RS Aggarwal Solutions Class 10 Ex 16D

Q1) Find the ratio in which the point P(m,6) divides the join of A(-4,3) and B(2,8). Also, find the value of m.

Let the point P(m, 6) divide the line AB in the ratio k : 1.

Then, by the section formula:

x = \(\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}\) , y = \(\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}\)

The coordinates of P are (m, 6).

m = \(\frac{2k-4}{k+1}\) , 6 = \(\frac{8k+3}{k+1}\)

\(\Rightarrow\) m (k+1) = 2k – 4 , 6k + 6 = 8k + 3

\(\Rightarrow\) m (k+1) = 2k – 4 , 6 – 3 = 8k – 6k

\(\Rightarrow\) m (k+1) = 2k – 4 , 2k = 3

\(\Rightarrow\) m (k+1) = 2k – 4 , k = \(\frac{3}{2}\)

Therefore, the point P divides the line AB in the ratio 3 : 2.

Now, putting the value of k in the equation m (k + 1) = 2k – 4 , we get

m(\(\frac{3}{2}+1\)) = 2(\(\frac{3}{2}\)) – 4

\(\Rightarrow\)m (\(\frac{3}{2}+1\)) = 3 – 4

\(\Rightarrow\) \(\frac{5m}{2}+1\) = -1

\(\Rightarrow\) 5m = -2

\(\Rightarrow\) m = \(\frac{-2}{5}\)

Therefore, the value of m = \(\frac{-2}{5}\)

So, the coordinates of P are (\(\frac{-2}{5},6\))

Q2) Find the ratio in which the point (-3,k) divides the join of A(-5,-4) and B(-2,3). Also, find the value of k.

Let the point P(-3, k) divide the line AB in the ratio s : 1.

Then, by the section formula:

x = \(\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}\) , y = \(\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}\)

The coordinates of P are (-3, k)

-3 = \(\frac{-2s-5}{s+1}\) , k = \(\frac{3s-4}{s+1}\)

\(\Rightarrow\)-3s-3 = -2s – 5, k(s+1) = 3s – 4

\(\Rightarrow\) -3s+2s = -5+3, k(s+1) = 3s – 4

\(\Rightarrow\) – s = – 2, k(s+1) = 3s – 4

\(\Rightarrow\) s = 2, k(s+1) = 3s – 4

Therefore, the point P divides the line AB in the ratio 2 : 1.

Now, putting the value of k in the equation k (s + 1) = 3s – 4, we get

k(2+1) = 3(2) – 4

\(\Rightarrow\) 3k = 6-4

\(\Rightarrow\) 3k = 2

Therefore, the value of k = \(\frac{2}{3}\)

So, the coordinates of P are (\(-3,\frac{2}{3}\))

Q3) In what ratio is the line segment joining A(2,-3) and B(5,6) divided by the X-Axis? Also, find the co-ordinates of the point of division.

Let AB be divided by the x-axis in the ratio k : 1 at the point P

Then, by the section formula the co-ordinates of P are

x = \(\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}\) , y = \(\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}\)

P = ( \(\frac{5k+2}{k+1}, \; \frac{6k-3}{k+1}\) )

But P lies on the x-axis, so its ordinate is 0

Therefore, \(\frac{6k-3}{k+1}\) = 0

\(\Rightarrow\) 6k- 3 = 0

\(\Rightarrow\) 6k = 3

\(\Rightarrow\) k = \(\frac{3}{6}\)

\(\Rightarrow\) k = \(\frac{1}{2}\)

Therefore, the required ratio is \(\frac{1}{2}:1\), which is same as 1:2

Thus, the x-axis divides the line AB in the ratio is 1:2 at the point P.

Applying k = \(\frac{1}{2}\), we get the co-ordinates of point:

P(\(\frac{5k+2}{k+1},0\))

= \(P\left (\frac{5*\frac{1}{2}+2}{\frac{1}{2}+1},0\right)\)

= \(P\left (\frac{\frac{5+4}{2}}{\frac{1+2}{2}},0\right )\)

= P\(\left (\frac{9}{3},0\right )\)

= P(3,0)

Hence, the point of intersection of AB and the x-axis is P(3,0)

Q4) In what ratio is the line segment joining the points A(-2, -3) and B(3,7) divided by the Y-axis ? Also, find the co-ordinates of the point of division.

Let AB be divided by the x-axis in the ratio k : 1 at the point P.

Then, by section formula the coordinates of P are

x = \(\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}\) , y = \(\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}\)

P = ( \(\frac{3k-2}{k+1} ,\; \frac{7k-3}{k+1}\) )

But P lies on the y-axis; so, its abscissa is 0.

