# CRE 1 - Basics | Geometry - Coordinate Geometry

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the distance between the points A (3, 5) & B (6, 1)

Answer: 5

**Explanation** :

Distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

∴ Distance between A (3, 5) & B (6, 1) = $\sqrt{{\left(6-3\right)}^{2}+{\left(1-5\right)}^{2}}$ = $\sqrt{{3}^{3}+{4}^{2}}$ = 5

Hence, 5.

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the equation of the line passing through A (4, -3) and parallel to the y-axis.

- A.
y = - 3

- B.
x = 4

- C.
y = 0

- D.
x = - 3

Answer: Option B

**Explanation** :

Equation of a line parallel to y-axis and passing through point (x_{1}, y_{1}) is x = x_{1}.

∴ Equation of the required line is x = 4.

Hence, option (b).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the equation of the line passing through A (4, -3) and parallel to the x-axis.

- A.
y = -3

- B.
x = 4

- C.
y = 0

- D.
x = -3

Answer: Option A

**Explanation** :

Equation of a line parallel to x-axis and passing through point (x_{1}, y_{1}) is y = y_{1}.

∴ Equation of the required line is y = -3.

Hence, option (a).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the value of ‘k’, if (3x + y - 4) - k(5x - 4y + 10) = 0 is parallel to x axis.

- A.
3/5

- B.
1/4

- C.
-1/4

- D.
None of these

Answer: Option A

**Explanation** :

For an equation to be parallel to x-axis the coefficient of x should be zero.

Given equation is (3 – 5k)x + (1 + 4k)y - 4 - 10k = 0

∴ 3 – 5k = 0

⇒ k = 3/5

Hence, option (a).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the center of the circle which has A (2, 9) & B (0, 11) as the extremities of its diameter.

- A.
(1, 10)

- B.
(2, 20)

- C.
(1, -1)

- D.
None of these

Answer: Option A

**Explanation** :

Center of a circle is located in the middle of the diameter.

Hence, the center is $\left(\frac{2+0}{2},\frac{9+11}{2}\right)$ = (1, 10)

Hence, option (a).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Compute the slope of the line passing through the points A (4, 1) & B (6, 2)

- A.
2

- B.
1/2

- C.
-1/2

- D.
None of these

Answer: Option B

**Explanation** :

Slope between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

∴ Slope between the points A (4, 1) & B (6, 2) is $\frac{2-1}{6-4}=\frac{1}{2}$

Hence, option (b).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the equation of the line which passes through (2, -2) and has a slope 2.

- A.
2x + y = 2

- B.
y - 2x = 6

- C.
2x - y = 6

- D.
None of these

Answer: Option C

**Explanation** :

Equation of a line passing through (x_{1}, y_{1}) with slope m is $\frac{y-{y}_{1}}{x-{x}_{1}}=m$

∴ Equation of the required line is $\frac{y-(-2)}{x-2}$ = 2

⇒ 2x – 4 = y + 2

⇒ 2x – y = 6

Hence, option (c).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the value of ‘k’ for which the three points A (-1, 10), B (k, 2) & C (17/5, 26/5) are collinear.

- A.
19/3

- B.
-19/3

- C.
6

- D.
None of these

Answer: Option A

**Explanation** :

For 3 or more points to be collinear they should have the same slope.

∴ Slope of A and B should be same as slope of A and C.

⇒ $\frac{10-2}{-1-k}=\frac{10-\frac{26}{5}}{-1-\frac{17}{5}}$

⇒ $-\frac{8}{k+1}=-\frac{24}{22}$

⇒ k + 1 = 22/3

⇒ k = 19/3

Hence, option (a).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the slope and the y-intercept of the line y = 3x - 15.

- A.
-3 and -15

- B.
3 and 15

- C.
3 and -15

- D.
None of these

Answer: Option A

**Explanation** :

A line with x and y intercepts as a and b respectively is $\frac{x}{a}+\frac{y}{b}$ = 1.

∴ Required line is $\frac{x}{2}+\frac{y}{-3}$ = 1

⇒ -3x + 2y = -6

⇒ 3x – 2y = 6

Hence, option (a).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the x-intercept & y-intercept of the line 3x - 2y - 24 = 0.

- A.
8 and -12

- B.
8 and 12

- C.
-8 and -12

- D.
None of these

Answer: Option A

**Explanation** :

To find the x-intercept of a line put y = 0 and to find y-intercept of a line put x = 0.

x-intercept,

∴ 3x – 2 × 0 – 24 = 0

⇒ x = 8

y-intercept,

∴ 3 × 0 – 2y – 24 = 0

⇒ y = -12

Hence, option (a).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Write the intercept form of the line whose general form is 4x + 3y – 24 = 0.

- A.
$\frac{x}{8}+\frac{y}{6}=1$

- B.
$\frac{x}{6}+\frac{y}{8}=1$

- C.
$\frac{x}{8}-\frac{y}{6}=1$

- D.
None of these

Answer: Option B

**Explanation** :

Intercept form of a line is $\frac{x}{a}+\frac{y}{b}=1$

Given line is 4x + 3y – 24 = 0.

⇒ 4x + 3y = 24

⇒ $\frac{x}{6}+\frac{y}{8}=1$

Hence, option (b).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the general form equation of the line joining the points A (3, 1) & B (0, -1).

