Samacheer Kalvi Class 10 Maths Solutions Chapter 5 Coordinate Geometry Exercise 5.5

Question 1. The area of triangle formed by the points (-5, 0), (0, -5) and (5, 0) is ..........
(a) 0 sq.units
(b) 25 sq.units
(c) 5 sq.units
(d) None of the options
Answer: (b) 25 sq.units
\( \text{Given points: } (x_1, y_1) = (-5, 0), (x_2, y_2) = (0, -5), (x_3, y_3) = (5, 0) \)
Now, we calculate the area of the triangle using the determinant formula. This method helps to find the area when coordinates are known.
\( \text{Area of a triangle} = \frac{1}{2} | (x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)) | \)
\( = \frac{1}{2} | (-5(-5 - 0) + 0(0 - 0) + 5(0 - (-5))) | \)
\( = \frac{1}{2} | (-5(-5) + 0(0) + 5(5)) | \)
\( = \frac{1}{2} | (25 + 0 + 25) | \)
\( = \frac{1}{2} | 50 | \)
\( = \frac{1}{2} \times 50 \)
\( = 25 \)
Therefore, the area of the triangle is 25 square units.
In simple words: To find the area of a triangle when you have its corner points, you can use a special formula. You put the x and y numbers into the formula, do the math, and the final answer will be the area in square units.

🎯 Exam Tip: Always remember the formula for the area of a triangle given three coordinate points and be careful with signs when substituting values.

Question 2. A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the Y axis. The path travelled by the man is ..........
(a) x = 10
(b) y = 10
(c) x = 0
(d) y = 0
Answer: (a) x = 10
The Y-axis is the vertical line where all x-coordinates are 0. If the wall is the Y-axis and the man walks 10 units away from it, his x-coordinate will always be 10. This creates a straight line parallel to the Y-axis. The path of the man is a straight line at a constant distance from the Y-axis.
In simple words: If the Y-axis is a wall, and you walk 10 steps away from it and keep walking straight, you will always be at the 'x = 10' position. So, your path is the line x = 10.

🎯 Exam Tip: Remember that the equation \( x = k \) represents a vertical line parallel to the Y-axis, and \( y = k \) represents a horizontal line parallel to the X-axis.

Question 3. The straight line given by the equation x = 11 is ..........
(a) parallel to X axis
(b) parallel to Y axis
(c) passing through the origin
(d) passing through the point (0,11)
Answer: (b) parallel to Y axis
An equation of the form \( x = \text{constant} \) always represents a vertical line. Since the Y-axis itself is a vertical line (where \( x = 0 \)), any line \( x = 11 \) will be parallel to the Y-axis. This line will pass through all points where the x-coordinate is 11, like (11,0), (11,1), (11,-2), etc.
In simple words: When a line's equation is just "x equals a number", it's a straight up-and-down line. This kind of line will always be parallel to the Y-axis, which is also an up-and-down line.

🎯 Exam Tip: Visualize the coordinate plane: \( x = k \) means all points have the same x-value, forming a vertical line; \( y = k \) means all points have the same y-value, forming a horizontal line.

Question 4. If (5,7), (3,p) and (6,6) are collinear, then the value of p is ..........
(a) 3
(b) 6
(c) 9
(d) 12
Answer: (c) 9
If three points are collinear, it means they lie on the same straight line. This implies that the area of the triangle formed by these three points must be zero. We use the area of triangle formula to solve for p.
Given points: \( (x_1, y_1) = (5, 7), (x_2, y_2) = (3, p), (x_3, y_3) = (6, 6) \)
Since the three points are collinear, the area of the triangle is 0.
\( \frac{1}{2} | (x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)) | = 0 \)
This means the expression inside the absolute value must be 0:
\( x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0 \)
\( 5(p - 6) + 3(6 - 7) + 6(7 - p) = 0 \)
\( 5p - 30 + 3(-1) + 42 - 6p = 0 \)
\( 5p - 30 - 3 + 42 - 6p = 0 \)
Combine the p terms and the constant terms.
\( (5p - 6p) + (-30 - 3 + 42) = 0 \)
\( -p + 9 = 0 \)
Now, solve for p:
\( -p = -9 \)
\( p = 9 \)
In simple words: When three points are in a straight line, they can't form a triangle, so the triangle's area is zero. We use this fact and a formula to find the missing number 'p' by setting the area to zero and solving the equation.

🎯 Exam Tip: The condition for collinearity of three points is that the area of the triangle formed by them is zero. This is a crucial concept in coordinate geometry.

