Coordinate Geometry - 1 - SSC CGL Quant Aptitude Free MCQ Test with solutions

According to the question,

Centroid of the triangle

Hence, the correct option is 

The equation of the line passing through the points, (1, 3) and (3, 5)
We know the two-point form of a line passing through these two points is: 

As the intercept form is represented as


Hence, the y-intercept is 2.

Coordinates of the point is (−4, 3).
Since, we need to find the reflection along x = −2, the y-coordinate remains unaffected.
Now, for reflection, the distance from the line of the reflection to the object should be equal to the distance to the image point.
So, for x-coordinate:
If the distance from the image to the line of reflection is 'A' and the distance from the object to the line of reflection be 'B'.
And, X₁ be the x coordinate of point (−4, 3).

Hence, the correct answer is (0, 3).

According to the problem,
The equation of the line is x − y = 5 or x − y − 5 = 0
Point(P) = (−2, 3)
Distance

Hence, the distance is 5√2.

Let the coordinates of the centroid be  (x, y).
x-axis coordinates of the centroid 

y-axis coordinates of the centroid 

So, Coordinates of centroid is 

Let the matrix be

The determinant is:
|A| = ad − bc
It is given that,
a + d = −6d = −6 − a  −(1)
For maximum value, bc = 0
Then,
|A| = ad
|A| = a(−6−a)|A| = −6a−a² 
Differentiating both sides w.r.t. a we get,

It has maximum value at a = −3
d = −6−ad = −6−(−3)d = −3
So, maximum value is:

Hence, the required answer is 9.

The given matrix is:

According to question,
(i) On adding the third row to the second row we get,

(ii) Subtract the third column from the first column we get,

It is clear that,
Third column=14×first column
So, the determinant will be zero as third coulmn is multiple of first column.
Hence, the required answer is 0.

The coordinates of A are (4, 2).
The coordinates of B are (2, 6).
The coordinates of C are (7, y).
x₁ = 2, x₂ = 7, y₁ = 6, y₂ = y, m₁ = 2, m₂ = 5
By using the section formula:

2x – 3y + 6 = 0

then it is in the format of 
The intercepts are -3, 2

According to Mid−Point Formula,

(x_m, y_m) are the coordinates of mid−point.

(x₁, y₁) are the coordinates of first point.

(x₂, y₂) are the coordinates of second point.
Given, coordinates of mid−point P = (3,1).
coordinates of second point B = (5, −4)
Let the coordinates of first point A (x₁, y₁).
Now, on using mid−point formula, we get

Therefore, 

Hence, the value of q is (5 − 2√6).

∠BMN + ∠DNM = 180^o (consecutive interior angles)

The slope is the measure of the rise (or change in the y-axis) over the run (or change in the x-axis).
Let slope of lines are m₁ and m₂ respectively

When lines are perpendicular, then m₁ × m₂ = −1

k = 5

It is given that,∠PQT = x°, ∠RQT = (x − 50)° & ∠TUR = (x + 25)°
We can write as,
∠PQT + ∠RQT = 180°
x+(x−50) = 180°
2x = 230
x = 115°
Also, we can write as,
∠TUR+∠URQ = 180°
(115+25)+∠URQ = 180°
∠URQ = 180 − 140 = 40°
Therefore,
∠URS = 180 −∠URQ
∠URS = 180 − 40 = 140°

Straight line = 3x + 4y = 12
General intercepted form of straight line 
This straight-line is written as-

Comparing (1) and (2),
a = 4 and b = 3
Therefore, the required length