DPP for JEE: Daily Practice Problems Ch10 Straight Lines (Solutions)

 Page 2


D lies on x – y – 5 = 0
?
? .
From the given condition
we get,
?     ? 
? Equation of the line is
3. (b) The equation of any line parallel to 2x + 6y + 7 = 0 is 2x + 6y + k =
0
This meets the axes at  and 
By hypothesis, AB = 10
10k
2
 = 3600 k = 
Hence there are two lines given by 
4. (c) We have the equation
y
2
 + xy + px
2
 – x – 2y + p = 0
We know any general equation
ax
2
 + by
2
 + 2hxy + 2gx + 2fy + c = 0 ... (1)
represents two straight lines if
abc + 2fgh – af 
2
 – bg
2
 – ch
2
 = 0 ... (2)
On comparing given equation with (1), we get
a = p, b = 1, , , f = – 1, c = p
Page 3


D lies on x – y – 5 = 0
?
? .
From the given condition
we get,
?     ? 
? Equation of the line is
3. (b) The equation of any line parallel to 2x + 6y + 7 = 0 is 2x + 6y + k =
0
This meets the axes at  and 
By hypothesis, AB = 10
10k
2
 = 3600 k = 
Hence there are two lines given by 
4. (c) We have the equation
y
2
 + xy + px
2
 – x – 2y + p = 0
We know any general equation
ax
2
 + by
2
 + 2hxy + 2gx + 2fy + c = 0 ... (1)
represents two straight lines if
abc + 2fgh – af 
2
 – bg
2
 – ch
2
 = 0 ... (2)
On comparing given equation with (1), we get
a = p, b = 1, , , f = – 1, c = p
Put these value in equation (2)
p × 1 × p + 2 × – 1 × –  ×  – p × (–1)
2
 – 1
× 
?  ? 
? 4p
2
 – 5p + 1 = 0  ? (4p – 1) (p – 1) = 0
5. (a) Let the point (h, k) lie on a line x + y = 4
then h + k = 4 ...(i)
and ...(ii)
and 4h + 3k = 5 ...(iii)
On solving (i) and (ii); and (i) and (iii), we get the required points (3, 1)
and (–7, 11).
Trick : Check with options. Obviously, points (3, 1) and (–7, 11)
lie on x + y = 4 and perpendicular distance of these points from 4x
+ 3y = 10 is 1.
6. (d) Slope of the line in the new position is , since it is  to the line
ax + by + c = 0 and it cuts the x-axis at (2,0). Hence, the required
line passes through (2, 0) and its slope is . Required eq. is
   
7. (b) We know that if m
1
 and m
2
 are the slopes of the lines represented by
ax
2
 + 2hxy + by
2
 = 0,
then sum of slopes = m
1
 + m
2
 =  and
Page 4


D lies on x – y – 5 = 0
?
? .
From the given condition
we get,
?     ? 
? Equation of the line is
3. (b) The equation of any line parallel to 2x + 6y + 7 = 0 is 2x + 6y + k =
0
This meets the axes at  and 
By hypothesis, AB = 10
10k
2
 = 3600 k = 
Hence there are two lines given by 
4. (c) We have the equation
y
2
 + xy + px
2
 – x – 2y + p = 0
We know any general equation
ax
2
 + by
2
 + 2hxy + 2gx + 2fy + c = 0 ... (1)
represents two straight lines if
abc + 2fgh – af 
2
 – bg
2
 – ch
2
 = 0 ... (2)
On comparing given equation with (1), we get
a = p, b = 1, , , f = – 1, c = p
Put these value in equation (2)
p × 1 × p + 2 × – 1 × –  ×  – p × (–1)
2
 – 1
× 
?  ? 
? 4p
2
 – 5p + 1 = 0  ? (4p – 1) (p – 1) = 0
5. (a) Let the point (h, k) lie on a line x + y = 4
then h + k = 4 ...(i)
and ...(ii)
and 4h + 3k = 5 ...(iii)
On solving (i) and (ii); and (i) and (iii), we get the required points (3, 1)
and (–7, 11).
Trick : Check with options. Obviously, points (3, 1) and (–7, 11)
lie on x + y = 4 and perpendicular distance of these points from 4x
+ 3y = 10 is 1.
6. (d) Slope of the line in the new position is , since it is  to the line
ax + by + c = 0 and it cuts the x-axis at (2,0). Hence, the required
line passes through (2, 0) and its slope is . Required eq. is
   