Question 5. The point of intersection of 3x - y = 4 and x + y = 8 is ..........
(a) (5,3)
(b) (2,4)
(c) (3,5)
(d) (4, 4)
Answer: (c) (3, 5)
We need to solve the given system of two linear equations simultaneously. The intersection point is where both equations are true. Let's label the equations:
(1) \( 3x - y = 4 \)
(2) \( x + y = 8 \)
Add equation (1) and equation (2) together. This method is effective when one variable has opposite coefficients.
\( (3x - y) + (x + y) = 4 + 8 \)
\( 4x = 12 \)
Now, divide by 4 to find x:
\( x = \frac{12}{4} \)
\( x = 3 \)
Substitute the value of \( x = 3 \) into equation (2) to find y:
\( 3 + y = 8 \)
Subtract 3 from both sides:
\( y = 8 - 3 \)
\( y = 5 \)
So, the point of intersection is (3, 5).
In simple words: To find where two lines cross, we solve their equations together. We can add the equations to get rid of one letter, find the value of the other letter, and then put that value back into an equation to find the first letter. The point (x, y) is where they meet.

🎯 Exam Tip: For simultaneous equations, choose the elimination or substitution method based on which variable is easiest to cancel or isolate.

Question 6. The slope of the line joining (12, 3), (4, a) is \( \frac { 1 }{ 8 } \). The value of 'a' is ..........
(a) 1
(b) 4
(c) -5
(d) 2
Answer: (d) 2
The slope of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). We are given the slope and the coordinates, so we can set up an equation to find 'a'.
Given points: \( (x_1, y_1) = (12, 3) \) and \( (x_2, y_2) = (4, a) \)
Given slope \( m = \frac{1}{8} \)
Substitute these values into the slope formula:
\( \frac{a - 3}{4 - 12} = \frac{1}{8} \)
\( \frac{a - 3}{-8} = \frac{1}{8} \)
Now, multiply both sides by -8 to isolate \( a - 3 \):
\( a - 3 = \frac{1}{8} \times (-8) \)
\( a - 3 = -1 \)
Add 3 to both sides to solve for a:
\( a = -1 + 3 \)
\( a = 2 \)
Therefore, the value of 'a' is 2.
In simple words: The slope tells us how steep a line is. If we know two points on a line and its slope, we can use the slope formula to find any missing number in the coordinates, like 'a' in this problem.

🎯 Exam Tip: Always remember the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \) and be careful with arithmetic, especially when dealing with negative numbers.

Question 7. The slope of the line which is perpendicular to a line joining the points (0, 0) and (-8, 8) is ..........
(a) -1
(b) 1
(c) \( \frac { 1 }{ 3 } \)
(d) -8
Answer: (b) 1
First, find the slope of the line joining the given points (0, 0) and (-8, 8). Then, use the relationship between slopes of perpendicular lines to find the required slope. Perpendicular lines have slopes that are negative reciprocals of each other.
Given points: \( (x_1, y_1) = (0, 0) \) and \( (x_2, y_2) = (-8, 8) \)
Slope of the line \( (m_1) = \frac{y_2 - y_1}{x_2 - x_1} \)
\( m_1 = \frac{8 - 0}{-8 - 0} \)
\( m_1 = \frac{8}{-8} \)
\( m_1 = -1 \)
Now, for two lines to be perpendicular, the product of their slopes must be -1. Let \( m_2 \) be the slope of the perpendicular line.
\( m_1 \times m_2 = -1 \)
\( (-1) \times m_2 = -1 \)
\( m_2 = \frac{-1}{-1} \)
\( m_2 = 1 \)
So, the slope of the line perpendicular to the given line is 1.
In simple words: First, we find how steep the first line is. Then, because we want a line that crosses it at a perfect right angle, we take the negative flip of that steepness. If the first line's steepness is -1, the perpendicular line's steepness will be 1.

🎯 Exam Tip: Remember that if the slope of a line is \( m \), the slope of a line perpendicular to it is \( -\frac{1}{m} \). If \( m = 0 \), the perpendicular line is vertical (undefined slope), and if \( m \) is undefined, the perpendicular line is horizontal (slope = 0).

