Chapter – 19 : Co-ordinate Geometry : Internal And External Division Of Straight Line Segment | Chapter Solution Class 9

Let us work out – 19

1. Find the Co-ordinate of the point which divides the line segment joining two points in the given ratio of the following:

(i) (6, -14) and (-8, 10); in the ratio 3 : 4 internally

(ii) (5, 3) and (-7, -2); in the ratio 2 : 3 internally

(iii) (-1, 2) and (4, -5); in the ratio 3 : 2 externally

(iv) (3, 2) and( (6, 5); in the ratio 2 : 1 externally

Solution:

(i) The Co-ordinate of the point which divides the line joining the points (6, -14) and (-8, 10) in the ratio of 3 : 4 internally is

=(\frac{3(-8)+4.6}{3+4}, \frac{3 \cdot 10+4(-14)}{3+4}) \\

=(\frac{-24+24}{7}, \frac{30-56}{7}) \\

=(\frac{0}{7},-\frac{26}{7}) \\

=(0, \frac{-26}{7})

(6, -14)

\therefore Required Co-ordinate is (0,-\frac{26}{7}) (Ans)

(ii) The Co-ordinate of the point which divides the line joining the points (5, 3) and (-7, -2) in the ratio 2 : 3 internally

=(\frac{2 .(-7)+3(5)}{2+3}, \frac{2 \cdot(-2)+33}{2+3}) \\

=(\frac{-14+15}{5}, \frac{-4+9}{5}) \\

=(\frac{-14+15}{5}, \frac{-4+9}{5}) \\

=(\frac{1}{5}, \frac{5}{5})\\

(\frac{1}{5}, 1) (Ans)

(iii) The Co-ordinate of the point which divides the lines joining the point (-1, 2) and (4, -5) in the ratio of 3 : 2 externally is

= (\frac{3.4-2 \cdot(-1)}{3-2}, \frac{3 \cdot(-5)-2(2)}{3-2})

=(\frac{12+2}{1}, \frac{-15-4}{1})

=(\frac{14}{1},-\frac{19}{1})

= (14, -19)

\therefore Required Co-ordinates is (14, -19) (Ans)

(iv) The Co-ordinates of the point which divides the line joining the point (3, 2) and (6, 5) in the ratio of 2 : 1 externally

=(\frac{2.6-1.3}{2-1}, \frac{2.5-1.2}{2-1})

=(\frac{12-3}{1}, \frac{10-2}{1})

=(9, 8)

\therefore Required Co-ordinate is (9, 8) (Ans)

2. Find the Co-ordinate of midpoint of line segment joining two points for the following

(i) (5, 4) and (3, -4)

(ii) (6, 0) and (0, 7)

Solution:

(i) Midpoint of line segment joining two points (5, 4) and (3, -4) is

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

=(\frac{8}{2}, \frac{0}{2})

= (4, 0)

\therefore Required Co-ordinate is (4, 0) (Ans)

(ii) Midpoint of the line segment joining two points (6, 0) and (0, 7)

=(\frac{6+0}{2}, \frac{0+7}{2})

=(\frac{6}{2}, \frac{7}{2})

=(3, \frac{7}{2})

\therefore Required co ordinate is (3, \frac{7}{2}) (Ans)

3. Let us calculate the ratio in which the point (1, 3) divides the line segment joining the point (4, 6) and (3, 5)

Solution:

Let the line segment joining the two points (4, 6) and (3, 5) divide the point (1, 3) in the ratio

\therefore The ordinate of the point (1, 3)

=\frac{m.5+n.6}{m+n}=\frac{5 m+6 n}{m+n} \\

\therefore \frac{5 m+6 n}{m+n}=3 \\

\text { or, } 5 m+6 n=3 m+3 n \\

\text { or, } 5 m-3 m=3 n-6 n \\

\text { or, } 2 m=-3 n \\

\frac{m}{n}=-\frac{3}{2} \\

\therefore m : n =-3 : 2

\therefore The line segment joining two points (4, 6) and (3, 5) divides the point (1, 3) in the ratio 3 : 2 externally.

