NCERT Solution For Class 10 Math Chapter – 7 Exercise – 7.4
Q1. Determine the ratio in which the line 2x + y – 4 = 0 divides the the line segment joining the points A(2,-2) and B(3,7).
Solution:-
X = (2+3k)/(k+1)
Y = (-2 + 7k)/(k+1)
Substituting the value of x and y given equation I.e 2x + y – 4 = 0 , we have
2[2+3k)/(k+1) + (-2 + 7k)/(k+1)]-4 = 0
-2 + 9k = 0
= k = 2/9
The ratio is 2:9
Q2. Find the relation between x and y if the points (x,y) (1,2) and (7,0) are collinear.
Solution:-
Area of triangle = 1/2 [x(2-0) + 1(o-y) +7 (y-2) ] = 0
2x – y + 7y – 14 = 0
2x + 6y – 14 = 0
X + 3y – 7 = 0
Which is the required result.
Q3. Find the center of a circle passing through points (6,-6) , (-3,-7) and (3,3).
Solution:-
If O is the center , then OA = OB = 0C (radii are equal)
If O = (x,y) then
OA = √(X-6)2 + (Y+6)2]
OB = √(X-3)2 + (Y+7)2]
OC = √(X-3)2 + (Y-3)2]
Choose O = OC we have
After simplifying above , we get -6x = 2y – 14 (I)
Similarly: OB = OC
(x-3)2 + (y+7)2 = (x-3)2 + (y-3)2
Y = -2
Substituting the value of y in equation(1) we get;
-6x = 2y – 14
X = -4 – 14 = -8
X = 3
Hence , the center of the circle located at point (3,-2).
Q4. The two opposite vertices of a square are (-1,2) and (3,2) Find the coordinate of the other two vertices.
Solution:-
AC = √(3+1)2 + (2-2)2] = 4
Coordinates of O can be calculated as follows:
X = (3-1)/2 = 1 y = (2+2)/2 = 2
So O(1,2)
A = 2√2
Each side of square = 2√2
AD = √(X1 + 1)2 + (Y1 – 1)2]
Squaring both sides
AD2 = √(X1 + 1)2 + (Y1 – 2)2]
AD2 = (X1 + 1)2 + (Y1 -2)2
CD2 = (X1 – 3)2 + (Y1– 2)2
AD = CD
(X1 + 1)2 + (Y1 – 2)2 = (X1 – 3)2 + (Y1 – 2)2
X1 = 1
CD2 = (X1 – 3)2 + (Y1 – 2)2
8 = (1-3)2 + (Y1 – 2)2
Y1 = 4
Hence D = (1,4)
O = (1,2)
BD
1 = (X2 + 1)/2
X2 = 1
And 2 = (y2 + 4)/2
Y2 = 0
Q5. 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 lawn in the plot The students are to sow the seeds of flowering plants on the remaining area of the plot.
- Taking A as origin , find the coordinates of the vertices of the triangle.
- What will be the coordinate of the vertices of triangle PQR if C is the origin?
Also calculate the areas of the triangle in these cases . What do you observe.
Solution:-
- Taking A as origin coordinates of the vertices P , Q and R are ,
From Figure: P (4,6) Q = (3,2) R = (6,5)
Here AD is x – axis and AB is the y – axis
- Taking C as origin,
Coordinate of vertices P , Q and R are (12,2) , (13,6) and (10,3) respectively.
Area of triangle = 1/2 [ 4(2-5) + 3(5-6) + 6(6-2)]
9/2 square units.
( ii ) Area of triangle PQR in case of origin C:
Area of triangle = 1/2 [12(6-3) + 13(3-2) + 10(2-6)]
= 9/2 square units
Areas is same in both case because right triangle remains the same no matter which point is considered as origin.
Q6. The vertices of a ∆ABC are A(4,6) B(1,5) and C(7,2) A line is drawn into intersect sides AB and AC at D and E respectively. Such that AD/AB = AE/AC = 1/4 Calculate the area of the ∆ADE and compare it with area of ∆ABC.
Solution:-
Consider the line segment AB which is divide by the point D at the ratio 1:3
X = [3(4) + 1(1)]/4 = 13/4
Y = [3(6) + 1(5)]/4 = 23/4
Similarly Coordinates of E can be calculated as follows:
X = [ 1(7) + 3(4)]/4 = 19/4
Y = [1(2) + 3(6)] /4 = 5
Area of ∆ABC can be calculated as follows:
= 1/2[4(5-2) + 1(2-6) +7(6-5)]
= 15/2 square units.
Area of ∆ ADE can be calculated as follows:
= 1/2 [4(23/4 – 5) + 13/4(5-6) + 19/4 (6-23/4)]
= 15/32 square units.
Hence ratio of area of triangle ADE to area of triangle ABC = 1:16
Q7. Let A (4,2) B(6,5) and C(1,4) be the vertices of ∆ABC.
- The medians from A meets BC at D. Find the coordinate of point D.
- Find the coordinate of the point P on AD such that AP : PD = 2:1
- Find the coordinate of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2:1
- What do you observe ?
- If A (x1,y1) B(x2,y2) and C (x3,y3) are the vertices of triangle ABC , find the coordinates of the centroid of the triangle.
Solution:-
- Coordinate of D = (6+1)/2 , (5+4)/2 = (7/2 , 9/2)
So D is (7/2 , 9/2)
- Coordinate of P = (11/3 , 11/3)
- Coordinate of E = (5/2 , 3)
Coordinate of Q = (11/3 , 11/3)
F is the mid – point of the sides AB
Coordinate of F = (5,7/2)
Point R divides the side CF in ratio 2:1.
Coordinate of R = (11/3 , 11/3)
- Coordinate of P , Q and R are same which shows the medians intersect each other at common point centroid of the triangles.
- If A (x1, y1) B(x2 , y2) and C (x3 , y3) are the vertices of triangle ABC the coordinate of centroid can be given as follows:
X = (x1 + x2 + x3)/3 and y = (y1 + y2 + y3)/3
Q8. ABCD is a rectangle formed by the points A (-1 , -1) , B(-1,4) C(5,4) and D (5,-1) P , Q , R and S are the midpoints of AB , BC , CD and DA respectively. Is the quadrilateral PQRS a square ? A rectangle ? or a rhombus ? Justify the answer.
Solution:-
P is the mid – point
Coordinate of P = (-1 -1)/2 , (-1+4)/2 = (-1 , 3/2)
Similarly , Q , R and S are ( As Q is mid – point of BC , R is mid – point of CD and S is midpoint of AD)
Coordinate of Q = (2,4)
Coordinate of R = (5 , 3/2)
Coordinate of S = (2 , -1)
Now ,
Length of PQ = √61/2
Length of SP = √61/2
Length of QR = √61/2
Length of RS = √61/2
Length of PR (diagonal) = 6
Length of QS (diagonal) = 5
The above values show that PQ = SP = QR = RS = √61/2 all sides are equal.
But PR ≠ QS diagonals are not of equal measure.
Hence , the given figure is a rhombus.