Question 8. If slope of the line PQ is \( \frac{1}{\sqrt{3}} \) then slope of the perpendicular bisector of PQ is ..........
(a) \( \sqrt { 3 } \)
(b) \( -\sqrt { 3 } \)
(c) 0
Answer: (b) \( -\sqrt { 3 } \)
The perpendicular bisector of a line segment is a line that is perpendicular to the segment and passes through its midpoint. For this question, we only need the perpendicular part, as the midpoint information is not needed to find the slope. The slope of a line perpendicular to a given line is the negative reciprocal of the given line's slope.
Given slope of line PQ \( (m_1) = \frac{1}{\sqrt{3}} \)
Let \( m_2 \) be the slope of the perpendicular bisector.
For perpendicular lines, \( m_1 \times m_2 = -1 \)
\( \frac{1}{\sqrt{3}} \times m_2 = -1 \)
Multiply both sides by \( \sqrt{3} \):
\( m_2 = -1 \times \sqrt{3} \)
\( m_2 = -\sqrt{3} \)
In simple words: The "steepness" of a line that cuts another line exactly in half and at a right angle (perpendicular bisector) is the negative flipped version of the original line's steepness. So, if the original slope is \( \frac{1}{\sqrt{3}} \), the new slope is \( -\sqrt{3} \).

🎯 Exam Tip: The term "perpendicular bisector" combines two conditions: perpendicularity (slopes are negative reciprocals) and bisector (passes through the midpoint). This question only tests the perpendicularity aspect for the slope.

Question 9. If A is a point on the Y axis whose ordinate is 8 and B is a point on the X axis whose abscissae is 5 then the equation of the line AB is ..........
(a) 8x + 5y = 40
(b) 8x - 5y = 40
(c) x = 8
(d) y = 5
Answer: (a) 8x + 5y = 40
A point on the Y-axis has its x-coordinate as 0. Since the ordinate (y-coordinate) of A is 8, point A is (0, 8). A point on the X-axis has its y-coordinate as 0. Since the abscissa (x-coordinate) of B is 5, point B is (5, 0). We can use the two-point form of a line or the intercept form.
Using the two-point form \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \):
\( (x_1, y_1) = (0, 8) \)
\( (x_2, y_2) = (5, 0) \)
Substitute these values into the formula:
\( \frac{y - 8}{0 - 8} = \frac{x - 0}{5 - 0} \)
\( \frac{y - 8}{-8} = \frac{x}{5} \)
Cross-multiply to eliminate the denominators:
\( 5(y - 8) = -8x \)
\( 5y - 40 = -8x \)
Rearrange the terms to the standard form \( Ax + By = C \):
\( 8x + 5y = 40 \)
This equation represents the line AB. The equation tells us the relationship between any x and y coordinate on the line.
In simple words: We find the exact locations of points A and B. Then, we use a formula that finds the straight line connecting them. This formula helps us write down the equation that all points on that line follow.

🎯 Exam Tip: When a point is on the Y-axis, its x-coordinate is 0. When a point is on the X-axis, its y-coordinate is 0. These are key facts for setting up coordinate geometry problems.

Question 10. The equation of a line passing through the origin and perpendicular to the line 7x - 3y + 4 = 0 is ..........
(a) 7x-3y +4 = 0
(b) 3x - 7y + 4 = 0
(c) 3x + 7y = 0
(d) 7x - 3y = 0
Answer: (c) 3x + 7y = 0
First, find the slope of the given line. Then, determine the slope of a line perpendicular to it. Finally, use the point-slope form of a linear equation, knowing that the required line passes through the origin (0, 0).
The given line is \( 7x - 3y + 4 = 0 \). To find its slope, rearrange it into the slope-intercept form \( y = mx + c \).
\( -3y = -7x - 4 \)
Divide by -3:
\( y = \frac{-7}{-3}x + \frac{-4}{-3} \)
\( y = \frac{7}{3}x + \frac{4}{3} \)
So, the slope of the given line \( (m_1) = \frac{7}{3} \).
The slope of a line perpendicular to this line \( (m_2) \) is the negative reciprocal:
\( m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{7}{3}} = -\frac{3}{7} \)
Now, we need the equation of a line with slope \( m_2 = -\frac{3}{7} \) that passes through the origin (0, 0). We can use the point-slope form \( y - y_1 = m(x - x_1) \).
\( y - 0 = -\frac{3}{7}(x - 0) \)
\( y = -\frac{3}{7}x \)
Multiply both sides by 7 to clear the fraction:
\( 7y = -3x \)
Rearrange into the standard form \( Ax + By = C \):
\( 3x + 7y = 0 \)
This equation represents the line that passes through the origin and is perpendicular to the given line.
In simple words: First, we find how steep the original line is. Then, we find the steepness of a line that would cross it at a right angle. Since the new line must pass through the center (origin), we use its steepness and the point (0,0) to write its equation.

