Q: Two pairs of lines have the equations 10x2+13xy−3y2=0 and ax2+2hxy+by2=0. The condition that any two lines among them are coincident is

The given equation is (2x+3y)(5x−y)=0 ⇒xy=3−2 or xy=5 ax2+2hxy+by2=0 ⇒a+2 h(xy)+b(xy)2=0 ⇒a+2 h3−2+b(94)=0 or a+2 h⋅5+b⋅25=0 ⇒9a−12 h+4 b=0 or a+10 h+25 b=0

Q: The point of intersection of the lines ax+by=1 and bx+ay=1 lies on the line

The lines are ax+by=1 bx+ay=1 Solving, point of intersection is (a+bab,a+bab) This point satisfies the equations in (a), (b), (c)

Q: The lines (λ−2)x+3λy+(λ+4)=0,x+y−1=0 and (2λ−5)x+(λ+1)y+2λ=0 are concurrent for

Lines are concurrent. ⇒⇒∣∣1λ−22λ−513λλ+1−1λ+42λ∣∣=0∣∣0(2λ+2)(4λ−5)0(4λ+4)(3λ+1)−1(λ+4)2λ∣∣=02(λ+1)(−5λ+11)=0∣=−1,511

Question 1

Statement 1 AB and AC are perpendicular lines with AB=4 and AC=3. The centroid of the triangle ABC is at a distance 35​ from A. and Statement 2 Centroid of a triangle divides any median in the ratio 2:1.

Accepted Answer

Statement 2 is true.
Consider Statement1.
Choosing A B and AC is the of coordinates, the coordinates of the vertices of the △ABC are (0,0),(4,0) and (0,3) centroid is at (34​,1).
If G denotes the centroid,
AG2=(34​)2+12 (Since A is the origin)
=3252​⇒AG=35​
Statement 1 is true.
Since we can get the coordinates of G by using statement 2 also,
Choice is (a)

Question 2

Statement 1 The pair of lines 2x2−3y2−x+19y−xy−28=0 intersect at P. The equation of the pair of lines through P perpendicular to the given pair of lines is 3x2−2y2+xy−3x +13y−18=0 and Statement 2 If the pair of lines represented by ax2+2hxy+by2+2gx+2fy+c=0 intersect at P the equation of the pair of lines through P perpendicular to the given pair of lines is bx2−2hxy+ay2+2gx+2fy+c=0

Accepted Answer

Statement 2 is false.
Consider statement 1
On factoring, the two lines are
x+y−4=0 and 2x−3y+7=0
The point of intersection of the lines is at (1,3).
The lines perpendicular to the above lines are of the form y−x+k=0 and 3x+2y+k2​=0 where k1​ and k2​ are obtained by using the fact that the lines pass through (1,3)
We get k1​=−2 and k2​=−9
The lines are y−x+2=0 and
3x+2y−9=0
Their combined equation is
3x2−2y2+xy−3x+13y−18=0
⇒ statement 1 is true
Choice (c)

Question 3

Triangle ABC is formed by the lines x−y+2=0,3x+2y−2=0 and 2x+y+1=0. Then,

Accepted Answer

Lines are
x−y+2=0- (1) ⇒ side BC
3x+2y−2=0- (2) ⇒ side CA
2x+y+1=0- (3) ⇒ side AB

Question 4

Two pairs of lines have the equations 10x2+13xy−3y2=0 and ax2+2hxy+by2=0. The condition that any two lines among them are coincident is

Accepted Answer

The given equation is (2x+3y)(5x−y)=0
⇒xy​=3−2​ or xy​=5
ax2+2hxy+by2=0
⇒a+2 h(xy​)+b(xy​)2=0
⇒a+2 h3−2​+b(94​)=0 or a+2 h⋅5+b⋅25=0
⇒9a−12 h+4 b=0 or a+10 h+25 b=0

