Coordinate Geometry Test - 1

AC² = (a - 3)² + (b - 4)²

100 = (a - 3)² + (b - 4)²

Check with the options. We find that the point (11, 10) satisfies both the conditions, i.e. this is also collinear with A and B.

According to the question:
Distance between (x, 0) and (6, 7) = Distance between (x, 0) and (4, -3)
(x - 6)² + 7² = (x - 4)² + (-3)²
x² + 36 - 12x + 49 = x² + 16 - 8x + 9
60 = 4x
x = 15
The required point is (15, 0).
Option 2 is correct.

Let the coordinates of the required point on the x-axis be (x, 0).
ATQ

(11 - x)² + (-8 - 0)² = 10²
121 + x² - 22x + 64 = 100
x² - 22x + 85 = 0

x² - 17x - 5x + 85 = 0
x(x - 17) - 5(x - 17) = 0

(x - 17)(x - 5) = 0
x = 17 or x = 5

Therefore, required coordinates = (17, 0) or (5, 0)

Q₄ as value of x is positive and value of y is negative.

The area of triangle formed by the vertices (p, q + r), (q, p + r), (r, p + q) is

x₁ = p , y₁ = q + r
x₂ = q , y₂ = p + r
x₃ = r , y₃ = p + q

The area of a triangle formed by joining the points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
½ |y₁ (x₂ – x₃) + y₂ (x₃ – x₁) + y₃ (x₁ – x₂)| sq. units
1/2 | (q + r)(q – r) + (p + r)(r – p) + (p + q)(p – q)|

= 1/2 |q² – qr + rq – r² + pr – p² + r² – rp + pq – pq + qp – q²|
= 2 |–2 qr – p² + qp|

Let the point be (x, y). As it is equidistant from (2, 0) and (0, 2),

∴ (x - 2)² + y² = x² + (y - 2)²

Checking with the options, only (x, y) = (2, 2) satisfies the equation.