KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).
Answer:
Let P(x₁, y₁) be common point of both lines and divide the line segment joining A(2, – 2) and B (3, 7) in ratio K : 1

Question 2.
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Answer:
Given point are A(x, y), B(1, 2) & C(7, 0) These points will be collinear if the area of the triangle formed by them is zero.
Area of ∆ ADC

0 = \(\frac{1}{2}\)[x(2 – 0)+1(0 – y)+7(y – 2)]
0 = x × 2 + 1 × – y + 7y – 14
0 = 2x – y + 7y – 14
2x + 6y – 14 = 0 divide by 2
x + 3y – 7 = 0
It is the relation between x & y

Question 3.
Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).
Answer:
Let A → (6,- 6), B → (3,- 7) and C → (3, 3)

Squaring
(x – 6)² + (y + 6)² = (x – 3)² + (y + 7)²
x² – 12x + 36 +y² +12y + 36
= x² – 6x + 9 + y² +14y + 49
x² – x² + y² – y² – 12x +6x + 12y – 14y + 72 – 58 = 0
– 6x – 2y = – 14 devide by – 2
3x + y = 7 → (1)
(x – 3)² +(y + 7)² = (x – 3)² +(y – 3)² (y + 7)² = (y – 3)²
y² +49 + 14y = y² + 9 – 6y
y² – y² + 14y + 6y = 9 – 49
20y = – 40
y = – 2
Putting y = – 2 in eqn (1) 3x + y = 7
3x – 2 = 7
3x = 7 + 2
3x = 9
x = \(\frac{9}{3}\) = 3
x = 3
Thus I(x, y) → (3, – 2)
Hence, the centre of a circle is (3, – 2)

Question 4.
The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of the other two vertices.
Answer:
Let ABCD is a square where two opposite vertices are A(- 1, 2) & C(3,2)

AB = BC = CD = AD [ABCD is a square]
AB = BC

Squaring both side
(x +1 )² + (y – 2)² = (3 – x )² + (2 – y )²
x² + 2x + 1 + y² – 4y + 4
= 9 + x² – 6x + 4 + y² – 4y
2x + 6x – 4y + 4y = 13 – 5
8x = 8
x = \(\frac{8}{8}\) = 1
x = 1
In ∆ ABC [B_ =90°

x² + 1 + 2x + y² + 4 – 4x + 9 + x²
– 6x + 4 + y² – 4y = (4)²
2x² – 4x + 2y² – 8y + 18 = 16
2x(1)² – 4(1) + 2y² – 8y + 18 – 16 = 0
2 – 4 + 2y² – 8y + 2 = 0
2y² – 8y = 0
2y(y – 4) = 0
2y = 0 & y – 4 = 0
y = 0 & y = 4
Hence, the other two vertices are (x, y) &(x₁, y₁) are (1, 0) & (1, 4)

Question 5.
The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the, boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the Fig.7.4 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 areas of the triangles in these cases. What do you observe?
Answer:
(i) Taking A as origin then AD is x – axis and AB is y – axis.
Co-ordinates of P, Q and R are
P → (4, 6), Q → (3, 2) & R → (6, 5)
area of ∆ PQR

area of ∆ PQR
= \(\frac{1}{2}\)[4(2 – 5) + 3(5 – 6) + (6 – 2)]
= \(\frac{1}{2}\)[4 × – 3 + 3 × – 1 + 6 × 4]
= \(\frac{1}{2}\)[- 12 – 3 + 24]
area of ∆ PQR
= \(\frac{1}{2}\)[- 15 + 24] = 9/2 sq units

(ii) Taking C as origin then CB is x – axis & CD is y – axis.
P(x₁, y₁) = (12, 2), Q(X₂, y₂) = (13, 6), and R(x₃, y₃) = (10, 3),
x₁ = 12, y₁ = 2, x₂ = 13, y₂ = 6, x₃ = 10 & y₃ = 3
area of ∆ PQR
= \(\frac{1}{2}\)[12(6 – 3) + 13(3 – 2) + 10(2 – 6)]
= \(\frac{1}{2}\)[12 x 3 + 13 x 1 + 10 x – 4]
= \(\frac{1}{2}\)[36 + 13 – 40]
= \(\frac{1}{2}\)[49 – 40] = 9/2 Sq units
Hence we observed that area of ∆ remains same in both case.

Question 6.
The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to 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 the area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6).
Answer:

D and E divide AB and AC respectively in the ration 1 : 3
Section Formula


area of ∆ ABC
= \(\frac{1}{2}\)[4(5 – 2) + 1(2 – 6) + 7(6 – 5)]
= \(\frac{1}{2}\)[4 × 3 + 1 × – 4 + 7 × 1]
= \(\frac{1}{2}\)[12 – 4 + 7]
= \(\frac{1}{2}\) × [19 – 4]
= \(\frac{15}{2}\)sq units

area of ADE : area of ABC =1:16

Question 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 yo observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.]
(v) If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ ABC, find the coordinates of the centroid of the triangle.
Answer:
(i)

median AD of the triangle ABC divide the side BC into two equal parts.
Therefore D is the mid – point of side BC Co-ordinates of mid Point

(ii) Form equation AP : PD = 2 : 1

Section Formula

(iii) BQ : QE = 2 : 1

co-ordinates of
\(\frac{A D}{A B}=\frac{A E}{E C}=\frac{1}{4}\)

(iv) The co-ordinates P, Q, & R are same (11/3, 11/3) All these points represents the same point which is called centriod.

(v) Point 0 is the centroid and AD is the midian. D is the mid point of BC and point 0 divide the AD into 2 : 1

Question 8.
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 mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Answer:
Given: A → (- 1, – 1), B → (- 1, 4), C → (5, 4) & D → (5, – 1) Co-ordinates of midpoint



∴ PQ = QR = RS = SP
∴ PQRS is either square (or) Rhomrus
Diagonal PR

and Diagonal QS

PR ≠ QS
∴ Diagonal are not equal
Therefore, PQRS is a rhombus.