Question 5

The point of intersection of the lines ax​+by​=1 and bx​+ay​=1 lies on the line

Accepted Answer

The lines are ax​+by​=1
bx​+ay​=1
Solving, point of intersection is (a+bab​,a+bab​)
This point satisfies the equations in (a), (b), (c)

Question 6

The lines (λ−2)x+3λy+(λ+4)=0,x+y−1=0 and (2λ−5)x+(λ+1)y+2λ=0 are concurrent for

Accepted Answer

Lines are concurrent.
⇒⇒​∣∣​1λ−22λ−5​13λλ+1​−1λ+42λ​∣∣​=0∣∣​0(2λ+2)(4λ−5)​0(4λ+4)(3λ+1)​−1(λ+4)2λ​∣∣​=02(λ+1)(−5λ+11)=0∣=−1,511​​

Question 7

The lines 2x+y=4 and 3x−y=1 intersect at M. A variable line is drawn through M meeting the x−w is at A and the y−w is at B
(a) The locus of the mid point of AB is the curve y=(x−1)x​
(b) The locus of the foot of perpendicular from the origin on AB is the curve.
x2+y2−x−2y=0
(c) Locus of the centroid of △OAB. Where, O is the origin, is the curve 3xy=2x−y
(d) Locus of the circumcentre of the △OAB where, O is the origin, is the curve y=(x−1)x​

Accepted Answer

The point M is (1,2)
Any line through M having slop is
Y−2=m(x−1)
⇒mx−y+(2−m)=0
 Put y=0 in (1)⇒A is ( mm−2​,0)
put x=0 in (1)⇒B is (0,2−m)
(a) Let P denote the mid-point of AB.
Then, x coordinate of
P=x=2 mm−2​=21​− m1​
Y coordinate of P=y=22−m​
=1−2m​
2m​=1−y
⇒x=21​−2(1−y)1​
⇒2x=1−1−y1​=1−y−y​
Locus of P is the curve X−xy+y=0 or
Y=x−1x​. 
(b)
Let P be the foot of the perpendicular from O. Slope of OP= m−1​.

Slove y= m−1​x and
mx+ m1​x=(m−2)
x=(m2+1)m(m−2)​
⇒y=(m2+1)−(m−2)​
Coordinates of P are (m2+1m(m−2)​,m2+1−(m−2)​) M=y−x​
Locus of P is the curve
y=1+(yx​)22+yx​​=(x2+y2)(2y+x)y​
x2+y2−x−2y=0
(c) Centroid of △OAB is the point (x,y) where,
x=3 mm−2​,y=32−m​
3mx=m−2
3y=2−m
3mx=−3y
m=−xy​
Locus is the curve
X=x−3y​x−y​−2​=3y2x+y​
3xy=2x+y
(d) circumcentre is the mid point of AB. Hence, the locus is the same as in (a)

Question 8

A ray of light is sent along the line 3x+y−9=0. Upon reaching the line 2x+7y−8=0, the ray gets reflected from it. Then,

Accepted Answer

2x+7y−8=0
(a) is true.
3x+y−9+λ(2x+7y−8)=0
⇒(2λ+3)x+(7λ+1)y−(8λ+9)=0→(1)
Slope of the line (1) is (7λ+1)−(2λ+3)​.
Slope of 3x+y−9=0 is →3
Slope of 2x+7y−8=0 is 0→7−2​
If θ is the angle between the above lines
tanθ=∣∣​1+76​−3+72​​∣∣​=∣∣​13−19​∣∣​
=1319​
For the reflected ray,
1+7(7λ+1)2(2λ+3)​−(7λ+1)2λ+3​)+72​​=±1319​
⇒λ=0,53−26​.
Hence the reflected ray corresponds to 53−26​
Equation of the reflected ray is
107x−129y−269=0
Solving 3x+y−9=0 and 2x+7y−8=0
We get the point P as (1955​,196​)

Question 9

The line x+y−5=0 intersects the pair of lines 2x2+3y2−5xy−5x+5y=0 at A and B. Then,

