ML Aggarwal Solution Class 10 Chapter 11 Section Formula Test
Test
Question 1
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the point C are (0, – 3). If origin is the mid-point of the base BC, find the coordinates of the points A and B
Sol :
Base BC of an equilateral ∆ABC lies on y-axis
co-ordinates of point C are (0, – 3),
origin (0, 0) is the mid-point of BC.
∴0=x+02⇒x2=0⇒x=0
y−32=0⇒y−3=0
⇒y=3
∴ Co-ordinates of B are (0,3)
Again let co-ordinates of A be (x, 0) as it lies on x-axis.
∵AB=AC=BC=6 units
=√(x−0)2+(0−3)2=√62
x2+(−3)2=62
x2+9=36⇒x2=36−9=27
⇒x=±3√3
∴ Co-ordiantes of A will be (±3√3,0)
Question 2
A and B have co-ordinates (4, 3) and (0, 1), Find
(i) the image A’ of A under reflection in the y – axis.
(ii) the image of B’ of B under reflection in the lineAA’.
(iii) the length of A’B’.
(i) Co-ordinates of A’, the image of A (4, 3)
reflected in y-axis will be ( – 4, 3).
(ii) Co-ordinates of B’ the image of B (0, 1)
reflected in the line AA’ will be (0, 5).
(iii) Length A’B’
Question 3
Find the co-ordinates of the point that divides the line segment joining the points P (5, – 2) and Q (9, 6) internally in the ratio of 3 : 1.
Sol :
Let R be the point whose co-ordinates are (x, y)
which divides PQ in the ratio of 3:1.
=3×9+1×53+1=27+54=324=8
y=m1y2+m2y1m1+m2=3×6+1×(−2)3+1
=18−24=164=4
∴ Co-ordinates of R will be (8,4)
Question 4
Find the coordinates of the point P which is three-fourth of the way from A (3, 1) to B ( – 2, 5).
Sol :
Co-ordinates of A (3, 1) and B ( – 2, 5)
P lies on AB such that
⇒AP=3 PB⇒AP:PB=3:1
Let co-ordinates of P be (x, y)
∴x=mx2+nx1m+n=3×(−2)+1×33+1
=−6+34=−34
y=my2+ny1m+n=3×5+1×13+1
=15+14=164=4
∴ Co-ordinates of P are (−34,4)
Question 5
P and Q are the points on the line segment joining the points A (3, – 1) and B ( – 6, 5) such that AP = PQ = QB. Find the co-ordinates of P and Q.
Sol :
Given
AP = PQ = QB
Let co-ordinates of P will be (x1,y1) and of Q will be (x2,y2)
∴x1=m1x2+m2x1m1+m2=1×(−6)+2(3)1+2
=−6+63=03=0
y1=m1y2+m2y1m1+m2
=1×5+2(−1)1+2=5−23=33=1
∴ Co-ordinates of P will be (0,1)
Again
x2=m1x2+m2x1m1+m2=2×(−6)+1(3)2+1
=−12+33=−93=−3
y2=m1y2+m2y1m1+m2=2×5+1(−1)∘2+1
=10−13=93=3
∴ Co-ordinates of Q will be (-3,3)
Question 6
The centre of a circle is (α + 2, α – 5). Find the value of a given that the circle passes through the points (2, – 2) and (8, – 2).
Sol :
Let A (2, -2), B (8, -2) and centre of the circle be
O (α + 2, α – 5)
OA=√(2−α−2)2+(−2−α+1)2
=√α2+(−1−α)2=√α2+(α+1)2..(i)
OB=√(8−α−2)2+(−2−α+1)2
=√(6−α)2+(−1−α)2
=√(6−α)2+(1+α)2...(ii)
From (i) and (ii)
√α2+(α+1)2=√(6−α)2+(1+α)2
Squaring both sides.
α2+(α+1)2=(6−α)2=(ˆ1+α)2
⇒α2=(6−α)2 [ dividing by (α+1)2]
⇒α2=36−12α+α2
=α2−α2+12α=36
⇒12α=36
∴α=3612=3
Question 7
The mid-point of the line joining A (2, p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.
Sol :
Given
(3, 5) is the mid-point of A (2, p) and B (q, 4)
∴q=4
and 5=p+42⇒p+4=10
⇒p=10−4=6
∴p=6
Hence p=6, q=4
Question 8
The ends of a diameter of a circle have the co-ordinates (3, 0) and ( – 5, 6). PQ is another diameter where Q has the coordinates ( – 1, – 2). Find the co-ordinates of P and the radius of the circle.
Sol :
Let AB be the diameter where co-ordinates of
A are (3, 0) and of B are (-5, 6).
