Chapter 19 Coordinate Geometry Internal And external Division Of A Straight Line Segment Exercise 19
Important Formulae :
1. Distance of (x, y) from the origin = \(\sqrt{x^2+y^2}\) units.
2. Distance between 2 points P(x1,y1) and Q(x2,y2)
= \(\overline{\mathrm{PQ}}=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2} \text { units. }\)
3. If two points A(x1,y1) and B(x2,y2) are internally divided by point P(x, y) in ratio mn, then the co-ordinates of P are
P = \(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\)
4. If two points A(x1,y1) and B(x2,y2) are externally divided by point P(x, y) in ratio mn, then the co-ordinates of P are
P = \(\left(\frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}\right)\)
Question 1. Find the co-ordinate of the point which divides the line segment joining two points in the given ratio for the following:
Read and Learn More WBBSE Solutions For Class 9 Maths
1. (6, -14) and (-8, 10) in the ratio 3 4 internally
Solution: The coordinates of the point
\(\begin{aligned}& =\left\{\frac{3 \times(-8)+4 \times 6}{3+4}, \frac{3 \times 10+4(-14)}{3+4}\right\} \\
& =\left(\frac{-24+24}{7}, \frac{30-56}{7}\right) \\
& =\left(0, \frac{-26}{7}\right)
\end{aligned}\)
∴ \(\left(0, \frac{-26}{7}\right)\)
2. (5, 3) and (-7, -2) in the ratio 2 3 internally
Solution: The coordinates of the point
& =\left\{\frac{2 \times(-7)+3 \times 5}{2+3}, \frac{2(-2)+3 \times 3}{2+3}\right\} \\
& =\left(\frac{-14+15}{5}, \frac{-4+9}{5}\right) \\
& =\left(\frac{1}{5}, 1\right)
\end{aligned}\)
∴ \(\left(\frac{1}{5}, 1\right)\)
3. (-1, 2) and (4, -5) in the ratio 3: 2 externally
Solution: The coordinates of the point
\(\begin{aligned}& =\left\{\frac{3 \times 4-2 \times(-1)}{3-2}, \frac{3(-5)-2 \times 2}{3-2}\right\} \\
& =\left(\frac{12+2}{1}, \frac{-15-4}{1}\right) \\
& =(14,-19)
\end{aligned}\)
= (14,-19)
∴ (14,-19)
4. (3, 2) and (6, 5) in the ratio externally
Solution: The coordinates of the point
=\(\left\{\frac{2 \times 6-1 \times 3}{2-1}, \frac{2 \times 5-1 \times 2}{2-1}\right\}\)
= \(\left(\frac{12-3}{1}, \frac{10-2}{1}\right)\)
= (9,8)
∴ (9,8)
Question 2. Find the co-ordinates of the mid-point of the line segment joining two points for the following:
1. (5, 4) and (3,-4)
Solution: The co-ordinates of the mid-point of the join of (5, 4) and (3,-4)
\(\begin{aligned}& =\left(\frac{5+3}{2}, \frac{4-4}{2}\right) \\
& =\left(\frac{8}{2}, 0\right)
\end{aligned}\)
= (4, 0)
∴ (4, 0)
2. (6, 0) and (0, 7)
Solution: The co-ordinates of the midpoint of the join of (6,0) and (0,7)
& =\left(\frac{6+0}{2}, \frac{0+7}{2}\right) \\
& =\left(3, \frac{7}{2}\right)
\end{aligned}\)
Question 3. Let us calculate the ratio in which point (1, 3) divides the line segment joining points (4, 6) and (3, 5).
Solution: Let the required ratio = m:n.
∴ \(1=\frac{m \times 3+n \times 4}{m+n}\)
or, m+n=3m + 4n.
or, m-3m = 4n-n
or, -2m=3n
or, \(\frac{m}{n}=\frac{-3}{2}\)
or, m: n= (-3):2
Question 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: Here the abscissa of the point P = x
∴ Point P point abscissa of (x co-ordinate) = \(\frac{m \times(-9)+n \times 7}{m+n}\)
∴ Point P point is situated on y-axis. ∴x=0.
∴ \(\frac{-9 m+7 n}{m+n}=0\)
or, -9m+7n = 0
or, -9m=-7n
or, 9m = 7n
or, \(\frac{m}{n}=\frac{7}{9}\)
∴ m:n = 7:9
Question 5. Prove that when the points A(7, 3), B(9, 6), C(10, 12) and D(8, 9) are joined in the order mentioned then they will form a parallelogram.