Accepted Answer

If is clear that one of the lines represented by 2x2+3y2−5xy−5x+5y=0
Factoring the expression,
2x2−3xy−2xy+3y2−5(x−y)=0
(2x−3y)(x−y)−5(x−y)=0
(x−y)(2x−3y−5)=0
The lines are x−y=0
2x−3y−5=0
The point of intersection of these lines is (−5,−5).
We can find A an B by solving the pairs
A is (2,25​) and B is (4,1)
Slope 0 A=1, Slope of 0 B=41​
tanθ=1+41​1−41​​=53​
AB2=(4−25​)2+(1−25​)2
=49​+49​=29​
AB=2​3​
If α is the angle between the given pair of lines
tanα=(2+3)2425​−6​​
=2×2×51​=51​

Question 10

Match the followings.

Accepted Answer

(a) The equation of the line joining the origin and the point of intersection of the line y=mx+2 and the curve x2+y2=1 is x2+y2=(2y−mx​)2 ⇒x2(4−m2)+2mxy+3y2=0Since these lines are at right angles,∴4−m2+3=0⇒m2=7∴m4=49(b) Equation of the bisectors of the angle between the lines 3x2+2pxy−3y2=0 is 3+3x2−y2​=−pxy​⇒x2+p6​xy−y2=0If this represents the lines5x2−2qxy−5y2=0i.e., x2−52​qxy−y2=0, thenp6​=5−2q​⇒pq=−15(c) Let the lines represented byax3+bx2y+cxy2+dy3=0 are y=m1​x,y=m2​x and y=m3​x. where, m1​m2​=−1The given equation can be written asd(xy​)3+c(xy​)2+b(xy​)+a=0⇒m1​m2​m3​=d−a​⇒m3​=da​m3​ is a root of the above equation∴d(da​)3+c(da​)2+b(da​)+a=0⇒ d2a3​+d2a2c​+dab​+a=0⇒a2+ac+bd+d=0(d) Let the slopes of the lines be m and 3 m Then m+3 m=−8⇒m=−2m.3 m=a⇒a=12

Question 11

Match the followings.

Accepted Answer

(a) Let the equation of the variable line be ax+by+c=0Given that a2+b2​2a+c​+a2+b2​2b+c​+a2+b2​−2a−2b+c​=0⇒c=0⇒ the line passes through the origin for all a and b(b) The given lines are x=0,y=0 and x+y=4. Clearly they form a right triangle, right angled at the origin.(c) 4a2+9 b2−c2+12ab=0⇒(2a+3 b)2=c2⇒2a+3 b=±c∴ either 2a+3 b+c=0 or 2a+3 b−c=0The given family of straight lines is ax+by+c=0 or −ax−by−c=0. They are concurrent at (2,3) or (−2,−3)(d) Let P′ be the reflection of P(1,3) on the line 3x+y+3=0Then the equation of PP′ is x−3y+k=0This passes through (1,3)⇒1−9+k=0⇒k=8The point of intersection of 3x+y+3=0 and x−3y +8=0 is M(10−17​,1021​)M is the mid-point of the line segment PP'.If P′ is (h,k) then21+h​=10−17​ and 23+k​=1021​⇒h=5−22​,k=56​

Question 12

Match the followings.

Accepted Answer

(a) length of the altitude p=2​∣x1​+y1​−2∣​=2​1​∴P=asin60⇒2​1​=2a3​​⇒a=32​​∴ Area =43​​×a2=43​​×32​=23​1​(b) Let C is the centre of the circle x2+y2=10 its radius AC=10​.line 5​x+2y=35​ intersect at A and B. CD⊥AB and D will be the midpoint of AB.∴CD=5+4​∣−35​∣​=5​Also AD=AC2−CD2​=10−5​=5​∴AB=2AD=25​∴ Area =21​AB×CD=21​×25​×5​=5(c) xy+2x+2y+4=0⇒(x+2)(y+2)=0∴ Required area =21​×2×2=2(d) Area =∣∣​am2−2hlm+bl2n2h2−ab​​∣∣​=2+5.0+3.04425​−6​​=1

Question 13

Obtain the coordinates of the centre of the circle x2+y2−4x+7y−9=0. Also find its radius.

