2
+ ?? 2
2
) + ( ?? 1
2
+ ?? 1
2
) - 2 ?? 2
? [using (i)]
? ? ???? = v( ?? 1
2
+ ?? 1
2
- ?? 2
) + ( ?? 2
2
+ ?? 2
2
- ?? 2
)
? ? ???? = v ?? 1
2
+ ?? 2
2

Page 5

JEE Solved Example on Straight Lines
JEE Mains
Q1: The opposite angular points of a square are ( ?? , ?? ) and ( ?? , - ?? ) . Then the co-ordinates of other
two points are
(A) D(
?? ?? ,
?? ?? ) , ?? ( -
?? ?? ,
?? ?? )
(B) ?? (
?? ?? ,
?? ?? ) , ?? (
?? ?? ,
?? ?? )
(C) ?? (
?? ?? ,
?? ?? ) , ?? ( -
?? ?? ,
?? ?? )
(D) None of these
Ans: (C)
Obviously, slope of AC = 5 / 2.
Let ?? be the slope of a line inclined at an angle of 45
°
to ???? , then tan ? 45
°
= ±
?? -
5
2
1 + ?? ·
5
2
? ?? = -
7
3
·
3
7
.
Thus, let the slope of ???? or ???? be 3/7and that of ???? or ???? be -
7
3
. Then equation of ???? is 3x -
7y + 19 = 0.
Also the equation of ???? is 7x + 3y - 4 = 0.
On solving these equations, we get, B ( -
1
2
,
5
2
).
Now let the coordinates of the vertex ?? be ( h , ?? ) . Since the middle points of ???? and ???? are same,
therefore
1
2
( ?? -
1
2
) =
1
2
( 3 + 1 ) ? ?? =
9
2
,
1
2
( ?? +
5
2
) =
1
2
( 4 - 1 )
? ? ?? =
1
2
. Hence, ?? = (
9
2
,
1
2
) .

Q2: If the lines ?? = ?? ?? + ?? and ?? ?? = ?? + ?? are equally inclined to the line ?? = ???? + ?? , then ?? =
(A)
?? + ?? v ?? ??
(B)
?? - ?? v ?? ??
(C)
?? ± ?? v ?? ??
(D)
?? ± ?? v ?? ??
Ans: (D)
?? 1
= 3 , ?? 2
=
1
2
and ?? 3
= ??
Let the angle between first and third line is ?? , and between second and third is ?? 2
, then
tan ? ?? 1
=
3 - ?? 1 + 3 ?? and tan ? ?? 2
=
?? -
1
2
1 +
?? 2

But ?? 1
= ?? 2
? ? ?
3 - ?? 1 + 3 ?? =
?? -
1
2
1 +
?? 2

? ? 7 ?? 2
- 2 ?? - 7 = 0 ? ?? =
1 ± 5 v 2
7

Q3: Let ?? ( - ?? , ?? ) , ?? ( ?? , ?? ) and ?? ( ?? , ?? v ?? ) be three points. Then the equation of the bisector of the
angle ???? ?? is
(A)
v ?? ?? ?? + ?? = ??
(B) ?? + v ?? ?? = ??
(C) v ?? ?? + ?? = ??
(D) ?? +
v ?? ?? ?? = ??

Ans: (C)
Slope of QR =
3 v 3 - 0
3 - 0
= v 3 i.e., ?? = 60
°

Clearly, ? ?????? = 120
°

???? is the angle bisector of the angle, so line ???? makes 120
°
with the positive direction of ?? -axis.
Therefore, equation of the bisector of ? ?????? is ?? = tan ? 120
°
?? or ?? = v 3 ?? i.e., v 3 ?? + ?? = 0.