🎯 Exam Tip: Remember that the slope of a line in the form \( Ax + By + C = 0 \) is \( -\frac{A}{B} \). This can save time compared to converting to slope-intercept form for every calculation.

Question 11. Consider four straight lines
(i) l₁ : 3y = 4x + 5
(ii) l₂ : 4y = 3x - 1
(iii) l₃ : 4y + 3x = 7
(iv) l₄ : 4x + 3y = 2
Which of the following statement is true?
(a) l₁ and l₂ are perpendicular
(b) l₂ and l₄ are parallel
(c) l₂ and l₄ are perpendicular
(d) l₂ and l₃ are parallel
Answer: (c) l₂ and l₄ are perpendicular
To check for parallel or perpendicular lines, we need to find the slope of each line first. For a line in the form \( Ax + By = C \), the slope is \( -\frac{A}{B} \). For \( y = mx + c \), the slope is \( m \).

(i) Line \( l_1 \): \( 3y = 4x + 5 \)
Rewrite as \( y = \frac{4}{3}x + \frac{5}{3} \).
Slope of \( l_1 \) is \( m_1 = \frac{4}{3} \).

(ii) Line \( l_2 \): \( 4y = 3x - 1 \)
Rewrite as \( y = \frac{3}{4}x - \frac{1}{4} \).
Slope of \( l_2 \) is \( m_2 = \frac{3}{4} \).

(iii) Line \( l_3 \): \( 4y + 3x = 7 \)
Rewrite as \( 4y = -3x + 7 \).
Rewrite as \( y = -\frac{3}{4}x + \frac{7}{4} \).
Slope of \( l_3 \) is \( m_3 = -\frac{3}{4} \).

(iv) Line \( l_4 \): \( 4x + 3y = 2 \)
Rewrite as \( 3y = -4x + 2 \).
Rewrite as \( y = -\frac{4}{3}x + \frac{2}{3} \).
Slope of \( l_4 \) is \( m_4 = -\frac{4}{3} \).

Now let's check the given statements:
(a) \( l_1 \) and \( l_2 \) are perpendicular?
\( m_1 \times m_2 = \frac{4}{3} \times \frac{3}{4} = 1 \). This is not -1, so they are not perpendicular. (False)

(b) \( l_2 \) and \( l_4 \) are parallel?
\( m_2 = \frac{3}{4} \) and \( m_4 = -\frac{4}{3} \). They are not equal, so they are not parallel. (False)

(c) \( l_2 \) and \( l_4 \) are perpendicular?
\( m_2 \times m_4 = \frac{3}{4} \times (-\frac{4}{3}) = -1 \). This is true, so they are perpendicular. (True)

(d) \( l_2 \) and \( l_3 \) are parallel?
\( m_2 = \frac{3}{4} \) and \( m_3 = -\frac{3}{4} \). They are not equal, so they are not parallel. (False)

The correct statement is that \( l_2 \) and \( l_4 \) are perpendicular. Calculating all slopes accurately is the most important step.
In simple words: We find the steepness (slope) for each of the four lines. Then, we check the pairs of lines mentioned in the options. Parallel lines have the same steepness. Perpendicular lines have steepness that multiplies to -1 (one is the negative flip of the other). We find the pair that fits the perpendicular rule.

🎯 Exam Tip: Parallel lines have equal slopes (\( m_1 = m_2 \)), while perpendicular lines have slopes that multiply to -1 (\( m_1 \times m_2 = -1 \)). Convert all equations to \( y = mx + c \) form to easily identify their slopes.

Question 12. A straight line has equation 8y = 4x + 21. Which of the following is true?
(a) The slope is 0.5 and the y intercept is 2.6
(b) The slope is 5 and the y intercept is 1.6<
(c) The slope is 0.5 and the y intercept is 1.6
(d) The slope is 5 and the y intercept is 2.6
Answer: (a) The slope is 0.5 and the y intercept is 2.6
To find the slope and y-intercept of a line, we need to convert its equation into the slope-intercept form, which is \( y = mx + c \), where 'm' is the slope and 'c' is the y-intercept. This standard form makes these values easy to read directly.
Given equation: \( 8y = 4x + 21 \)
Divide the entire equation by 8 to isolate y:
\( y = \frac{4x}{8} + \frac{21}{8} \)
Simplify the fractions:
\( y = \frac{1}{2}x + \frac{21}{8} \)
Now, compare this with \( y = mx + c \):
Slope \( m = \frac{1}{2} = 0.5 \)
Y-intercept \( c = \frac{21}{8} \)
To express the y-intercept as a decimal, divide 21 by 8:
\( 21 \div 8 = 2.625 \)
Rounding to one decimal place as in the options, the y-intercept is approximately 2.6. Thus, the slope is 0.5 and the y-intercept is 2.6.
In simple words: To know how steep a line is and where it crosses the 'y' axis, we change its equation to a special form: "y equals some number times x, plus another number". The number with 'x' is the steepness (slope), and the other number is where it crosses the 'y' axis (y-intercept).