Therefore, \(\frac{3k-2}{k+1}\) = 0

\(\Rightarrow\) 3k- 2 = 0

\(\Rightarrow\) 3k = 2

\(\Rightarrow\) k = \(\frac{2}{3}\)

\(\Rightarrow\) k = \(\frac{2}{3}\)

Therefore, the required ratio is \(\frac{2}{3}:1\) which is same as 2 : 3

Thus, the x-axis divides the line AB in the ratio 2:3 at the point P.

Applying k = \(\frac{2}{3}\),we get the coordinates of point P.

P(\(0,\frac{7k-3}{k+1}\))

= \(P\left (0,\frac{7*\frac{2}{3}-3}{\frac{2}{3}+1}\right)\)

= \(P\left (0,\frac{\frac{14-9}{3}}{\frac{2+3}{3}},0\right )\)

= P\(\left (0,\frac{5}{5}\right )\)

= P(0,1)

Hence, the point of intersection of AB and the x-axis is P(0,1)

Q5) In what ratio does the line x-y-2 = 0 divide the line segment joining the points A(3, -1) and B(8,9) ?

Let the line x – y – 2 = 0 divide the line segment joining the points A(3,-1) and B(8,9) in the ratio k:1 at P

Then, by section formula the coordinates of P are

x = \(\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}\) , y = \(\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}\)

P = (\(\frac{8k+3}{k+1}) – (\frac{9k-1}{k+1}\))

Since, P lies on the line x – y – 2 = 0, we have.

\((\frac{8k+3}{k+1})-( \frac{9k-1}{k+1})-2 = 0\)

\(\Rightarrow\) 8k + 3 – 9k + 1 – 2k – 2 = 0

\(\Rightarrow\) 8k – 9k -2k + 3 + 1 – 2 = 0

\(\Rightarrow\) -3k + 2 = 0

\(\Rightarrow\) -3k = -2

\(\Rightarrow\) k = \(\frac{2}{3}\)

So, the required ratio is \(\frac{2}{3}:1\), which is equal to 2 : 3

Q6) Find the lengths of the medians of \(\Delta\;ABC\) whose vertices are A(0,-1), B(2,1) and C(0,3).

The vertices of \(\Delta\;ABC\) are A(0,-1), B(2,1) and C(0,3)

Let AD, BE and CF be the medians of \(\Delta\;ABC\)

Let D be the midpoint of BC. So, the co-ordinates of D are

\(D(\frac{2+0}{2},\frac{1+3}{2})\) = D\((\frac{2}{2},\frac{4}{2})\) = D(1,2)

Let E be the midpoint of AC. So, the co-ordinates of E are

\(D(\frac{0+0}{2},\frac{-1+3}{2})\) = D\((\frac{0}{2},\frac{2}{2})\) = E(0,1)

Let F be the midpoint of AB. So, the co-ordinates of F are

\(D(\frac{0+2}{2},\frac{-1+1}{2})\) = D\((\frac{2}{2},\frac{0}{2})\) = F(1,0)

AD = \(\sqrt{(1-0)^{2}+(2-(-1))^{2}}\)

= \(\sqrt{(1)^{2}+(3)^{2}}\)

= \(\sqrt{1+9}\)

= \(\sqrt{10}\)units

BE = \(\sqrt{(0-2)^{2}+(1-1)^{2}}\)

= \(\sqrt{(-2)^{2}+(0)^{2}}\)

= \(\sqrt{4+0}\)

= 2 units

CF = \(\sqrt{(1-0)^{2}+(0-3)^{2}}\)

= \(\sqrt{(1)^{2}+(-3)^{2}}\)

= \(\sqrt{1+9}\)

= \(\sqrt{10}\)units

Therefore, the lengths of the medians: AD = \(\sqrt{10}\)units , BE = 2 units , CF = \(\sqrt{10}\)units

Q7) Find the centroid of \(\Delta\;ABC\) whose vertices are A(-1,0) ,B(5, -2) and C(8,2).

Here (x1 = -1 , y1 = 0) , (x2 = 5 , y2 = -2) and (x3 = 8 , y3 = 2)

Let G(x, y) be the centroid of the \(\Delta\;ABC\). Then,

x = \(\frac{1}{3}(x_{1}+x_{2}+x_{3})\)

= \(\frac{1}{3}(-1+5+8)\)

= \(\frac{1}{3}(12)\)

= 4

y = \(\frac{1}{3}(y_{1}+y_{2}+y_{3})\)

= \(\frac{1}{3}(0-2+2)\)

= \(\frac{1}{3}(0)\)

= 0

Hence, the centroid of \(\Delta\;ABC\) is G(4,0)

Q8) If G(-2,1) is the centroid of a \(\Delta\;ABC\) and two of its vertices are A(1,-6) and B(-5, 2), find the third vertex of the triangle.