4. Let us calculate in what ratio is the line segment joining the points (7, 3) and (-9, 6) divided by the y-axis.

Solution:

Let the line segment joining two points (7, 3) and (-9, 6) is divided by y-axis at the point P in the ratio m : n

The abscissa of P =\frac{m \cdot(-9)+n \cdot7}{m+n}

=\frac{-9 m+7 n}{m+n}

Since P is the point is on y- axis, so x = 0

\quad \frac{-9 m+7 n}{m+n}=0 \\

\text { or }-9 m+7 n=0 \\

\text { or, }-9 m=-7 n \\

\text { or, } \frac{m}{n}=\frac{7}{9} \\

\therefore m : n = 7 : 9

The line segment joining two points (7, 3) and (-9, 6) is divided by the y-axis internally in the ratio 7 : 9. (Ans)

5. Prove that when the points A (7, 3), B (9, 6), C (10, 12) and D (8, 9) are joined in order, then they will form a parallelogram.

Solution:

We have,

A (7, 3), B (9, 6), C (10, 12), D (8, 9)

Plotting the points on the cartesian plane, we see that if form a Quadrilateral ABCD.

The Co-ordinate of mid point of diagban AC.

=(\frac{7+10}{2}, \frac{3+12}{2})=(\frac{17}{2}, \frac{15}{2})

The co-ordinate of mid point of diagonat BD

=(\frac{9+8}{2}, \frac{6+9}{2})=(\frac{17}{2}, \frac{15}{2})

Diagonal AC and BD of Quadrilateral ABCD bisect each other

\therefore ABCD is a parallelogram.(proved)

6. If the points (3, 2),(6, 3),(x, y) and (6, 5) then joined in order and form a parallelogram then let us calculate the point (x, y)

Solution:

Let \mathrm{A=(3,2), B=(6,3), C=(x, y), D(6,5)}

The co-ordinate of midpoint of diagonal AC

=(\frac{6+6}{2}, \frac{3+5}{2})=(\frac{12}{2}, \frac{1}{2})=(6,4)

\therefore ABCD is a parallelogram

\therefore Diagonal AC and BD of quadrilateral ABCD bisects each other:

\therefore \frac{x+3}{2}=6, \quad \frac{y+2}{2}=4

or, x+3=12, \quad or, y+2=8

\therefore she required point is (9, 6) (Ans)

7. If \mathrm{(x_1, y_1),(x_2 y_2),(x_3, y_3) \ and \ (x_4, y_4)} points are joined in order to form parallelogram then prove that \mathrm{x_1+x_3 = x_2+x_4 \ and \ y_1+y_3=y_2+y_4.}

Solution:

Let \mathrm{A=(x_1, y_1), B=(x_2, y_2), C=(x_3, y_3), D=(x_4, y_4)}

\because ABCD is a parallelogram

\therefore The co-ordinate of midpoint of diagonal AC

=(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2})

The co-ordinate of midpoint of diagonal BD

=(\frac{x_2+x_4}{2}, \frac{y_2+y_4}{2})

\because ABCD is a parallelogram

\therefore \frac{x_1+x_3}{2}=\frac{x_2+x_4}{2} \ \frac{y_1+y_3}{2}=\frac{y_2+y_4}{2}

or, x_1+x_3=x_2+x_4 \ y_1+y_3=y_2+y_4 (proved)

8. The co-ordinate of vertices of A, B, C of a triangle ABC are (-1, 3) (1, -1) and (5, 1) respectively, let us calculate the length of Median A

Solution:

AD is median of \triangle A B C

\therefore D is mid point of BC

\therefore The co-ordinate of mid point of diagonal BC =(\frac{5+1}{2}, \frac{1-1}{2})

=(\frac{6}{2}, 0) \\

= (3, 0)

\therefore A D =\sqrt{(-1-3)^2+(3-0)^2} \\

=\sqrt{(-4)^2+(3)^2} \\

=\sqrt{16+9} \\

=\sqrt{25} \\

= 5 units (Ans)

9. The co-ordinates of vertices of triangle are (2, -4)(6, -2) and (-4, 2) respectively. Let us find the length of three medians of triangle.