🎯 Exam Tip: Always transform the given equation into the \( y = mx + c \) form to directly identify the slope (m) and y-intercept (c). Be careful with decimal conversions and rounding if required.

Question 13. When proving that a quadrilateral is a trapezium, it is necessary to show
(a) Two sides are parallel.
(b) Two parallel and two non-parallel sides.
(c) Opposite sides are parallel.
(d) All sides are of equal length.
Answer: (b) Two parallel and two non-parallel sides.
A trapezium (also known as a trapezoid in some regions) is defined as a quadrilateral with at least one pair of parallel sides. However, the stricter definition often used in geometry problems requires exactly one pair of parallel sides. Therefore, to prove a quadrilateral is a trapezium, you must show that it has two parallel sides and that the other two sides are not parallel. This distinction prevents squares, rectangles, and parallelograms from being classified as trapeziums.
In simple words: To show a shape is a trapezium, you need to prove that only two of its sides run in the same direction (are parallel), and the other two sides are definitely not parallel.

🎯 Exam Tip: Understand the precise definitions of quadrilaterals. For a trapezium, having *exactly* one pair of parallel sides is the key distinction from a parallelogram (which has two pairs of parallel sides).

Question 14. When proving that a quadrilateral is a parallelogram by using slopes you must find ..........
(a) The slopes of two sides
(b) The slopes of two pair of opposite sides
(c) The lengths of all sides
(d) Both the lengths and slopes of two sides
Answer: (b) The slopes of two pair of opposite sides
A parallelogram is a quadrilateral where both pairs of opposite sides are parallel. To prove that lines are parallel using slopes, you must show that their slopes are equal. Therefore, to prove a quadrilateral is a parallelogram using slopes, you need to calculate the slopes of all four sides and then confirm that the slopes of one pair of opposite sides are equal, and the slopes of the other pair of opposite sides are also equal. This ensures that both pairs of opposite sides are indeed parallel.
In simple words: To prove a shape is a parallelogram using slopes, you need to check the steepness of all four sides. If the opposite sides have the same steepness, then they are parallel, and the shape is a parallelogram.

🎯 Exam Tip: For proving a parallelogram using slopes, remember to find the slopes of all four sides. It's not enough to check just one pair of opposite sides; both pairs must be parallel.

Question 15. Which of the following pairs of lines has (2,1) as its point of intersection?
(a) x - y - 3 = 0; 3x - y - 7 = 0
(b) x + y = 3; 3x + y = 7
(c) 3x + y = 3; x + y = 7
(d) x + 3y - 3 = 0; x - y - 7 = 0
Answer: (b) x + y = 3; 3x + y = 7
To find which pair of lines intersects at (2,1), substitute \( x=2 \) and \( y=1 \) into each equation in every option. The correct pair will be the one where (2,1) satisfies both equations simultaneously. This means that when you put x=2 and y=1 into each equation, both equations should become true statements.

Let's check option (b):
Equation 1: \( x + y = 3 \)
Substitute \( x=2, y=1 \): \( 2 + 1 = 3 \)
\( 3 = 3 \) (This is true)

Equation 2: \( 3x + y = 7 \)
Substitute \( x=2, y=1 \): \( 3(2) + 1 = 7 \)
\( 6 + 1 = 7 \)
\( 7 = 7 \) (This is true)

Since (2,1) satisfies both equations in option (b), this is the correct answer. You would find that (2,1) does not satisfy both equations for options (a), (c), and (d). For example, for option (a), \( x - y - 3 = 0 \Rightarrow 2 - 1 - 3 = -2 \neq 0 \), so option (a) is incorrect.
In simple words: We are given a point and several pairs of line equations. We need to check which pair of lines passes through this point. We do this by putting the x and y values of the point into each equation. If the point makes both equations in a pair true, then that's the pair of lines that crosses at that point.

🎯 Exam Tip: For "point of intersection" questions in multiple-choice format, the fastest way to solve is often to substitute the given point into each option and see which pair of equations holds true. Always check both equations in a pair.