Two vertices of \(\Delta\;ABC\) are A(1,-6) and B(-5,2). Let the third vertex be C (a, b)

Then the co-ordinates of its centroid are

C\((\frac{1-5+a}{3},\frac{-6+2+b}{3})\)

C\((\frac{-4+a}{3},\frac{-4+b}{3})\)

But it is given that G(-2,1) is the centroid. Therefore

-2 = \(\frac{-4+a}{3}\) , 1 = \(\frac{-4+b}{3})\)

\(\Rightarrow\) -6 = – 4 + a, 3 = – 4 + b

\(\Rightarrow\) -6+4 = a, 3 + 4 = b

\(\Rightarrow\) a = -2, b = 7

Therefore, the third vertex of \(\Delta\;ABC\) is C(-2,7)

Q9) Find the third vertex of \(\Delta\;ABC\) and two of its vertices are B( -3,1 ) and C(0, -2) , and its centroid is at the origin.

Two vertices of \(\Delta\;ABC\) are B(-3,1) and C(0,-2). Let the third vertex be A (a, b)

Then the co-ordinates of its centroid are

C\((\frac{-3+0+a}{3},\frac{1-2+b}{3})\)

C\((\frac{-3+a}{3},\frac{-1+b}{3})\)

But it is given that the centroid is at the origin, that is G(0,0). Therefore

0 = \(\frac{-3+a}{3}\) , 0 = \(\frac{-1+b}{3})\)

\(\Rightarrow\) 0 = – 3 + a, 0 = – 1 + b

\(\Rightarrow\) 3 = a, 1 = b

\(\Rightarrow\) a = 3, b = 1

Therefore, the third vertex of \(\Delta\;ABC\) is A(3,1)

Q10) Show that the points A(3, 1), B(0, -2), C(1,1) and D(4, 4) are the vertices of parallelogram ABCD.

The points are A (3, 1), B (0, -2), C (1, 1) and D (4, 4)

Join AC and BD, intersecting at O

We know that the diagonals of a parallelogram bisect each other

Midpoint of AC = \((\frac{3+1}{2},\frac{1+1}{2})\)

= \((\frac{4}{2},\frac{2}{2})\)

= (2, 1)

Midpoint of BD = \((\frac{0+4}{2},\frac{-2+4}{2})\)

= \((\frac{4}{2},\frac{2}{2})\)

= (2, 1)

Thus, the diagonals AC and BD have the same midpoint.

Therefore, ABCD is a parallelogram

Q11) If the points P (a, -11), B (5, b), C (2, 15) and D (1, 1) are the vertices of a parallelogram PQRS, find the values of a and b.

The points are P (a, -11), B (5, b), C (2, 15) and D (1, 1)

Join PQ and QS, intersecting at O

We know that the diagonals of a parallelogram bisect each other

Therefore, O is the midpoint of PR as well as QS

Midpoint of PQ = \((\frac{a+2}{2},\frac{-11+15}{2})\)

= \((\frac{a+2}{2},\frac{4}{2})\)

= \( (\frac{a+2}{2},2)\)

Midpoint of QS = \( (\frac{5+1}{2},\frac{b+1}{2})\)

= \( (\frac{6}{2},\frac{b+1}{2})\)

= \( (3,\frac{b+1}{2})\)

Therefore, \(\frac{a+2}{2}\) = 3 , \(\frac{b+1}{2}\) = 2

\(\Rightarrow\) a + 2 = 6 , b + 1 = 4

\(\Rightarrow\) a = 6 – 2, b = 4 – 1

\(\Rightarrow\) a = 4 and b = 3

Q12) If three consecutive vertices of a parallelogram ABCD are A(1, -2), B(3, 6) and C(5, 10), find its fourth vertex D.

Let A (1, -2), B (3, 6), C (5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D (a, b)

Join AC and BD, intersecting at O

We know that the diagonals of a parallelogram bisect each other

Therefore, O is the midpoint of AC as well as BD

Midpoint of AC = \((\frac{1+5}{2},\frac{-2+10}{2})\)

= \((\frac{6}{2},\frac{8}{2})\)

= (3, 4)

Midpoint of BD = \((\frac{3+a}{2},\frac{6+b}{2})\)

= \((\frac{3+a}{2},\frac{6+b}{2})\)

Therefore, \(\frac{3+a}{2}\) = 3 , \(\frac{6+b}{2}\) = 4

\(\Rightarrow\) 3 + a = 6 , 6 + b = 8

\(\Rightarrow\) a = 6 – 3, b = 8 – 6

\(\Rightarrow\) a = 3 and b = 2

Therefore, the fourth vertex is D (3,2)

Q13) In what ratio does y-axis divide the line segment joining the points (-4, 7) and (3, -7) ?

