Test Level 2: Coordinate Geometry - 1 (September 9) - CAT MCQ

Since AB = BC = CA, hence triangle is equilateral. Therefore, it is an acute angled triangle.

To go from the point (5, 5) to the point (9, 2), we must move over to the right by 4 units and down by 3 units.
Since we are dealing with a rectangle, the same must be true for (a, 13) and (15, b).
Thus, a + 4 = 15 and 13 - 3 = b.
From this, a = 11 and b = 10.
So a - b = 11 - 10 = 1.

The intersection point of x - 3y + 1 = 0 and 2x + 5y - 9 = 0 is (2, 1).
And, m = 1/0.
So, the required line is y - 1 = (1/0)(x - 2)
⇒ x = 2.

Intercept form of line = x/y + y/b = 1
In case of equal intercepts, a = b
Therefore, 

 As it passes through (1, - 2),
⇒ 1 - 2 = a, a = -1
The equation of line is x + y + 1 = 0.

As line intersects x-axis at A (10, 0)
Length of intercept on x-axis, a = 10
Similarly length of intercept on y-axis, b = 10
∴ Using intercept form, equation of line is x/10 + y/10 = 1
or x + y = 10.

The point of intersection of the lines 4x - 3y - 1 = 0 and 5x - 2y - 3 = 0 is (1, 1).
The equation of line parallel to 2y - 3x + 2 = 0 is 2y - 3x + k = 0.
It also passes through (1, 1), so k = 1.
Hence, the required equation is 2y - 3x + 1 = 0 or 3x - 2y = 1.

Slope of PQ: [1 - (-3)]/[6 - 0] = 2/3 
Slope of RS: [2 - (-4)]/[5 - (-4)] = 2/3
Since the slopes of the two lines are equal, the lines are parallel. 

The equation of the y-axis is x = 0.
So, the line 2x + 3y = 7 will cut the y-axis at the point 3y = 7, or y = 7/3.
So, k = 7/3, or 

where r = radius = 1 unit
Area = πr² = π1² = π

1(0 + 2) - 1 (9k - 2) - 4 (-3 - 0) = 0
9k = 16
or k = 16/9