Co-ordinates of its origin O will be
Now PQ is another diameter in which co-ordinates of Q are (-1,-2)
Let co-ordinates of P be (x, y) Then co-ordinates of centre O will be
(−1+x2,−2+y2)
∴−1+x2=−1
⇒−1+x=−2
⇒x=−2+1=−1
and −2+y2=3
⇒−2+y=6
⇒y=6+2=8
∴ Co-ordinates of P will be (-1,8)
Now length of radius OP
=√(−1+1)2+(8−3)2
=√(0)2+(5)2=√0+25
=√25=5 units
Question 9
In what ratio does the point ( – 4, 6) divide the line segment joining the points A( – 6, 10) and B (3, – 8) ?
Sol :
Let the point (-4, 6) divides the line segment joining the points
A (-6, 10) and B (3, -8), in the ratio m : n
−4=3m−6nm+n⇒−4m−4n=3m−6n
⇒−4n+6n=3m+4m⇒7m=2n
⇒mn=27
∴ Ratio =2 :7
Question 10
Find the ratio in which the point P ( – 3, p) divides the line segment joining the points ( – 5, – 4) and ( – 2, 3). Hence find the value of p.
Sol :
Let P (-3, p) divides AB in the ratio of m1 : m2 coordinates of
A (-5, -4) and B (-2, 3)
⇒−3=m1(−2)+m2(−5)m1+m2
⇒−3=−2m1−5m2m1+m2
⇒−3m1−3m2=−2m1−5m2
⇒−3m1+2m1=−5m2+3m2
⇒−m1=−2m2⇒2m2=m1
⇒m1m2=21⇒m1:m2=2:1
Again,
p=m1y2+m2y1m1+m2
=2×3+1×(−4)2+1=6−43=23
Here, p=23
Question 11
In what ratio is the line joining the points (4, 2) and (3, – 5) divided by the x-axis? Also find the co-ordinates of the point of division.
Sol :
Let the point P which is on the x-axis, divides the line segment
joining the points A (4, 2) and B (3, -5) in the ratio of m1 : m2.
and let co-ordinates of P be (x, 0)
⇒−5m1+2m2m1+m2=0⇒−5m1+2m2=0
⇒−5m1=−2m2⇒5m1=2m2
⇒m1m2=25⇒m1:m2=2:5
Again,
x=m1x2+m2x1m1+m2=2(3)+5(4)2+5=6+207=267
∴ Co-ordinates of P will be (267,0)
Question 12
If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points ( – 4 – 3) and (6, 3). Hence, find the co-ordinates of P.
Sol :
Let co-ordinates of A be (-4, 3) and of B (6, 3) and of P be (2, y)
Let the ratio in which the P divides AB be m1:m2
∵x=m1x2+m2x1m1+m2
⇒2=m1×6+m2×(−4)m1+m2
⇒2=6m1−4m2m1+m2
⇒2m1+2m2=6m1−4m2
⇒6m1−2m1=2m2+4m2
⇒4m1=6m2
⇒m1m2=64=32
∴m1:m2=3:2
∴y=m1y2+m2y1m1+m2=3×3+2×33+2=9+65=155=3
∴ co-ordinates of P will be (2,3)
Question 13
Determine the ratio in which the line 2x + y – 4 = 0 divide the line segment joining the points A (2, – 2) and B (3, 7). Also find the co-ordinates of the point of division.
Sol :
Points are given A (2, -2), B (3, 7)
and let the line 2x + y – 4 = 0 divides AB in the ratio m1 : m2
at P and let co-ordinates of
and y=m1×7+m2(−2)m1+m2=7m1−2m2m1+m2
∴P lies on the line 2x+y-4=0, then
2(3m1+2m2)m1+m2+7m1−2m2m1+m2−4=0
⇒6m1+4m2+7m1−2m2−4m1−4m2=0
⇒9m1−2m2=0⇒9m1=2m2
⇒m1m2=29 or m1:m2=2:9
∴x=2×3+2×92+9=6+1811=2411
and y=2×7−2×92+9=14−1811=−411
∴ Co-ordinates of P will be (2411,−411)
Question 14
The point A(2, – 3) is reflected in the v-axis onto the point A’. Then the point A’ is reflected in the line x = 4 onto the:point A”.
(i) Write the coordinates of A’ and A”.
(ii) Find the ratio in which the line segment AA” is divided by the x-axis. Also find the coordinates of the point of division.
Sol :
A’ is the reflection of A(2, -3) in the x-axis
(i) ∴ Co-ordinates of A’ will be (2, 3)
Draw a line x = 4 which is parallel to y-axis
A” is the reflection of A’ (2, 3)
∴Co-ordinates OA” will be (6, 3)
(ii) Join AA” which intersects x-axis at P whose
co-ordinate are (4, 0)
Let P divide AA” in the ratio in m1:m2
⇒3m1−3m2=0⇒3m1=3m2
⇒m1m2=33=11
∴m1:m2=1:1
Hence P(4, 0) divides AA” in the ratio 1 : 1
Question 15
ABCD is a parallelogram. If the coordinates of A, B and D are (10, – 6), (2, – 6) and (4, – 2) respectively, find the co-ordinates of C.