Solution: Mid-point of the diagonal AC is E.
\(\begin{aligned}& =\left(\frac{7+10}{2}, \frac{3+12}{2}\right) \\
& =\left(\frac{17}{2}, \frac{15}{2}\right)
\end{aligned}\)
& the mid-point of the diagonal BD is F = \(\left(\frac{9+8}{2}, \frac{6+9}{2}\right)\)
= \(=\left(\frac{17}{2}, \frac{15}{2}\right)\)
Diagonals of the quadrilateral, AC & BD intersect each other at
= \(=\left(\frac{17}{2}, \frac{15}{2}\right)\)
∴ ABCD is a parallelogram. Proved
Question 6. If the points (3, 2), (6, 3), (x, y) and (6, 5) when joined in the order mentioned and form a parallelogram, then let us calculate the point (x, y).
Solution:
Given
If the points (3, 2), (6, 3), (x, y) and (6, 5) when joined in the order mentioned and form a parallelogram,
Let A(3,2), B(6,3), C(x,y) & D(6,5) be the four points of the parallelogram ABCD. E and F are the mid-points of the diagonals AC & BD.
∴ E is the mid-point of AC.
∴ Co-ordinate of E= \(\left(\frac{3+x}{2}, \frac{2+y}{2}\right)\)
∵ F is the mid-point of BD.
∴ Co-ordinates of F = \(\left(\frac{6+6}{2}, \frac{3+5}{2}\right)\) = (6,4)
∴ ABCD is a parallelogram.
∴ AC and BD will bisect each other, i.e., E and F will be the same point.
∴ \(\frac{3+x}{2}=6\)
or, 3+ x = 12
or, x = 12-3=9
or, 2+ y = 8
or, y=8-2=6
∴ The required point is (9,6).
Question 7. If ((x1,y1), (x2,y2), (x3,y3) and (x4,y4) points are joined in order to form a parallelogram, then prove that x1 +x3 = x2 +x4 and y1 + y3 = y2+y4
Solution:
Given
If ((x1,y1), (x2,y2), (x3,y3) and (x4,y4) points are joined in order to form a parallelogram
Let A((x1,y1), B(x2,y2), C(x3,y3) & D(x4,y4) are the four points of the parallelogram ABCD.
∴ E is the mid-point of AC.
Co-ordinates of E = \(\left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}\right)\)
F is the mid-point of BD.
Co-ordinates of F = \(\left(\frac{\mathrm{x}_2+\mathrm{x}_4}{2}, \frac{\mathrm{y}_2+\mathrm{y}_4}{2}\right)\)
∵ ABCD is a parallelogram.
Diagonals AC & BD intersect at a point if E & F are the same points.
∴ \(\frac{x_1+x_3}{2}=\frac{x_2+x_4}{2} \text { or, } x_1+x_3=x_2+x_4\)
& \(\frac{y_1+y_3}{2}=\frac{y_2+y_4}{2} \text { or, } y_1+y_3=y_2+y_4\)
Question 8. The coordinates of vertices A, B, and C of a triangle ABC are (-1, 3), (1, -1) and (5, 1) respectively, let us calculate the length of the median AD.
Solution:
Given
The coordinates of vertices A, B, and C of a triangle ABC are (-1, 3), (1, -1) and (5, 1) respectively
Let D is the mid-point of BC.
Co-ordinates of D = \(\left(\frac{1+5}{2}=\frac{-1+1}{2}\right)\) = (3,0)
∴ Length of median AD = \(\sqrt{(-1-3)^2+(3-0)^2} \text { unit }\)
\(\begin{aligned}& =\sqrt{(-4)^2+(3)^2} \text { unit } \\
& =\sqrt{16+9} \text { unit } \\
& =\sqrt{25} \text { unit } \\
& =5 \text { unit }
\end{aligned}\)
Question 9. The coordinates of the vertices of the triangle are (2, 4), (6, -2) and (-4, 2) respectively. Let us find the length of three medians of a triangle.
Solution:
Given
The coordinates of the vertices of the triangle are (2, 4), (6, -2) and (-4, 2) respectively.
Let D, E, & F be the mid-points of BC, CA & AB respectively.