7. (b) We know that if m
1
 and m
2
 are the slopes of the lines represented by
ax
2
 + 2hxy + by
2
 = 0,
then sum of slopes = m
1
 + m
2
 =  and
product of slopes = m
1
 m
2
 = 
Consider the given equation which is
x
2
 + 2hxy + 2y
2
 = 0
On comparing this equation with ax
2
 + 2hxy + by
2
 = 0,
we have a = 1, 2h = 2h and b = 2
Let the slopes be m
1
 and m
2
.
Given : m
1
: m
2
 = 1: 2
Let   m
1
 = x  and  m
2
 = 2x
? m
1
 +  m
2
 =  ? h = – 3x  ...(i)
and   m
1
m
2
 =  ? x . 2x =  ? x = ...(ii)
? From eqs. (i) and (ii), we have h = 
8. (c) The equation of the line through (–1, 3) and having the slope 1 is 
.
Any point on this line at a
distance r from P (–1, 3) is
This point is on the line 2x + y = 3 if
...(i)
But tan ? = 1; 
Page 5


D lies on x – y – 5 = 0
?
? .
From the given condition
we get,
?     ? 
? Equation of the line is
3. (b) The equation of any line parallel to 2x + 6y + 7 = 0 is 2x + 6y + k =
0
This meets the axes at  and 
By hypothesis, AB = 10
10k
2
 = 3600 k = 
Hence there are two lines given by 
4. (c) We have the equation
y
2
 + xy + px
2
 – x – 2y + p = 0
We know any general equation
ax
2
 + by
2
 + 2hxy + 2gx + 2fy + c = 0 ... (1)
represents two straight lines if
abc + 2fgh – af 
2
 – bg
2
 – ch
2
 = 0 ... (2)
On comparing given equation with (1), we get
a = p, b = 1, , , f = – 1, c = p
Put these value in equation (2)
p × 1 × p + 2 × – 1 × –  ×  – p × (–1)
2
 – 1
× 
?  ? 
? 4p
2
 – 5p + 1 = 0  ? (4p – 1) (p – 1) = 0
5. (a) Let the point (h, k) lie on a line x + y = 4
then h + k = 4 ...(i)
and ...(ii)
and 4h + 3k = 5 ...(iii)
On solving (i) and (ii); and (i) and (iii), we get the required points (3, 1)
and (–7, 11).
Trick : Check with options. Obviously, points (3, 1) and (–7, 11)
lie on x + y = 4 and perpendicular distance of these points from 4x
+ 3y = 10 is 1.
6. (d) Slope of the line in the new position is , since it is  to the line
ax + by + c = 0 and it cuts the x-axis at (2,0). Hence, the required
line passes through (2, 0) and its slope is . Required eq. is
   
7. (b) We know that if m
1
 and m
2
 are the slopes of the lines represented by
ax
2
 + 2hxy + by
2
 = 0,
then sum of slopes = m
1
 + m
2
 =  and
product of slopes = m
1
 m
2
 = 
Consider the given equation which is
x
2
 + 2hxy + 2y
2
 = 0
On comparing this equation with ax
2
 + 2hxy + by
2
 = 0,
we have a = 1, 2h = 2h and b = 2
Let the slopes be m
1
 and m
2
.
Given : m
1
: m
2
 = 1: 2
Let   m
1
 = x  and  m
2
 = 2x
? m
1
 +  m
2
 =  ? h = – 3x  ...(i)
and   m
1
m
2
 =  ? x . 2x =  ? x = ...(ii)
? From eqs. (i) and (ii), we have h = 
8. (c) The equation of the line through (–1, 3) and having the slope 1 is 
.
Any point on this line at a
distance r from P (–1, 3) is
This point is on the line 2x + y = 3 if
...(i)
But tan ? = 1; 
(i) becomes,
 
Hence the required distance = .
9. (c) Let the equation of the line be .
So, the coordinates of A and B are (a, 0) and (0, b) respectively.
Since the point (–5, 4) divides AB in the ratio 1 : 2
? – 5 = 
and 4 =  ? a = –  and b = 12
So the line is – = 1, i.e. – 8x + 5y = 60
10. (a) Let Q(a, b) be the reflection of P(4, – 13) in the line5x + y + 6 = 0
Then the mid-point  lies on 5x + y + 6 = 0
 5a + b + 19 = 0...(i)
Also PQ is perpendicular to 5x + y + 6 = 0
Therefore = – 1 a – 5b – 69 = 0..(ii)
Solving (i) and (ii), we get a = – 1, b = – 14.
11. (b) We have the equation