CRE 1 - Basics

Distance between two points (x₁, y₁) and (x₂, y₂) 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.

Equation of a line parallel to y-axis and passing through point (x₁, y₁) is x = x₁.

∴ Equation of the required line is x = 4.

Hence, option (b).

Equation of a line parallel to x-axis and passing through point (x₁, y₁) is y = y₁.

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

Hence, option (a).

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).

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).

Slope between two points (x₁, y₁) and (x₂, y₂) 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).

Equation of a line passing through (x₁, y₁) 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).

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).

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).

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).

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).

General equation of a line joining points (x₁, y₁) and (x₂, y₂) 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).

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).

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).

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).

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).

Distance of a point (x₁, y₁) 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).

Coordinates of a point dividing (x₁, y₁) and (x₂, y₂) 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).

Coordinates of a point dividing (x₁, y₁) and (x₂, y₂) 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).