∴ Now co-ordinates of
D= \(\begin{aligned}
& D=\left(\frac{6-4}{2}=\frac{-2+2}{2}\right) \\
& =(1,0)
\end{aligned}\)
Co-ordinates of E = \(\left(\frac{2-4}{2}=\frac{-4+2}{2}\right)\) = (-1,-1)
Co-ordinates of F = \(\left(\frac{2+6}{2}=\frac{-4-2}{2}\right)\) = (4,-3)
∴ Length of median BE = \(\sqrt{(2-1)^2+(-4-0)^2} \text { unit }\)
\(\begin{aligned}& =\sqrt{(1)^2+(-4)^2} \\
& =\sqrt{1+16} \text { unit } \\
& =\sqrt{17} \text { unit }
\end{aligned}\)
Length of median BE = \(\sqrt{(6+1)^2+(-2+1)^2} \text { unit }\)
\(\begin{aligned}& =\sqrt{(7)^2+(-1)^2} \\
& =\sqrt{50} \text { unit } \\
& =5 \sqrt{2} \text { unit }
\end{aligned}\)
Length of median CF = \(\sqrt{(-4-4)^2+(2+3)^2} \text { unit }\)
\(\begin{aligned}& =\sqrt{(-8)^2+(5)^2} \text { unit } \\
& =\sqrt{64+25} \text { unit } \\
& =\sqrt{89} \text { unit }
\end{aligned}\)
Question 10. The coordinates of the midpoints of sides of a triangle are (4, 3), (-2, 7) and (0, 11). Let us calculate the coordinates of its vertices.
Solution:
Given
The coordinates of the midpoints of sides of a triangle are (4, 3), (-2, 7) and (0, 11).
Let A (x1,y1) B (x2,y2) & C(x3,y3) are three vertices of the ΔABC
& P(4, 3); Q (-2, 7) & R (0, 11) are the three mid-points of the AB, BC & CA respectively.
\(\frac{x_1+x_2}{2}=4, \text { or, } x_1+x_2=8\) ….(1)
\(\frac{y_1+y_2}{2}=3, \text { or, } y_1+y_2=6\) …(2)
\(\frac{x_2+x_3}{2}=-2, \text { or, } x_2+x_3=-4\) ….(3)
\(\frac{y_2+y_3}{2}=7, \text { or, } y_2+y_3=14\) ….(4)
\(\frac{x_3+x_1}{2}=0, \text { or, } x_3+x_1=0\) ….(5)
\(\frac{y_3+y_1}{2}=11, \text { or, } y_3+y_1=22\) …(6)
Adding (1), (3) & (5), we get 2(x1+ x2+x3)= 4
or, x1+ x2+x3 = 4/2
or, x1+ x2+x3 = 2 ….(7)
(6) – (1) => x3 =-6
(6) – (3) => x1= 6
(6) – (5) => x = 2
Again, by adding (2), (4) & (5), we get
2(y1+ y2+y3)= 42
∴ y1+ y2+y3 = 21 …… (8)
(8) – (2) => y3= 15
(8) – (4) => y1 = 7
(8) – (4) => y2 = -1
∴ The coordinates of the vertices of the triangle are (6, 7) (2, -1), and (-6, 15).
Question 11. Multiple choice questions
1. The mid-point of line segment joining two points (1, 2m), and (-1 + 2m, 21 – 2m) is
1. (1, m)
2. (1, -m)
3. (m, -1)
4. (m, 1)
Solution; Mid-point
= \(\left(\frac{1-1+2 m}{2}, \frac{2 m+21-2 m}{2}\right)\) = (m, 1)
∴ 4. (m + 1)
2. The abscissa at point P which divides the line segment joining two points A(1, 5) and B(-4, 7) internally in the ratio 2:3 is
1. -1
2. 11
3. 1
4. -11
Solution:
Co-ordinates of P = \(\left(\frac{2 \times(-4)+3 \times 1}{2+3}, \frac{2 \times 7+3 \times 5}{2+3}\right)\)
= \(\left(-1, \frac{29}{5}\right)\)
∴ The abscissa is -1.
∴ – 1
3. The coordinates of the end points of a diameter of a circle are (7, 9) and (-1,-3). The coordinates of the centre of the circle is
1. (3, 3)
2. (4, 6)
3. (3, -3)
4. (4, -6)
Solution: Co-ordinates of the centre
\(=\left(\frac{7-1}{2}, \frac{9-3}{2}\right)\)
= (3, 3)
∴ (3, 3)
4. A point which divides the line segment joining two points (2,-5) and (-3,-2) externally in the ratio 4: 3. The ordinate of the point is
1. -18
2. -7
3. 18
4. 7
Solution: The coordinates of the point
=\(\left(\frac{4 \times(-3)-3 \times 2}{4-3}, \frac{4 \times(-2)-3 \times(-5)}{4-3}\right)\)
= (-18, 7)
∴ The ordinate of the point is 7.