Solution:

Let A = (2, -4), B = (6, -2), C = (-4, 2)

D is the mid point of BC

E is the mid point of AC

F is the mid point of AB

\therefore The co-ordinate of mid point of B C=(\frac{6-4}{2}, \frac{-2+2}{2})=(\frac{2}{2}, \frac{0}{2})=(1,0)

\therefore The co-ordinate of mid point of A C=(\frac{2-4}{2}, \frac{-4+2}{2})=(\frac{-2}{2}, \frac{-2}{2})=(-1,-1)

\therefore The co-ordinate of mid point of A B=(\frac{6+2}{2}, \frac{-4-2}{2})=(\frac{8}{2},-\frac{6}{2})=(4,-3)

AD =\sqrt{(2-1)^2+(-4-0)^2} \\

=\sqrt{1+16}=\sqrt{17} \text { units (Ans) } \\

BE =\sqrt{(6+1)^2+(-2+1)^2} \\

=\sqrt{49+1}=\sqrt{50} \text { units (Ans) } \\

CF =\sqrt{(4+4)^2+(-3-2)^2} \\

=\sqrt{64+25}=\sqrt{89} \text { units (Ans) } \\

10. The co-ordinate of mid point of sides of a triangle are (4, 3),(-2, 7) and (0, 11). Let us calculate the co-ordinate of its vertices.

Solution:

Let D = (4, 3), E = (-2, 7), F = (0, 11)

\therefore D, E and F are the mid points of BC, AC and AB respectively.

Let \mathrm{A=(x_1, y_1), \quad B=(x_2, y_2), \quad C =(x_3, \mathrm{y}_3)}

D is the mid point of BC

\frac{x_2+x_3}{2}=4 \quad, \quad \frac{y_2+y_3}{2}=3 \\

or, x_2+x_3 = 8 -------- (i) \quad y_2+y_3 = 6 --------- (ii)

E is the mid point of AC

\frac{x_1+x_3}{2}=-2 \quad, \quad \frac{y_1+y_3}{2}=7 \\

x_1+x_3=-4---------(iii) \quad y_1+y_3=14---------(iv)

F is mid point of AB,

=\frac{x_1+x_2}{2}=0 \quad \frac{y_1+y_2}{2}=11

x_1+x_2=0 ------ (v) \quad y_1+y_2=22 -------- (vi)

(i) + (iii) + (v) gives

2(x_1+x_2+x_3)=4

\text { or, } x_1+x_2+x_3=2 -------- (vii)

(ii) + (iv) + (vi) gives

2(y_1+y_2+y_3)=42

\text { or, } y_1+y_2+y_3=21----------(viii )

\mathrm{(vii) - (i) gives, \quad (viii) - (ii) gives}

x_1=-6, \quad y_1=15

\mathrm{(vii) - (iii) gives, \quad (viii) - (iv) gives}

x_2=6, \quad y_2=7

\mathrm{(vii) - (v) gives, \quad (viii) - (vi) gives}

x_3 = 2, \quad y_3=-1

\therefore A = (-6, 15), B = (6, 7), C = (2, -1)

\therefore These are the three vertices of the triangle (Ans)

11. M.C.Q

(i) The mid point of the line segment joining two points (l, 2 m), and (-l + 2m, 2l – 2m) is

(a) (l, m)

(b) (2, -m)

(c) (m, -l)

(d) (m, l)

Solution:

Mid point of line segment joining two point (l, 2m) and (-l + 2m, 2l – 2m)

=(\frac{l-l+2 m}{2}, \frac{2 m+2 l-2 m}{2}) \\

= (m, l)

\therefore(d) is correct option.

(ii) The abscissa at the point p which divides the line segment joining two point A (1, 5), B (-4, 7) internally in the ratio 2 : 3 is

(a) -1

(b) 11

(c) 1

(d) -11

Solution:

The abscissa at the point P

P =\frac{2 \cdot(-4)+3 \cdot 1}{2+3} \\

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

\therefore (a) is correct answer.

(iii) The co-ordinates of end points of a diameter of a circle are (7, 9) and (-1, -3). The co-ordinate of centre of circle is

(a) (3, 3)

(b) (4, 6)

(c) (3, -3)

(d) (4, -6)

Solution:

The co ordinate of mid point of end of a diameter i.e Co-ordinate of centre of circle is

=(\frac{7-1}{2}, \frac{9-3}{2})=(\frac{6}{2}, \frac{6}{2})=(3,3)

\therefore(a) is correct answer.

