15 extra multiple-choice questions for Coordinate Geometry (10th Standard Mathematics, Samacheer Kalvi), beyond the ones printed in the textbook — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
If the points \((0, 0)\), \((a, 0)\) and \((0, b)\) are collinear, then which of the following must be true?
- A. \(a + b = 0\)
- B. \(ab = 0\)Correct
- C. \(a^2 + b^2 = 0\)
- D. \(a = b\)
Explanation. The points are collinear if and only if the area of the triangle formed by them is zero. The area of the triangle with vertices \((0, 0)\), \((a, 0)\), and \((0, b)\) is \(\\frac{1}{2}|ab|\). Setting this to zero gives \(ab = 0\).
Q2
The centroid of a triangle with vertices \((1, -2)\), \((3, 6)\), and \((-4, 5)\) is given by which point?
- A. \((0, 3)\)Correct
- B. \((0, 9)\)
- C. \((1, 3)\)
- D. \((0, 2)\)
Explanation. The coordinates of the centroid of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are \((\\frac{x_1 + x_2 + x_3}{3}, \\frac{y_1 + y_2 + y_3}{3})\). For the given vertices, this is \((\\frac{1+3-4}{3}, \\frac{-2+6+5}{3}) = (0, 3)\).
Q3
If the inclination of a straight line is \(150^\circ\), then what is its slope?
- A. \(\\sqrt{3}\)
- B. \(-\\sqrt{3}\)
- C. \(\\frac{1}{\\sqrt{3}}\)
- D. \(-\\frac{1}{\\sqrt{3}}\)Correct
Explanation. The slope of a line with inclination \(\theta\) is given by \(m = \tan \theta\). Here, \(\theta = 150^\circ\), so the slope is \(m = \tan 150^\circ = \tan(180^\circ - 30^\circ) = -\tan 30^\circ = -\frac{1}{\sqrt{3}}\).
Q4
If a line passing through the points \((a, 3)\) and \((4, 7)\) is perpendicular to a line with slope \(2\), then what is the value of \(a\)?
- A. \(12\)Correct
- B. \(-12\)
- C. \(6\)
- D. \(-4\)
Explanation. Since the lines are perpendicular, the product of their slopes is \(-1\). Thus, the slope of the line passing through \((a, 3)\) and \((4, 7)\) must be \(-\\frac{1}{2}\). Solving \(\\frac{7-3}{4-a} = -\\frac{1}{2}\) gives \(4-a = -8\), which simplifies to \(a = 12\).
Q5
What is the equation of the straight line passing through the point \((-5, 8)\) and parallel to the Y-axis?
- A. \(y = 8\)
- B. \(x = -5\)Correct
- C. \(y = -5\)
- D. \(x = 8\)
Explanation. A line parallel to the Y-axis has the equation \(x = c\). Since the line passes through the point \((-5, 8)\), the \(x\)-coordinate is constant and equal to \(-5\). Thus, the equation of the line is \(x = -5\).
Q6
Find the slope of the straight line given by the equation \(5x + 2y - 9 = 0\).
- A. \(\\frac{5}{2}\)
- B. \(-\\frac{5}{2}\)Correct
- C. \(\\frac{2}{5}\)
- D. \(-\\frac{2}{5}\)
Explanation. The general equation of a straight line is \(ax + by + c = 0\), and its slope is given by \(m = -\\frac{a}{b}\). For the line \(5x + 2y - 9 = 0\), \(a = 5\) and \(b = 2\), so the slope is \(-\\frac{5}{2}\).
Q7
A straight line passes through the point \((4, 3)\) and has equal intercepts on the coordinate axes. What is the equation of this line?
- A. \(x + y = 7\)Correct
- B. \(x - y = 1\)
- C. \(x + y = 1\)
- D. \(3x + 4y = 25\)
Explanation. The intercept form of a line is \(\\frac{x}{a} + \\frac{y}{b} = 1\). Since the intercepts are equal, \(a = b\), so the equation is \(x + y = a\). Substituting the point \((4, 3)\) gives \(4 + 3 = a \\implies a = 7\). Thus, the line is \(x + y = 7\).
Q8
What is the Y-intercept of the straight line whose equation is \(3x - 4y + 12 = 0\)?
- A. \(-3\)
- B. \(3\)Correct
- C. \(-4\)
- D. \(4\)
Explanation. The Y-intercept can be found by setting \(x = 0\) in the equation of the line. Substituting \(x = 0\) into \(3x - 4y + 12 = 0\) yields \(-4y + 12 = 0\), which simplifies to \(y = 3\).