Accepted Answer

We have g=−2,f=27​,c=−9
Centre of the circle is at (2,2−7​) and its radius = 4+449​+9​=2101​​

Question 14

Obtain the coordinates of the centre of the circle 3x2+3y2 +6x−8y+1=0. Also find its radius.

Accepted Answer

The equation of the circle may be rewritten as
x2+y2+2x−38​y+31​=0
We have g=1,f=−34​ and c=31​
Centre of the circle is at (−1,34​) and radius
=1+916​−31​​=322​​

Question 15

Find the equation of the circle on the line segment joining the points (1,−3) and (2,5) as diameter.

Accepted Answer

Equation of the circle is (x−1)(x−2)+(y+3)(y−5)=0 or x2+y2−3x−2y−13=0

Question 16

Find the equation of the circle whose centre is at the point (5,2) and which passes through the point (1,−1).

Accepted Answer

Since the circle passes through (1,−1), its radius is (5−1)2+(2+1)2​=5. The equation of the circle is therefore, (x−5)2+(y−2)2=25
⇒x2+y2−10x−4y+4=0

Question 17

Find the equation of the circle passing through the points (1,2),(2,−1) and (3,5)

Accepted Answer

Let the equation of the circle be x2+y2+2gx+2fx+c=0 Since it passes through (1,2),1+4+2g+4f+c=0 or 2g+4f+c=−5
Similarly, using the fact that the circle passes through the points (2,−1) and (3,5), we get the equations 4 g−2f+c=−5
6 g+10f+c=−34−(3)
Solving (1), (2) and (3) we obtain the values of g,f and c as 6−29​,18−29​ and 9100​ respectively.
Thus, the equation of the circle is x2+y2−658​x−1858​y+9100​=0 or 9(x2+y2)−87x−29y+100=0

Question 18

Find the equation of the circle whose centre is on the line x=2y and which passes through the points (−1,2) and (3,−2)

Accepted Answer

Let the equation of the circle be x2+y2+2gx+2fy+c=0
Since the centre lies on the line x=2y,−g=−2f or g=2f
Since the circle passes through the points (−1,2) and (3,−2), we have
1+4−2 g+4f+c=0
and 9+4+6 g−4f+c=0
or −2 g+4f+c=5
and 6g−4f+c=−13
Since g=2f, on solving the equations we get g=−2, f=−1 and c=−5.
The equation of the circle is x2+y2−4x−2y−5=0

Question 19

Find the equation of the circle which touches the coordinate axes and passes through the point (2,1).

Accepted Answer

Since the circle has to pass through the point (2,1) and touch the coordinate axes, the circle or circles must be in the first quadrant.
We may assume the equation of the circle as (x−h)2+(y−h)2=h2
P(2,1) lies on the circle ⇒(2−h)2+(1−h)2=h2
⇒h2−6 h+5=0
⇒h=1,5
The two circles are (x−1)2+(y−1)2=1

and (x−5)2+(y−5)2=25
or x2+y2−2x−2y+1=0
and x2+y2−10x−10y+25=0

Question 20

Find the equation of the circle which touches the lines x=0,y=0 and x=c

Accepted Answer

The coordinates of the centre of the circle are easily seen as (2c​,2c​) and its radius is 2c​.

The equation of the circle is (x−2c​)2+(y−2c​)2=(2c​)2 ⇒4(x2+y2)−4cx−4cy+c2=0

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Solutions for IIT-JEE Super Course in Mathematics - Coordinate Geometry and Vector Algebra by Trishna Knowledge Systems

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