(iv) A point which divides the line segment joining two points (2, -5) and (-3, -2) externally in the ratio 4 : 3. The ordinate of point is

(a) -18

(b) -7

(c) 18

(d) 7

Solution:

The ordinate of point is =\frac{4 .(-2)-3 \cdot(-5)}{4-3}

=\frac{-8+15}{1}=7

\therefore(d) is correct answer.

(v) If the points P(1, 2), Q(4, 6), R(5, 7) and S(x, y) are the vertices of a parallelogram PQRS, then

(a) x = 2, y = 4

(b) x = 3, y = 4

(c) x = 2, y = 3

(d) x = 2, y = 5

Solution:

\therefore Co-ordinate of Mid point of diagonal PR = co-ordinate of mid point of diagonal QS

or, (\frac{1+5}{2}, \frac{2+7}{2})=(\frac{4+x}{2}, \frac{6+y}{2})

or, (3, \frac{9}{2})=(\frac{4+x}{2}, \frac{6+y}{2})

or, \frac{4+x}{2}=3 \quad, \frac{6+y}{2}=\frac{9}{2}

or, 4+x=6 ; 6+y=9

or, x=6-4, y=9-6

\therefore x=2 \quad, \quad \therefore y=3

\therefore(c) is correct answer

12. Short answer type question

(i) C is the centre of a circle and AB is the diameter; the co-ordinate. of A and C are (6, -7) and (5, -2), Let us calculate the co-ordinate of B

Solution:

\therefore C is mid point of AB.

\therefore Let B = (x, y)

Then \frac{x+6}{2}=5 \quad \frac{y-7}{2}=-2

\text { or, } x+6=10 \text { or, } y-7=-4 \\

\text { or; } x=10-6 \text { or, } y=-4+7 \\

\therefore x = 4,  y = 3

\therefore The co-ordinate of B = (4, 3) (Ans)

(ii) The points P and Q lie on 1 st and 3rd quadrent respectively.

(ii) The distance of the two points from x-axis and y-axis are 6 units and 4units respectively . Let us write the co-ordinate of mid point of line segment PQ.

Solution:

\therefore P lies in the 1st quadrant

\therefore the co-ordinate of P = (6, 4)

\therefore Q lies in 3rd quadrent

\therefore The co-ordinate of Q = (-6, -4)

The co-ordinate of mid point of line segment PQ

=(\frac{6-6}{2}, \frac{4-4}{2}) \\

= (0, 0) (Ans)

(iii) The points A and B lie on 2nd and 4th quadrent repectively and distance of each point from x-axis and y-axis are 8 units and 6 units respecyively Let us write the co-ordinate of mid point of the line segment.

Solution:

\therefore A lies on 2nd Quadrant

\therefore The co-ordinate of A = (-8, 6)

\therefore B lies on 4th quadrant

\therefore The co-ordinate of B=(8,-6)

\therefore The co-ordinate of mid point of line segment AB

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

=(0, 0) (Ans)

(iv) The point P lies on the line segment AB and AP = PB; the c0-ordinate of A and B are (3, -4) and (-5, 2) respectively. Let us write the co-ordinate of point P.

Solution:

\therefore The point P lies on line segment AB and AP = PB

\therefore P is the mid point of AB

\therefore The co-ordinate of point P=(\frac{3-5}{2}, \frac{-4+2}{2})

=(\frac{-2}{2}, \frac{-2}{2})=(-1,-1) (\text { Ans })

(v) The sides of Rectangle ABCD are parallel to Co-ordinate axes. Co-ordinate of B and D are (7, 3) and (2, 6); Let us write the co-ordinate of A and C and mid point of diagonal AC.

Solution:

\therefore Mid point of diagonal AC

Mid point of Diagonal BD

=(\frac{7+2}{2}, \frac{3+6}{2})=(\frac{9}{2}, \frac{9}{2}) \\

\therefore ABCD are parallel to the co-ordinate ones

\therefore The co-ordinate of A and C be (2, 3) and (7, 6) (Ans)