Q4: If for a variable line
?? ?? +
?? ?? = ?? , the condition
?? ?? ?? +
?? ?? ?? =
?? ?? ?? ? ( ?? is a constant) is satisfied, then
locus of foot of perpendicular drawn from origin to the line is
(A) ?? ?? + ?? ?? = ?? ?? / ??
(B) ?? ?? + ?? ?? = ?? ?? ??
(C) ?? ?? + ?? ?? = ?? ??
(D) ?? ?? - ?? ?? = ?? ??
Ans: (C)
Equation of perpendicular drawn from origin to the line
?? ?? +
?? ?? = 1 is ?? - 0 =
?? ?? ( ?? - 0 )
[ ? m of given line =
- ?? ?? , ? m of perpendicular =
?? ?? ]
? by - ax = 0
?
?? ?? -
?? ?? = 0
Now, the locus of foot of perpendicular is the intersection point of line
?? ?? +
?? ?? = 1.
and
?? ?? -
?? ?? = 0.
To find locus, squaring and adding (i) and (ii)
(
?? ?? +
?? ?? )
2
+ (
?? ?? -
?? ?? )
2
= 1
? ?? 2
(
1
?? 2
+
1
?? 2
) + ?? 2
(
1
?? 2
+
1
?? 2
) = 1
? ?? 2
(
1
?? 2
) + ?? 2
(
1
?? 2
) = 1 , [ ?
1
?? 2
+
1
?? 2
=
1
?? 2
]
? ?? 2
+ ?? 2
= ?? 2
.

Q5: If the circle ?? ?? + ?? ?? - ?? ?? - ?? ?? + ?? = ?? does not touch or intersect the axes and the point
( ?? , ?? ) lies inside the circle, then
(A) ?? < ?? < ??
(B) ?? < ?? < ????
(C) ?? < ?? < ????
(D) none of these
Ans: (C)
The centre of the given circle is ( 2 , 3 ) and the radius = v 4 + 9 - ?? i.e. v 13 - k.
Since the given circle does not touch or intersect the coordinate axes and the point
( 2 , 2 ) lies inside the circle
? ?? -cooridnate of centre > radius i.e. 2 > v 13 - k,
?? -coordinate of centre > radius i.e. 3 > v 13 - k and
4 + 4 - 8 - 12 + ?? < 0
? 4 > 13 - ?? , 9 > 13 - ?? and - 12 + ?? < 0
? ?? > 9 , ?? > 4 and ?? < 12 ? 9 < ?? < 12

Q6: If ?? ?? , ?? ?? be the inclination of tangents drawn from the point ?? to the circle ?? ?? + ?? ?? = ?? ?? with
?? -axis, then the locus of ?? , it is given that ?? ???? ? ?? ?? + ?? ???? ? ?? ?? = ?? , is
(A) ?? ( ?? ?? - ?? ?? ) = ?? ????
(B) ?? ( ?? ?? - ?? ?? ) = ?? ?? - ?? ??
(C) ?? ( ?? ?? - ?? ?? ) = ?? ????
(D) none of these
Ans: (C)
Any tangent is ?? = ???? ± ?? v 1 + ?? 2

If it passes through ?? ( h , ?? ) , then ( ?? - ?? h )
2
= ?? 2
( 1 + ?? 2
) or ?? 2
( h
2
- ?? 2
) - 2 ?? h ?? +
( ?? 2
- ?? 2
) = 0, where ?? = tan ? ??
c ot ? ?? 1
+ c ot ? ?? 2
= ?? ?
1
?? 1
+
1
?? 2
= ?? ?
?? 1
+ ?? 2
?? 1
?? 2
= ?? ? ?
2 h ?? ( ?? 2
- ?? 2
)
= ?? ? ?? ( ?? 2
- ?? 2
) = 2 ????

Q7: The chord of contact of tangents from a point ?? to a circle passes through ?? . If ?? ?? and ?? ?? are
the lengths of the tangents from ?? and ?? to the circle, then ???? is equal to
(A)
?? ?? + ?? ?? ??
(B)
?? ?? - ?? ?? ??
(C) v ?? ?? ?? + ?? ?? ??
(D) v ?? ?? ?? - ?? ?? ??
Ans: (C)
Let ?? ( ?? 1
, ?? 1
) and ?? ( ?? 2
, ?? 2
) be two points and ?? 2
+ ?? 2
= ?? 2
be the given circle
then the chord of contact of tangents drawn from ?? to the given circle is
xx
1
+ yy
1
= a
2

It will pass through ?? ( ?? 2
, ?? 2
) , if ? ?? 2
?? 1
+ ?? 2
?? 1
= ?? 2

Now, ?? 1
= v ?? 1
2
+ ?? 1
2
- ?? 2
, ? ?? 2
= v ?? 2
2
+ ?? 2
2
- ?? 2

? ? ???? = v ( ?? 2
- ?? 1
)
2
+ ( ?? 2
- ?? 1
)
2
? ? ???? = v( ?? 2
2
+ ?? 2
2
) + ( ?? 1
2
+ ?? 1
2
) - 2 ( ?? 1
?? 2
+ ?? 1
?? 2
)
? ? ???? = v( ?? 2
2
+ ?? 2
2
) + ( ?? 1
2
+ ?? 1
2
) - 2 ?? 2
? [using (i)]
? ? ???? = v( ?? 1
2
+ ?? 1
2
- ?? 2
) + ( ?? 2
2
+ ?? 2
2
- ?? 2
)
? ? ???? = v ?? 1
2
+ ?? 2
2