Let y-axis divides the line segment joining the points (-4, 7) and (3, -7) in the ratio k: 1.Then

\(0=\frac{3k-4}{k+1}\)

\(\Rightarrow\) 3k = 4

\(\Rightarrow\) \(k=\frac{4}{3}\)

Hence, the required ratio is 4:3

Q14) If the point P(\(P(\frac{1}{2},y)\), y) lies on the line segment joining the points A(3, -5) and B(-7, 9) then find the ratio in which P divides AB. Also, find the value of y.

Let the point \(P(\frac{1}{2},y)\)divides the line segment joining the points (3,-5) and ( -7, 9) in the ratio k: 1.Then

\((\frac{1}{2},y)=\left (\frac{k(-7)+3}{k+1},\frac{k(9)-5}{k+1}\right )\)

\(\Rightarrow\frac{-7k+3}{k+1}=\frac{1}{2}\) and \(\Rightarrow\frac {9k-5}{k+1}= y\)

\(\Rightarrow\) k + 1 = -14k + 6

\(\Rightarrow\) 15k = 5

\(\Rightarrow k = \frac{1}{3}\)

Now, substituting k = \(\frac{1}{3}\) in \(\frac{9k-5}{k+1}= y\), we get

\(\frac{\frac{9}{3}-5}{\frac{1}{3}+1}=y\)

\(\Rightarrow \frac{9-15}{1+3}=\frac{-3}{2}\)

Hence the required ratio is 1: 3 and \(y=\frac{-3}{2}\)

Q15) Find the ratio in which the line segment joining the points A(3, -3) and B(-2, 7) is the divided x-axis. Also, find the point of division.

The line segment joining the points A (3, -3) and B (-2, 7) is divided by x-axis. Let the required ratio be k: 1. So

\(0=\frac{k(7)-3}{k+1}\)

\(\Rightarrow k =\frac {3}{7}\)

Now

\(Point of division=\left(\frac{k(-2)+3}{k+1},\frac{k(7)-3}{k+1}\right )\)

= \(\left (\left(\frac{\frac{3}{7}*(-2)+3}{\frac{3}{7}+1}\right),\left (\frac{\frac{3}{7}*(7)-3}{\frac{3}{7}+1}\right)\right)\)

= \(\left (\left(\frac{-6+21}{3+7}\right),\left(\frac{21-21}{3+7}\right ) \right )\)

= \((\frac{3}{2},0)\)

Hence the required ratio is 3: 7 and the point of division is \((\frac{3}{2},0)\)

Q16) the base QR of an equilateral triangle PQR lies on x-axis. The co-ordinates of the point Q are (-4, 0) and the origin is the midpoint of the base. Find the co-ordinates of the points P and R.

Let (x, 0) be the coordinates of R. Then

\(0=\frac{-4+x}{2}\)

\(\Rightarrow\) x = 4

Thus, the coordinates of R are (4, 0).

Here, PQ = QR = PR and the coordinates of P lies on y-axis. Let the coordinates of P be (0, y). Then

PQ = QR

\(\Rightarrow\) PQ2 = QR2

\(\Rightarrow\) (0 + 4)2 + (y – 0)2 = 82

\(\Rightarrow\) y2 = 64 – 16 = 48

\(\Rightarrow\)\(y=\pm 4\sqrt{3}\)

Hence, the required coordinates are R(4, 0) and P(\(0, 4\sqrt{3}\)) or P(\(0, -4\sqrt{3}\))

Q17) The base BC of an equilateral triangle ABC lies on Y-axis. The co-ordinates of points C are (0, -3). the origin is the midpoint of the base. Find the coordinates of the points A and B. also, find the co-ordinates of another point D such that ABCD is a rhombus.

Let (0, y) be the coordinates of B. Then

\(0=\frac{-3+y}{2}\)

\(\Rightarrow\) y = 3

Thus, the coordinates of B are (0, 3).

