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Coordinate Geometry Questions and Answers

Important Instructions
11.

Which of the following will be a point on a line that contains (−1, −1) and (2, 5)?

Exam Year(s)

Kerala PSC KAS - 2020,

Answer: C

The point on the line is (100, 201).

Find the equation of the line through (−1, −1) and (2, 5).

\[ \begin{aligned} m &= \frac{5 - (-1)}{2 - (-1)} = \frac{6}{3} = 2 \\ y - 5 &= 2(x - 2) \\ y &= 2x + 1 \end{aligned} \]

For \(x = 100\), \(y = 2(100) + 1 = 201\), so (100, 201) lies on the line.

The other points fail the test: for \(x = 30\) the line gives 61, not 41; for \(x = 80\) it gives 161, not 91; and for \(x = 110\) it gives 221, not 121. Those options add 11 to \(x\), a pattern the two given points do not follow.

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Answer: B

The Cartesian coordinate system is divided into four quadrants.

  • Quadrant I: (+x, +y)
  • Quadrant II: (-x, +y)
  • Quadrant III: (-x, -y)
  • Quadrant IV: (+x, -y)

Since the point has a negative x-coordinate (-3) and a positive y-coordinate (5), it lies in the Second Quadrant.

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13.

What is the distance between the points (3, 2) and (6, 6)?

Answer: C

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).

Distance = \(\sqrt{(6-3)^2 + (6-2)^2}\)

= \(\sqrt{3^2 + 4^2}\)

= \(\sqrt{9 + 16}\)

= \(\sqrt{25} = 5\) units.

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14.

What are the coordinates of the midpoint of the line segment joining the points (1, 5) and (7, 3)?

Answer: B

The midpoint averages the coordinates.

\[ \left(\frac{1+7}{2},\ \frac{5+3}{2}\right) = (4,\ 4) \]

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15.

What is the slope of the line passing through the points (2, 3) and (4, 7)?

Answer: B

\[ m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{7-3}{4-2} = 2 \]

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18.

Find the coordinates of the point which divides the line segment joining (-1, 7) and (4, -3) in the ratio 2:3.

Answer: A

Using the section formula, the coordinates (x, y) are given by:

x = \(\frac{m_1x_2 + m_2x_1}{m_1+m_2}\) and y = \(\frac{m_1y_2 + m_2y_1}{m_1+m_2}\)

Here, \((x_1, y_1) = (-1, 7)\), \((x_2, y_2) = (4, -3)\), and \(m_1:m_2 = 2:3\).

x = \(\frac{2(4) + 3(-1)}{2+3} = \frac{8-3}{5} = \frac{5}{5} = 1\)

y = \(\frac{2(-3) + 3(7)}{2+3} = \frac{-6+21}{5} = \frac{15}{5} = 3\)

The coordinates are (1, 3).

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19.

What is the area of the triangle with vertices (1, -1), (-4, 6) and (-3, -5)?

Answer: B

The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by the formula:

Area = \(\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|\)

Area = \(\frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|\)

Area = \(\frac{1}{2} |1(11) + (-4)(-4) + (-3)(-7)|\)

Area = \(\frac{1}{2} |11 + 16 + 21| = \frac{1}{2} |48| = 24\) sq units.

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20.

What is the slope of a line that is perpendicular to the line y = -3x + 4?

Answer: C

The given line is in the form y = mx + c, so its slope (m₁) is -3.

If two lines are perpendicular, the product of their slopes is -1. Let the slope of the perpendicular line be m₂.

m₁ × m₂ = -1

-3 × m₂ = -1

m₂ = \(\frac{-1}{-3} = \frac{1}{3}\).

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Tags: Coordinate Geometry Questions and Answers || Coordinate Geometry MCQ || Geometry and Coordinates GK || Distance Formula Quantitative Aptitude || Coordinate Geometry MCQs