Sol :
Let the co-ordinates of C be (x, y) and other three vertices
of the given parallelogram are A (10, – 6), B, (2, – 6) and D (4, – 2)
∴ ABCD is a parallelogram
Its diagonals bisect each other.
Let AC and BD intersect each other at O.
∴O is mid-points of BD
∴ Co-ordinates of O will be
Again O is the mid-point of AC then
3=10+x2⇒10+x=6
⇒x=6−10=−4
and −4=−6+y2
⇒6+y=−8⇒y=−8+6
∴y=−2
Hence Co-ordinates of C will be (-4,-2)
Question 16
ABCD is a parallelogram whose vertices A and B have co-ordinates (2, – 3) and ( – 1, – 1) respectively. If the diagonals of the parallelogram meet at the point M(1, – 4), find the co-ordinates of C and D. Hence, find the perimeter of the parallelogram. find the perimeter of the parallelogram.
Sol :
ABCD is a || gm , m which co-ordinates of A are (2, -3) and B (-1, -1)
Its diagonals AC and BD bisect each other at M (1, -4)
∴ M is the midpoint of AC and BD
Let co-ordinates of C be(x1,y1) and of D be (x2,y2)
when M is the midpoint of AC then
⇒2+x1=2
⇒x1=2−2=0
and −8=−3+y1
⇒y1=−8+3=−5
∴ Co-ordinates of C are (0,-5)
Again M is mid-point of BD, then
1=−1+x22,−4=−1+y22
⇒−1+x2=2
⇒x2=2+1=3
and −1+y2=−8
⇒y2=−8+1=−7
∴ co-ordinates of D are (3,-7)
Now, length of AB
=√[2−(−1)]2+[−3−(−1)]2
=√(2+1)2+(−3+1)2
=√(3)2+(−2)2=√9+4=√13
and Length of BC
=√[0−(−1)]2+[−5−(−1)]2
=√(1)2+(−5+1)2
=√1+(−4)2=√1+16=√17
∴ Perimeter of ||g, ABCD=2(AB+BC)
=2(√13+√17)
Question 17
In the adjoining figure, P (3, 1) is the point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of A and B.
Again ,
1=m1y2+m2y1m1+m2=2×y+3×02+3
=2y+05=2y5
⇒2y=5⇒y=52
∴Co-ordinates of A will be (5,0) and of B
will be (⋅0,52)
Question 18
Given, O, (0, 0), P(1, 2), S( – 3, 0) P divides OQ in the ratio of 2 : 3 and OPRS is a parallelogram.
Find : (i) the co-ordinates of Q.
(ii)the co-ordinates of R.
(iii) the ratio in which RQ is divided by y-axis.
⇒2y′5=2
⇒2y′=10⇒y′=5
∴ Co-ordinates of Q will be (52,5)
∵the diagonals of a parallelogram bisect each other
∴In ||gm OPRS, diagonals OR and PS bisect each other at M
∵M is the mid-point of PS
∴Co-ordinates of M will be
(ii) ∵M is the mid-point of OR also
∴−1=0+x′′2
⇒x′′=−2
and 1=0+y′′2⇒y′′=2
∴ Co-ordinates of R will be (-2,2)
(iii) RQ is dividing by y-axis in N
Let the ratio in which N divides RQ in m1:m2
∵ N lies on y-axis
∴ its abscissia (x)=0
0=\frac{m_{2} x^{\prime}+m_{2} x^{\prime \prime}}{m_{1}+m_{2}} $
$\Rightarrow 0=\frac{m_{1} \times \frac{5}{2}+m_{2}(-2)}{\cdot m_{1}+m_{2}}
⇒5m12−2m2m1+m2=0
⇒52m1−2m2=0
⇒52m1=2m2
⇒m1m2=2×25=45
∴m1:m2=4:5
Question 19
If A (5, – 1), B ( – 3, – 2) and C ( – 1, 8) are the vertices of a triangle ABC, find the length of the median through A and the co-ordinates of the centroid of triangle ABC.
Sol :
A (5, -1), B (-3, -2) and C (-1, 8) are the vertices of ∆ABC
D, E and F are the midpoints of sides BC, CA and AB respectively and G is the centroid of the ∆ABC
(x1+x22,y1+y22) or (−3−12,−2+82)
(−42,62) or (-2,3)
∵G is the centroid
∴ Co-ordinates of G will be
(x1+x2+x33,y1+y2+y33)
or (5−3−13,−1−2+83) or (13,53)
Length of AD=√(x2−x1)2+(y2−y1)2
=√(−2−5)2+(3+1)2=√(−7)2+(4)2
=√49+16=√65 units
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