4. 7
5. If the points P(1, 2), Q(4, 6), R(5, 7) and S(x, y) are the vertices of a parallelogram PQRS, then
1. x = 2, y = 4
2. x = 3, y=4
3. x = 2, y = 3
4. x = 2, y = 5
Solution: The mid-point of the diagonal PR is
= \(\left(\frac{1+5}{2}, \frac{2+7}{2}\right)=\left(3, \frac{9}{2}\right)\)
= \(\left(3, \frac{9}{2}\right)\)
The mid-point of the diagonal QS = \(\left(\frac{4+x}{2}, \frac{6+y}{2}\right)\)
∴ As mid-points PR & QS are the same point.
∴ \(\frac{4+x}{2}=3\)
or, 4+ x = 6
or, x = 2
or, \(\frac{6+y}{2}=\frac{9}{2}\)
or, 12+2y= 18
or, 2y= 18-12
or, y = 6/2=3
∴ 3. x = 2, y = 3
Question 12. Short answer type questions:
1. C is the centre of a circle and AB is its diameter; the coordinates of A and C are (6, -7) and (5, -2). Let us calculate the coordinates of B.
Solution:
Given
C is the centre of a circle and AB is its diameter; the coordinates of A and C are (6, -7) and (5, -2).
Let the co-ordinate of B = (x, y)
∴ The mid-point of AB = \(\left(\frac{6+x}{2}, \frac{-7+y}{2}\right)\)
∴As centre is the mid-point of the diameter.
∴ \(\frac{6+x}{2}=5[\latex]
or, 6+x=10
or, x = 4
and [latex]\frac{-7+y}{2}=-2\)
or, – 7+ y = -4
or, y = -4+7
or, y = 3
∴ The coordinates of B = (4,3).
2. The points P and Q lie on 1st and 3rd quadrants respectively. The distance of the two points from the x-axis and y-axis is 6 units and 4 units respectively. Let us write the coordinates of the mid-point of line segment PQ.
Solution:
Given
The points P and Q lie on 1st and 3rd quadrants respectively. The distance of the two points from the x-axis and y-axis is 6 units and 4 units respectively.
As P lies in 1st quadrant
∴ The abscissa & the ordinate both are positive.
∴ The coordinates of P = (4, 6).
Again, Q lies in the 3rd quadrant.
∴ The abscissa & the ordinate both are negative.
∴ The coordinates of Q = (-4, -6).
∴ The mid point of PQ is = \(\left(\frac{4-4}{2}, \frac{6-6}{2}\right)=(0,0) .\)
= (0, 0).
3. points A and B lie in the 2nd and 4th quadrants respectively and the distance of each point from x-axis and y-axis are 8 units and 6 units respectively. Let us write the coordinate of the mid-point of line segment AB.
Solution:
Given
points A and B lie in the 2nd and 4th quadrants respectively and the distance of each point from x-axis and y-axis are 8 units and 6 units respectively.
As lies in the 2nd quadrant.
∴ The abscissa is negative, but the ordinate is positive.
∴ The coordinates of A = (-6, 8).
Again, B lies in the 4th quadrant.
∴ The abscissa is positive, but the ordinate is negative.
∴ The coordinates of B = (6, -8).
∴ The mid-point of A and B is \(\left(\frac{-6+6}{2}, \frac{8-8}{2}\right)=(0,0)\)
= (0, 0).
4. The point Plies on the segment AB and AP = PB; the coordinates of A and B are (3,-4) and (-5, 2) respectively. Let us write the coordinates of point P.
Solution:
Given
The point Plies on the segment AB and AP = PB; the coordinates of A and B are (3,-4) and (-5, 2) respectively.
As P lies on AB and AP = PB
∴ P is the mid-point of AB.
∴ The co-ordinates of P = \(\left(\frac{3-5}{2}, \frac{-4+2}{2}\right)\) =(-1,-1).
5. The sides of rectangle ABCD are parallel to the coordinate axes. The coordinates of B and D are (7, 3) and (2, 6). Let us write the coordinates of A and C and the mid-point of diagonal AC.
Solution:
Given
The sides of rectangle ABCD are parallel to the coordinate axes. The coordinates of B and D are (7, 3) and (2, 6).
ABCD is a rectangle, AB II x-axis & AD II y-axis.
∴The coordinates of A = (2, 3).
∴ The coordinates of C
∴ The midpoint of AC = \(\left(\frac{2+7}{2}, \frac{3+6}{2}\right)=\left(\frac{9}{2}, \frac{9}{2}\right)\)