Coordinate Geometry Questions and Answers

Q: Find the value of 'a' for which the points (a, 1), (1, -1), and (11, 4) are collinear.

Correct answer: 5The two fully known points fix the line. Its slope is\[m = \frac{4 - (-1)}{11 - 1} = \frac{5}{10} = \frac{1}{2}\]For (a, 1) to lie on the same line, the slope from it to (1, -1) must also be \(\tfrac{1}{2}\):\[\frac{-1 - 1}{1 - a} = \frac{1}{2} \quad\Longrightarrow\quad \frac{-2}{1-a} = \frac{1}{2} \quad\Longrightarrow\quad -4 = 1 - a \quad\Longrightarrow\quad a = 5\]Checking with the third point: the slope from (5, 1) to (11, 4) is \(\tfrac{3}{6} = \tfrac{1}{2}\), as required.

Q: What is the distance between the points (3, 4) and (-3, -4) ?

Correct answer: 10\[d = \sqrt{(-3-3)^2 + (-4-4)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\]Both points lie 5 units from the origin on the same line through it, so the distance is twice that, which is a useful check.

Q: What is the x-intercept of the line 4x - 5y = 20?

The x-intercept is where \( y = 0 \).\[ 4x = 20 \quad\Longrightarrow\quad x = 5 \]

Q: Find the value of 'k' if the distance between the points (4, k) and (1, 0) is 5.

\[ \sqrt{(4-1)^{2} + (k-0)^{2}} = 5 \]\[ 9 + k^{2} = 25 \quad\Longrightarrow\quad k^{2} = 16 \]\[ k = \pm 4 \]Both signs satisfy the distance, since squaring loses the sign - the point may lie above or below.

Important Instructions

31.

Find the value of 'a' for which the points (a, 1), (1, -1), and (11, 4) are collinear.

A.5
B.-5
C.-3
D.3

Answer: B

Correct answer: 5

The two fully known points fix the line. Its slope is

\[m = \frac{4 - (-1)}{11 - 1} = \frac{5}{10} = \frac{1}{2}\]

For (a, 1) to lie on the same line, the slope from it to (1, -1) must also be \(\tfrac{1}{2}\):

\[\frac{-1 - 1}{1 - a} = \frac{1}{2} \quad\Longrightarrow\quad \frac{-2}{1-a} = \frac{1}{2} \quad\Longrightarrow\quad -4 = 1 - a \quad\Longrightarrow\quad a = 5\]

Checking with the third point: the slope from (5, 1) to (11, 4) is \(\tfrac{3}{6} = \tfrac{1}{2}\), as required.

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32.

What is the distance between the points (3, 4) and (-3, -4) ?

A.8
B.10
C.12
D.14

Answer: A

Correct answer: 10

\[d = \sqrt{(-3-3)^2 + (-4-4)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\]

Both points lie 5 units from the origin on the same line through it, so the distance is twice that, which is a useful check.

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33.

What is the equation of a line that passes through the origin and has a slope of -4?

A.Y = -4x
B.Y = x - 4
C.Y = 4x
D.X = -4y

Answer: A

A line through the origin has zero intercept, so \( y = mx \).

\[ y = -4x \]

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34.

The points (1, 1), (3, 4), and (5, 1) are the vertices of a/an:

A.Equilateral triangle
B.Isosceles triangle
C.Scalene triangle
D.Right-angled triangle

Answer: B

Let's find the lengths of the sides using the distance formula.

Side 1 (between (1,1) and (3,4)): \(\sqrt{(3-1)^2 + (4-1)^2} = \sqrt{2^2 + 3^2} = \sqrt{4+9} = \sqrt{13}\).

Side 2 (between (3,4) and (5,1)): \(\sqrt{(5-3)^2 + (1-4)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}\).

Side 3 (between (1,1) and (5,1)): \(\sqrt{(5-1)^2 + (1-1)^2} = \sqrt{4^2 + 0^2} = 4\).

Since two sides have the same length (\(\sqrt{13}\)), the triangle is isosceles.

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35.

Find the coordinates of the circumcenter of the triangle whose vertices are (8, 6), (8, -2) and (2, -2).

A.(5, 2)
B.(3, 2)
C.(5, 4)
D.(4, 2)

Answer: A

First, notice the vertices. The side from (8, 6) to (8, -2) is a vertical line (x=8). The side from (8, -2) to (2, -2) is a horizontal line (y=-2). This means they are perpendicular, and the triangle is a right-angled triangle with the right angle at (8, -2).

In a right-angled triangle, the circumcenter is the midpoint of the hypotenuse.

The hypotenuse connects the vertices (8, 6) and (2, -2).

Midpoint = \((\frac{8+2}{2}, \frac{6+(-2)}{2}) = (\frac{10}{2}, \frac{4}{2}) = (5, 2)\).

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36.

What is the x-intercept of the line 4x - 5y = 20?