Coordinate Geometry Questions and Answers

Answer: A

Parallel lines have the same slope. The slope of the given line is m=2.

The new line also has a slope of 2 and passes through (1, 7).

Using the point-slope form y - y₁ = m(x - x₁):

y - 7 = 2(x - 1)

y - 7 = 2x - 2

y = 2x + 5

Answer: A

In a parallelogram, the diagonals bisect each other. This means the midpoint of one diagonal is the same as the midpoint of the other diagonal.

Let the vertices be A(-1, 0), B(3, 1), C(2, 2), and D(x, y).

Midpoint of diagonal AC = \((\frac{-1+2}{2}, \frac{0+2}{2}) = (1/2, 1)\).

Midpoint of diagonal BD = \((\frac{3+x}{2}, \frac{1+y}{2})\).

Equating the midpoints: \(\frac{3+x}{2} = \frac{1}{2} \Rightarrow 3+x=1 \Rightarrow x=-2\).

\(\frac{1+y}{2} = 1 \Rightarrow 1+y=2 \Rightarrow y=1\).

The fourth vertex is (-2, 1).

Answer: B

The area of a rhombus can be calculated as half the product of its diagonals.

Let the vertices be A(3,0), B(4,5), C(-1,4), D(-2,-1).

Length of diagonal AC = \(\sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}\).

Length of diagonal BD = \(\sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}\).

Area = \(\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24\) sq units.

Answer: A

First, find the slope (m) of the line.

m = \(\frac{9-5}{4-2} = \frac{4}{2} = 2\).

Now use the point-slope form with one of the points, say (2, 5):

y - 5 = 2(x - 2)

y - 5 = 2x - 4

y = 2x + 1.

This is in the slope-intercept form y = mx + c. The y-intercept (c) is 1.

Answer: D

The y-axis is where \( x = 0 \).

\[ 3y = 6 \quad\Longrightarrow\quad y = 2 \]

\[ (0,\ 2) \]

Answer: B

The distance from the x-axis is the magnitude of the \( y \)-coordinate.

\[ |12| = 12 \]

The \( x \)-coordinate gives the distance from the y-axis instead.

Answer: C

For lines to be parallel, their slopes must be equal.

Slope of the first line \(kx+3y-1=0 \Rightarrow 3y = -kx+1 \Rightarrow y = (-k/3)x+1/3\). Slope m₁ = -k/3.

Slope of the second line \(2x+y+5=0 \Rightarrow y = -2x-5\). Slope m₂ = -2.

Set m₁ = m₂:

-k/3 = -2

k = 6.

Answer: D

Let the third vertex be (x, y). Since it lies on y = x+3, its coordinates are (x, x+3).

Using the area formula: Area = \(\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|\)

5 = \(\frac{1}{2} |2(-2-(x+3)) + 3((x+3)-1) + x(1-(-2))|\)

10 = |2(-x-5) + 3(x+2) + x(3)|

10 = |-2x - 10 + 3x + 6 + 3x| = |4x - 4|

This gives two possibilities:

1) 4x - 4 = 10 => 4x = 14 => x = 14/4 = 7/2. Then y = 7/2 + 3 = 13/2. Vertex is (7/2, 13/2).

2) 4x - 4 = -10 => 4x = -6 => x = -6/4 = -3/2. Then y = -3/2 + 3 = 3/2. Vertex is (-3/2, 3/2).

Answer: C

Reflecting in the x-axis flips the sign of \( y \) and leaves \( x \) alone.

\[ (3,\ -5) \ \rightarrow \ (3,\ 5) \]

Answer: A

The intercept form of a line is

\[ \frac{x}{a} + \frac{y}{b} = 1 \]

Setting \( y = 0 \) gives \( x = a \), and \( x = 0 \) gives \( y = b \) - which is the check that the intercepts are right.