Q8: The circles ?? ?? + ?? ?? + ?? ?? ?? ?? - ?? ?? = ?? and ?? ?? + ?? ?? + ?? ?? ?? ?? - ?? ?? = ?? cut each other
orthogonally. If ?? ?? , ?? ?? are perpendicular from ( ?? , ?? ) and ( ?? , - ?? ) on a common tangent of these
circle, then ?? ?? ?? ?? is equal to
(A)
?? ?? ??
(B) ?? ??
(C) ?? ?? ??
(D) ?? ?? + ??
Ans: (B)
Given ?? 1
?? 2
+ a
2
= 0
If I + my = 1 is common tangent of these circle, then
- | ?? 1
- 1
v 1
1
+ ?? 2
= ± v ?? 1
2
+ ?? 2
= ( ?? ?? 1
+ 1 )
2
= ( 1
2
+ ?? 2
) ( ?? 1
2
+ ?? 2
)
? ? ?? 2
?? 1
2
- 2 ?? ?? 1
+ ?? 2
( 1
2
+ ?? 2
) - 1 = 0

Similarly, ?? 2
?? 2
2
- 21 ?? 2
+ ?? 2
( 1
2
+ ?? 2
) - 1 = 0
So that g
1
, g
2
are the roots of the equation
?? 2
?? 2
- 2 ???? + ?? 2
( ?? 2
+ ?? 2
) - 1 = 0
? ? ?? 1
?? 2
=
?? 2
( ?? 2
+ ?? 2
) - 1
?? 2
= - ?? 2
? ? ?? 2
( ?? 2
+ ?? 2
) = 1 - ?? 2
?? 2
? …

Now, ?? 1
?? 2
=
| ???? - 1 |
v 1
2
+ ?? 2
·
| - ???? - 1 |
v ?? 2
+ ?? 2
=
| 1 - ?? 2
?? 2
|
1
2
+ ?? 2
= ?? 2
[ from (ii)]

Q9: The equation of the line touching both the parabolas ?? ?? = ?? ?? and ?? ?? = - ???? ?? is
(A) ?? ?? + ?? - ?? = ??
(B) ?? + ?? ?? + ?? = ??
(C) ?? - ?? ?? + ?? = ??
(D) ?? - ?? ?? - ?? = ??
Ans: (C)
Any tangent to ?? 2
= 4 ?? is ?? = ???? +
1
??
Solving with ?? 2
= - 32 ??
?? 2
= - 32 ( ???? +
1
?? ) ? ?? 2
+ 32 ???? +
32
?? = 0
But the line is tangent to 2 nd parabola.
Discriminant = 0
? ( 32 )
2
?? 2
= 4 ·
32
?? ? ? ? ?? 3
=
4 . 32
( 32 )
2
=
1
8
? ? ?? =
1
2

```

Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

FAQs on Solved Example: Straight Lines - Mathematics (Maths) for JEE Main & Advanced

 1. How to find the equation of a straight line passing through two given points?
Ans. To find the equation of a straight line passing through two given points, first calculate the slope using the formula (y2 - y1)/(x2 - x1). Then, substitute one of the points and the slope into the point-slope form equation y - y1 = m(x - x1).
 2. What is the condition for two lines to be parallel?
Ans. Two lines are parallel if they have the same slope but different y-intercepts. This means that the slopes of the two lines are equal, but the y-intercepts are not equal.
 3. How to determine if two lines are perpendicular to each other?
Ans. Two lines are perpendicular if the product of their slopes is -1. In other words, if the slopes of the two lines are negative reciprocals of each other, then they are perpendicular.
 4. How to find the distance between a point and a line?
Ans. To find the distance between a point (x1, y1) and a line Ax + By + C = 0, use the formula |Ax1 + By1 + C| / √(A^2 + B^2).
 5. What is the significance of the slope of a straight line?
Ans. The slope of a straight line represents the rate at which the line is rising or falling. It also indicates the direction of the line - a positive slope means the line is increasing, while a negative slope means the line is decreasing.

Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

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