Here, AB = BC= AC and by symmetry the coordinates of A lies on x-axis. Let the coordinates of A be (x, 0). Then

AB = BC

\(\Rightarrow\) AB2 = BC2

\(\Rightarrow\) (x – 0)2 + (0 – 3)2 = 62

\(\Rightarrow\) x2 = 36 – 9 = 27

\(\Rightarrow\) \(y=\pm 3\sqrt{3}\)

If the co-ordinates of point A are (\(3\sqrt{3},0\)), then the co-ordinates of D are (\(-3\sqrt{3},0\))

If the co-ordinates of point A are (\(-3\sqrt{3},0\)), then the co-ordinates of D are (\(3\sqrt{3},0\))

Hence, the required coordinates are A(\(3\sqrt{3},0\)), B(0, 3) and D(\(-3\sqrt{3},0\))

or

A(\(-3\sqrt{3},0\)), B(0, 3) and D(\(3\sqrt{3},0\))

Q18) Find the ratio in which the point P(-1, y) lying on the line segment joining points A(-3,10) and B(6, -8) divides it. Also, find the value of y.

Let k be the ratio in which P(-1, y) divides the line segment joining the points A(-3, 10) and B( 6, -9).

Then

\((-1, y) = \left (\frac{k(6)-3}{k+1},\frac{k(-8)+10}{k+1}\right )\)

\(\Rightarrow\frac {k(6)-3}{k+1}= – 1\) and

\(\Rightarrow y = \frac {k(-8)+10}{k+1}\)

\(\Rightarrow k = \frac{2}{7}\)

Now, substituting k = \(\frac{2}{7}\) in \(\frac{k(-8)+10}{k+1}\), we get

\(\frac{\frac{-8*2}{7}+10}{\frac{2}{7}+1}=y\)

\(\Rightarrow \frac{-16+70}{9}= 6\)

Hence the required ratio is 2: 7 and y = 6

Q19) ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). if P,Q, R and S be the midpoints of AB, BC, CD and DA respectively, show that PQRS is a rhombus.

Here, the points P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Then

Co-ordinates of P = \((\frac{-1-1}{2},\frac{-1+4}{2})\)

= \((-1,\frac{3}{2})\)

Co-ordinates of Q = \((\frac{-1+5}{2},\frac{4+4}{2})\)

= (2, 4)

Co-ordinates of R = \((\frac{5+5}{2},\frac{4-1}{2})\)

= \((5,\frac{3}{2})\)

Co-ordinates of S = \((\frac{-1+5}{2},\frac{-1-1}{2})\)

= (2, -1)

Now,

\(PQ=\sqrt{(2+1)^{2}+\left (4-\frac{3}{2}\right)^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\)

\(QR =\sqrt{(5-2)^{2}+\left (\frac{3}{2}-4\right)^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\)

\(RS =\sqrt{(5-2)^{2}+\left (\frac{3}{2}+1\right)^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\)

\(SP = \sqrt{(2+1)^{2}+\left (-1-\frac{3}{2}\right)^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2}\)

\(PR=\sqrt{(5+1)^{2}+\left (\frac{3}{2}-\frac{3}{2}\right)^{2}}=\sqrt{36}= 6\)

\(QS = \sqrt{(2-2)^{2}+\left (-1-4\right)^{2}}=\sqrt{25}= 5\)

Thus PQ = QR = RS = SP and PR is not equal to QS.

Therefore PQRS is a rhombus

Q20) The midpoint P of the line segment joining the points A(-10, 4) and B(-2, 0) lies on the line segment joining the points C(-9, -4) and D(-4, y). Find the ratio in which P divides CD. Also find the value of y.

The midpoint of AB is \(\left (\frac{-10-2}{2},\frac{4+0}{2}\right)=P(-6,2)\)

Let k be the ratio in which P divides CD. So

\((-6, 2) = \left (\frac{k(-4)-9}{k+1},\frac{k(y)-4}{k+1}\right )\)

\(\Rightarrow\frac {k(-4)-9}{k+1}= – 6 and y = \frac {k(y)-4}{k+1}\)

\(\Rightarrow k = \frac{3}{2}\)

Now, substituting k = \(\frac{3}{2}\) in \(\frac{k(y)-4}{k+1}=2\), we get

\(\frac{y*\frac{3}{2}-4}{\frac{3}{2}+1}=2\)

\(\Rightarrow \frac{3y-8}{5}= 2\)

\(\Rightarrow y = \frac{10+8}{3}\)

\(\Rightarrow y = 6\)

Hence the required ratio is 3: 2 and y = 6


Practise This Question

The diagram below represents a section of undisturbed layers of sedimentary rock. It shows the location of fossils of several closely related species. According to currently accepted evolutionary theory, which is the most probable assumption about species A, B, and C?