Q9
Find the slope of the line perpendicular to the line segment joining \((2, 5)\) and \((-4, 9)\).
- A. \(-\\frac{2}{3}\)
- B. \(\\frac{3}{2}\)Correct
- C. \(-\\frac{3}{2}\)
- D. \(\\frac{2}{3}\)
Explanation. The slope of the segment joining \((2, 5)\) and \((-4, 9)\) is \(m_1 = \\frac{9-5}{-4-2} = -\\frac{2}{3}\). Since the required line is perpendicular, its slope \(m_2\) satisfies \(m_1 \\times m_2 = -1\), which gives \(m_2 = \\frac{3}{2}\).
Q10
Which of the following lines is parallel to the line \(3x - y + 5 = 0\)?
- A. \(3x + y - 2 = 0\)
- B. \(x - 3y + 4 = 0\)
- C. \(6x - 2y + 7 = 0\)Correct
- D. \(x + 3y - 1 = 0\)
Explanation. Two lines are parallel if and only if their slopes are equal. The slope of \(3x - y + 5 = 0\) is \(3\). The line \(6x - 2y + 7 = 0\) can be rewritten as \(2(3x - y) + 7 = 0\), which also has a slope of \(3\), making them parallel.
Q11
What is the equation of the line passing through \((0, 0)\) and perpendicular to the line \(x + 2y - 3 = 0\)?
- A. \(2x - y = 0\)Correct
- B. \(x - 2y = 0\)
- C. \(2x + y = 0\)
- D. \(x + 2y = 0\)
Explanation. The slope of the line \(x + 2y - 3 = 0\) is \(-\\frac{1}{2}\). A line perpendicular to it must have a slope of \(2\). Since this perpendicular line passes through the origin \((0, 0)\), its equation is \(y = 2x\), which can be written as \(2x - y = 0\).
Q12
If the vertices of a quadrilateral are \((0, 0)\), \((a, 0)\), \((a, b)\), and \((0, b)\) in order, where \(a > 0\) and \(b > 0\), then what is the area of this quadrilateral?
- A. \(ab\)Correct
- B. \(\\frac{1}{2}ab\)
- C. \(2ab\)
- D. \(a^2 + b^2\)
Explanation. The vertices \((0, 0)\), \((a, 0)\), \((a, b)\), and \((0, b)\) describe a rectangle with adjacent sides of lengths \(a\) and \(b\) along the axes. The area of a rectangle is simply the product of its length and breadth, which is \(ab\).
Q13
At what point do the straight lines \(x - y + 2 = 0\) and \(2x + y - 5 = 0\) intersect?
- A. \((1, 3)\)Correct
- B. \((3, 1)\)
- C. \((-1, 1)\)
- D. \((2, 4)\)
Explanation. To find the point of intersection, solve the system of linear equations. Adding \(x - y + 2 = 0\) and \(2x + y - 5 = 0\) gives \(3x - 3 = 0\), so \(x = 1\). Substituting \(x = 1\) into the first equation yields \(y = 3\).
Q14
What is the equation of the straight line passing through the points \((1, -1)\) and \((3, 5)\)?
- A. \(3x - y - 4 = 0\)Correct
- B. \(3x + y - 2 = 0\)
- C. \(x - 3y - 4 = 0\)
- D. \(x + 3y + 2 = 0\)
Explanation. The equation of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(\\frac{y - y_1}{y_2 - y_1} = \\frac{x - x_1}{x_2 - x_1}\). Substituting \((1, -1)\) and \((3, 5)\) yields \(\\frac{y + 1}{6} = \\frac{x - 1}{2}\), which simplifies to \(3x - y - 4 = 0\).
Q15
If the line joining the points \((2, k)\) and \((k, 11)\) has a slope of \(2\), then what is the value of \(k\)?
- A. \(3\)
- B. \(4\)
- C. \(5\)Correct
- D. \(6\)
Explanation. The slope of the line is given by \(m = \\frac{y_2 - y_1}{x_2 - x_1}\). Setting this equal to \(2\) gives \(\\frac{11 - k}{k - 2} = 2\). Cross-multiplying, we get \(11 - k = 2(k - 2)\), which simplifies to \(11 - k = 2k - 4 \\implies 3k = 15 \\implies k = 5\).