- A.
2x – 3y + 3 = 0

- B.
2x + 3y – 3 = 0

- C.
2x – 3y – 3 = 0

- D.
None of these

Answer: Option C

**Explanation** :

General equation of a line joining points (x_{1}, y_{1}) and (x_{2}, y_{2}) is $\frac{y-{y}_{2}}{x-{x}_{2}}=\frac{{y}_{1}-{y}_{2}}{{x}_{1}-{x}_{2}}$

∴ $\frac{y-(-1)}{x-0}=\frac{1-(-1)}{3-0}$

⇒ 3y + 3 = 2x

⇒ 2x – 3y – 3 = 0

Hence, option (c).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Which of the following best describes the relation between the lines 6x - 9y - 4 = 0 & -4x + 6y - 13 = 0?

- A.
Prallel

- B.
Intersection, not perpendicular

- C.
Perpendicular

- D.
None of these

Answer: Option A

**Explanation** :

Slope of a line ax + by + c = 0 is -a/b.

∴ Slope of the line 6x - 9y – 4 = 0 is -6/-9 i.e., 2/3

∴ Slope of the line -4x + 6y – 13 = 0 is –(-4)/6 i.e., 2/3

Since slope of both these lines is same, these lines are parallel lines.

Hence, option (a).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Which of the following best describes the relation between the lines 6x + 10y - 19 = 0 & 5x - 3y + 17 = 0?

- A.
Parallel

- B.
Intersectiong, not perpendicular

- C.
Perpendicular

- D.
None of these

Answer: Option C

**Explanation** :

Slope of a line ax + by + c = 0 is -a/b.

∴ Slope of the line 6x + 10y – 19 = 0 is -6/10 i.e., -3/5

∴ Slope of the line 5x - 3y + 17 = 0 is –5/-3 i.e., 5/3

Here we see that the product of the slopes of two lines is -1 (-3/5 × 5/3), hence the two given lines are perpendicular to each other.

Hence, option (c).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the equation of line passing through (2,-3) and parallel to x + 3y + 12 = 0

- A.
3x - y + 7 = 0

- B.
x + 3y + 7 = 0

- C.
x + 3y - 7 = 0

- D.
None of these

Answer: Option B

**Explanation** :

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

∴ The required line is x + 3y + d = 0

Since this line passes through (2, -3)

⇒ 2 + 3 × -3 + d = 0

⇒ d = 7.

∴ The required line is x + 3y + 7 = 0

Hence, option (b).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the equation of line passing through (2,-3) and perpendicular to x + 2y + 12 = 0.

- A.
2x - y – 7 = 0

- B.
x + 2y – 7 = 0

- C.
2x - y + 7 = 0

- D.
None of these

Answer: Option A

**Explanation** :

Equation of a line parallel to ax + by + c = 0 will be bx - ay + d = 0

∴ The required line is 2x - y + d = 0

Since this line passes through (2, -3)

⇒ 2 × 2 – (-3) + d = 0

⇒ d = -7.

∴ The required line is 2x - y – 7 = 0

Hence, option (a).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the perpendicular distance of the point (1, 1) from the line 3x + 4y - 4 = 0.

- A.
4/5

- B.
2/5

- C.
3/5

- D.
None of these

Answer: Option C

**Explanation** :

Distance of a point (x_{1}, y_{1}) from a line ax + by +c = 0 is $\left|\frac{a{x}_{1}+b{y}_{1}+c}{\sqrt{{a}^{2}+{b}^{2}}}\right|$

∴ The required distance = $\frac{3\times 1+4\times 1-4}{\sqrt{{3}^{2}+{4}^{2}}}$ = 3/5

Hence, option (c).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the co-ordinates of the point P which divides the line-segment joining the points A (7, 9) & B (8, 2) in the ratio 3 : 4 internally.

- A.
(7, 6)

- B.
(6, 6)

- C.
(52/7, 42/7)

- D.
None of these

Answer: Option C

**Explanation** :

Coordinates of a point dividing (x_{1}, y_{1}) and (x_{2}, y_{2}) in the ratio a : b internally is given by $\left[\frac{b{x}_{1}+a{x}_{2}}{a+b},\frac{b{y}_{1}+a{y}_{2}}{a+b}\right]$

∴ The coordinates of point P are $\left[\frac{4\times 7+3\times 8}{3+4},\frac{4\times 9+3\times 2}{3+4}\right]$= (52/7, 42/7)

Hence, option (c).

Workspace:

**CRE 1 - Basics | Geometry - Coordinate Geometry**

Find the co-ordinates of the point P which divides the line-segment joining the points A (2, -6) & B (3, 5) in the ratio 2:1 externally.

- A.
(8/3, 16/3)

- B.
(4, 6)

- C.
(4, 16)

- D.
None of these

Answer: Option A

**Explanation** :

Coordinates of a point dividing (x_{1}, y_{1}) and (x_{2}, y_{2}) in the ratio a : b externally is given by $\left[\frac{b{x}_{1}-a{x}_{2}}{a-b},\frac{b{y}_{1}-a{y}_{2}}{a-b}\right]$

∴ The coordinates of point P are $\left[\frac{2\times 3-1\times 2}{2-1},\frac{2\times 5-1\times -6}{2-1}\right]$ = (4, 16)

Hence, option (a).

Workspace:

## Feedback

**Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.**