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This point also lies on 2x + y - 4 = 0
⇒
⇒
⇒ 9k - 2 = 0
⇒ k = 2/9
Therefore, the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the points A(2, −2) and B(3, 7) is 2:9.
2. Find a relation between x and y if the points (x,y),(1, 2)and (7, 0)are collinear.
Area =
0 = 1/2 [2x - y + 7y - 14]
0 =1/2 [2x + 6y - 14]
[2x + 6y - 14] = 0
x + 3y - 7 = 0
This is the required relation between x & y.
3. Find the centre of a circle passing through the points (6,-6), B(3,-7)and (3, 3).
However OA = OB (Radii of same circle)
⇒
.png)
=>x² + 36 - 12x + y² + 36 + 12y = x² + 9 - 6x + y² + 49 -14y
⇒ -6x + 2y + 14 = 0
⇒ 3x + y = 7 ....1
Similary OA = OC (Radii of same circle)
=>x² + 36 - 12x + y² + 36 + 12y = x² + 9 - 6x + y² + 9 - 6y
⇒ -6x + 18y + 54 = 0
⇒ -3x + 9y = -27 .....(2)
On adding equation (1) and (2), we obtain
10y = - 20
y = - 2
From equation (1), we obtain
3x − 2 = 7
3x = 9
x = 3
Therefore, the centre of the circle is (3, −2).
4. The two opposite vertices of a square are (-1,2) and (3,2). Find the coordinates of the other two vertices.
Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively. Let (x,y), (x
1
,y
1
) be the coordinate of vertex B and D respectively.
We know that the sides of a square are equal to each other.
∴ AB = BC
.png)
=>x² + 2x + 1 + y² -4y + 4 = x² + 9 -6x + y² + 4 - 4y
⇒ 8x = 8
⇒ x = 1
We know that in a square, all interior angles are of 90°.
In ΔABC,
AB² + BC² = AC²
.png)
⇒4 + y² + 4 − 4y + 4 + y² + 4 − 4y = 16
⇒2y² + 16 − 8 = 16
⇒2y² - 8 = 0
⇒y(y - 4) = 0
⇒y = 0 or 4
We know that in a square, the diagonals are of equal length and bisect each other at 90°. Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD
Coordinate of point O = ((-1+3)/2, (2+2)/2)
⇒1 + x1/2 = 1
⇒1 + x1 = 2
⇒x1 = 1
and y + y
1 /2 = 2
⇒ y + y1= 4
⇒If y = 0
⇒y1= 4
⇒If y = 4
⇒y1= 0
Therefore, the required coordinates are (1, 0) and (1, 4).
5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of △ PQR if C is the origin? Also calculate the area of the triangle in these cases. What do you observe?
.png)
= 1/2 [4(2-5) + 3(5-6) + 6(6-2)]
= 1/2 [-12 -3 +24]
= 9/2 sq. units
(ii) Taking C as origin, CB as x- axis, and CD as y- axis, the coordinates of vertices P, Q, and R are (12, 2), (13, 6), and (10, 3) respectively.
area of the triangle (PQR) =
Given that (AD)/(AB) = (AE)/(AC) = 1/4
(AD)/(AD+DB) = (AE)/(AE+EC) = 1/4
(AD)/(DB)=(AE)/(EC) = 1/3
Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3
Coordinates of Point D =
= (13/4, 23/4)
Coordinates of Point E =
= (19/4, 20/4)
Area of triangle =
=
Area of ΔABC = 1/2[4(5-2) + 1(2-6) + 7(6-5)]
= 1/2[12 - 4 + 7]
= 15/2
Clearly, the ratio between the areas of ΔADE and ΔABC is 1:16.
7. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of △ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP: PD = 2: 1.
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ: QE = 2: 1 and CR : RF = 2 : 1.
(iv) What do you observe?
(Note: The point which is common to all the three medians is calledcentroid and this point divides each median in the ratio 2: 1)
(v) If A(x
1
,y
1
), B(x
2
,y
2
),and C(x
3
,y
3
),are the vertices of △ ABC, find the coordinates of the centroid of the triangle.
(i) Median AD of the triangle will divide the side BC in two equal parts.
Therefore, D is the mid-point of side BC
Coordinate of D = (6 +1/2, 5 + 4/2) =(7/2,9/2)
(ii) Point P divides the side AD in a ratio 2:1.
Coordinate of P =
= (11/3 , 11/3)
(iii) Median BE of the triangle will divide the side AC in two equal parts.
Therefore, E is the mid-point of side AC.
Coordinate of E = (4+1/2, 2+4/2) = (5/2,3)
Point Q divides the side BE in a ratio 2:1.
Coordinate of Q =
= (11/3 , 11/3)
Median CF of the triangle will divide the side AB in two equal parts. Therefore, F is the mid-point of side AB
Coordinate of F = (4+6/2, 2+5/2) = (5,7/2)
Point R divides the side CF in a ratio 2:1.
Coordinate of R =
Let the centroid of this triangle be O.
Point O divides the side AD in a ratio 2:1.
Coordinate of O =
=
8. ABCD is a rectangle formed by joining points A(-1,-1), B (-1,4), C (5,4) and D (5,-1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? Or a rhombus? Justify your answer.
P is the mid point of side AB
Therefore the coordinates of P are ((-1-1)/2,(-1+4)/2) = (-1, 3/2)
Similary the coordinates of Q , R and S are (2,4),(5, 3/2), and (2, -1) respectilvely
Length of PQ =
Length of QR =
Length of RS =
Length of SP =
Length of PR =
= 6
Length of QS =
= 5
It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths. Therefore, PQRS is a rhombus.
