Question

[Maths]
Topic:- Co-ordinate Geometry

Q1) Sub-Topic :- Reflection
The points (6, 2), (3, -1) and (-2, 4) are the vertices of a right angled triangle. Check whether it remains a right angled triangle after reflection in the y-axis.

Q2) Sub-Topic :- Equation of a line
If the coordinates of the vertex A of a square ABCD are (3,-2) and the equation of the diagonal BD is 3x – 7y + 6 = 0, find the equation of the diagonal AC. Also find the coordinates of the centre of the square.

Spams/plagiarism strictly prohibited.​

Answer 1

Answer:

Let A (6, 2), B (3, -1) and C (-2, 4) be the points of a right-angled triangle

Then,

The co-ordinates of the images of A, B, C reflected in y-axis will be:

A’ (-6, 2), B’ (-3, -1) and C’ (2, 4).

Hence, by joining these points

We see that ∆A’B’C’ is also a right-angled triangle.

Answer 2

Topic :-

Coordinate Geometry

Answer 1 :-

Let us name points (6, 2), (3, -1) and (-2, 4) as A,B and C respectively.

When a point (a, b) gets reflected in y-axis then coordinate of its image is (-a, b).

So, image of A,B and C are

A' ≡ (-6, 2)

B' ≡ (-3, -1)

C' ≡ (2, 4)

Calculating square of distance between these points using distance formula,

(A'B')² = (-3 - (-6))² + (-1 - 2)²

(A'B')² = (-3 + 6)² + (-3)²

(A'B')² = (3)² + (-3)²

(A'B')² = 9 + 9

(A'B')² = 18

(B'C')² = (2 - (-3))² + (4 - (-1))²

(B'C')² = (2 + 3)² + (4 + 1)²

(B'C')² = (5)² + (5)²

(B'C')² = 25 + 25

(B'C')² = 50

(C'A')² = (-2 - 6)² + (4 - 2)²

(C'A')² = (-8)² + (2)²

(C'A')² = 64 + 4

(C'A')² = 68

We can observe that,

(C'A')² = (B'C')² + (A'B')²

Hence, it follows Pythagoras Theorem which means triangle A'B'C' will be also a right angled triangle.

Answer 2 :-

Equation of a line perpendicular to line ax + by + c = 0 will be

bx - ay + k = 0, where k is a parameter which can be obtained with some information from question.

Diagonals of square bisect each other at right angles.

As per above statement, BD ⊥ AC.

Equation of diagonal AC perpendicular to diagonal BD : 3x + (-7)y + 6 = 0 will be

-7x - 3y + k = 0 or

7x + 3y - k = 0

Point A (3, -2) passes through diagonal AC, hence

7(3) + 3(-2) - k = 0

21 - 6 - k = 0

15 - k = 0

k = 15

So, equation of diagonal AC : 7x + 3y - 15 = 0

Coordinates of the center of square can be obtained by solving the two equations of diagonals as diagonals bisect at the center of square.

(3x - 7y = -6) × 3

9x - 21y = -18

(7x + 3y = 15) × 7

49x + 21y = 105

Adding both equations,

(9x - 21y) + (49x + 21y) = -18 + 105

(9x + 49x) + (21y - 21y) = 87

58x = 87

x = 3/2 = 1.5

Substituting value of 'x' in any equation,

3x - 7y = -6

3(3/2) - 7y = -6

(9/2) -7y = -6

7y = (9/2) + 6

7y = ((9+12)/2)

7y = 21/2

y = 3/2 = 1.5

Hence, coordinates of the center of the square is (1.5, 1.5).